Projectile Motion: Kinematics Theory of V_1y and V_2y of two objects dropped from same height

A golf ball is dropped straight down, and a bowling ball is rolled off a horizontal tabletop. If the golf ball and bowling ball are dropped from the same height at the same time, then how do the vertical components of their final velocities compare at the ground? (Ignoring air resistance) Solution

1. v_{2 golf} > v_{2 bowling}
2. v_{2 golf} = v_{2 bowling}
3. v_{2 golf} < v_{2 bowling}
4. It cannot be determined from the given information.

Projectile Motion: Kinematics, Uniform and Non-Uniform Motion

A little kid throws a ball horizontally off a cliff, from 20 meters above the ground below, and it hits the ground with a horizontal distance of 5.0 m from the cliff.

The initial, vertical component of velocity (v_{1 y}) is equal to the initial, horizontal component of velocity (v_{1 x}).. Solution

1. True
2. False

The ball is thrown horizontally so the initial velocity is purely in the horizontal direction (and zero in the vertical direction).

The initial, vertical component of velocity (v_{1 y}) = 0 m/s

If the ball's trajectory is separated into components, the time it takes for the ball to travel the horizontal distance (d_x) equals the time it takes for the ball to travel the vertical distance (d_y). Solution

1. True
2. False

t_x = t_y

The time it takes for the ball to travel in the horizontal direction is the same as the time it takes to travel vertically.

Determine the initial (horizontal) speed that the ball was thrown, neglecting air resistance. Solution

Parabolic Motion

The path of a projectile that has an approximate parabolic trajectory from point A to E is shown below where A and E are on the same horizontal plane (and B and D are on the same horizontal plane). Neglect air resistance.

The total velocity is constant. Solution

1. True
2. False

The total velocity is constantly changing due to gravity accelerating the projectile at ±9.81 m/s², in the vertical component.

At which point is the magnitude of the vertical component of velocity greatest ? Solution

1. A
2. B
3. C
4. D
5. All the same

The vertical velocity (V_y) is always changing direction and magnitude, due to the constant influence of gravity's acceleration in the vertical direction.

Interesting Points:
A = – E
B = – D
C = 0 m/s

The horizontal or vertical components of velocity are never zero. Solution

1. True
2. False

The vertical component of velocity is 0 m/s at point C.

Given the horizontal component of velocity at point A is 5 m/s, determine the horizontal components of velocity at the other 4 points, (B, C, D, and E) Solution

1. 5, 4, 3, and 2 m/s
2. 3, 4, 2, and 1 m/s
3. 6, 7, 8, and 9 m/s
4. 5, 5, 5, and 5 m/s

There is no acceleration in the horizontal direction.

Velocity is uniform in the horizontal direction.

The horizontal component of velocity is constant, 5 m/s.

Velocity is uniform in the horizontal (x) direction, so the formula below is used for it. Solution d = v · t

1. True
2. False

The horizontal velocity (V_x) is constant!

(It's the vertical velocity that changes due to gravity because it is in the same vertical direction as gravity. This vertical change is what causes the overall trajectory to have non-uniform motion, even though the horizontal velocity is constant).

Projectile Motion: Kinematics, Uniform and Non-Uniform Motion

A toy rocket with no thrusters is fired straight (horizontally) at a target with an initial speed of 700 m/s. As the rocket travels through the air towards the target, it drops a vertical distance of 0.05m from the centre of the target.

If the rocket trajectory is separated into components, the time it takes for the rocket to travel the horizontal distance (d_x) equals the time it takes for the rocket to travel the vertical distance (d_y). Solution

1. True
2. False

t_x = t_y

The time it takes for the rocket to travel in the horizontal direction is the same as the time it takes to travel vertically.

Which two of the following statements is true about the rocket's motion after it is launched? Solution

1. The vertical component of motion is uniform because there is no acceleration
2. The vertical component of motion is non-uniform because there is acceleration
3. The horizontal component of motion is uniform because there is no acceleration
4. The horizontal component of motion is non-uniform because there is acceleration

On Earth's surface there is always acceleration in the vertical direction, it's gravity!

Acceleration equals zero in the horizontal direction because there are no thrusters and we can neglect air resistance, so motion is uniform.

Calculate the horizontal distance from the rocket launcher to the target. Solution

d =

Hint Clear Info

Incorrect Attempts:

CHECK

m

Hint Unavailable

Projectile Motion: Kinematics, Uniform and Non-Uniform Motion

A ball rolls 10m/s down a hill angled at 30˚ to the horizontal and then off a cliff that's 25m high.

If the positive direction is defined as up and to the right, determine the magnitude of the vertical component of the initial velocity. Solution

1. 10cos30˚
2. -10sin30˚
3. 10cos30˚
4. -10sin60˚

Use trigonometry to solve for the vertical component of velocity.

In this case, the vertical component of velocity uses sin (while horizontal uses cos, not shown here). v_{1 y} is down and negative, since up is defined as positive. $\begin{align} sin\,30 & = \dfrac{opposite}{hypotenuse} \\ \\ sin\,30 & = \dfrac{-\vec{v}_{1y}}{10\ m/s} \\ \\ (10\ m/s)sin\,30 & = \dfrac{-\vec{v}_{1y}}{1} \\ \\ \vec{v}_{1y} & = -10sin\,30 \\ \\ \end{align}$

Calculate the horizontal distance travelled by the ball before hitting the ground. Solution

d =

Hint Clear Info

Incorrect Attempts:

CHECK

m

Hint Unavailable

Projectile Motion: Kinematics Theory of what goes up at V_1y must come down at -V_1y

How does initial vertical velocity (V_1y) compare to final vertical velocity (V_2y), on the same horizontal plane, in the projectile motion of an object thrown straight upwards? [2] Solution Video

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

Projectile Motion: Kinematics, Uniform and Non-Uniform Motion, Find v

A projectile is launched in a parabolic trajectory at an unknown angle from the horizontal, and lands 100m horizontally from where it was launched.

The final vertical velocity where the projectile lands is equal in magnitude and opposite in direction to the initial vertical velocity where it was launched. Solution

1. True
2. False

Since the projectile lands at the same horizontal plane from where it was launched:

v_2y = -v_2y

The quantities have opposite signs because they are in opposite directions.

The velocity in the vertical (y-component) direction is non-uniform, while the velocity in the horizontal (x-component) direction is uniform. Solution

1. True
2. False

Since there is gravity acting in the vertical (y-component) direction, then it is non-uniform.

Since there is no acceleration in the horizontal (x-component), then the motion is uniform and d = vt.

If the projectile was in the air for a total of 4.0 seconds, find the initial velocity of the projectile at launch. Solution

Hint Clear Info

Incorrect Attempts:

CHECK

m/s

Hint Unavailable

Projectile Motion: Kinematics, Uniform and Non-Uniform Motion, Find t without v

A frog jumps at an angle of 50˚ to the horizontal and lands 0.9m horizontally from its initial position. Find the time the frog was in the air during its parabolic flight. Solution

t =

Hint Clear Info

Incorrect Attempts:

CHECK

s

This one is trickier because we don't have 'v', so we use the components in terms of 'v'

Projectile Motion: Kinematics, Uniform and Non-Uniform Motion

A PGA golfer can hit a hole-in-one shot from the fairway a certain distance from the green, and 20m vertically above the fairway. Lets say the golf ball leaves the club at 70 m/s at an angle of elevation 65˚ to the horizontal.

When the golf ball first reaches 20m above the fairway, the vertical component of velocity (v_y) is equal in magnitude and opposite in direction to the v_y at the green. Solution

1. True
2. False

True

For how much time was the ball in the air? Solution

t =

Hint Clear Info

Incorrect Attempts:

CHECK

s

It helps to split the motion into two phases, with the separation located at the horizontal line level with the plane of the green.

Kinematic System of Equations

A baseball is projected downward at 10 m/s from 12m above the ground at the same time that a golf ball is released from rest from 10 m above the ground. The balls travel vertically and this question neglects air resistance.

When the balls meet at the same horizontal plane in the air, their distance travelled and time travelled will be the same for each ball. Solution

1. True
2. False

When the balls meet at the same horizontal plane in the air, their distance above the ground and time travelled will be the same for each ball.

Determine the distance the golf ball travels in the air when the two balls meet at the same horizontal level. Solution

Hint Clear Info

Incorrect Attempts:

CHECK

m

Hint Unavailable

Let x represent the distance the golf ball travels until the balls meet...

Given for Golf Ball:
v₁ = +0 m/s downward
a = +9.81 m/s² downward
distance from ground = 10 m
distance until balls meet = x
Given for Baseball:
v₁ = +10 m/s downward
a = +9.81 m/s² downward
distance from ground = 12 m
distance until balls meet = x + 2 Make an equation ① for distance the golf ball travels until the two balls meet $\begin{align} d & = v_1 t + \dfrac{1}{2}at^2 \\ \\ x & = \cancelto{0}{0(t)} + \dfrac{1}{2}(9.81)t^2 \\ \\ x & = 4.905t^2 \\ \\ \end{align}$ Make an equation ② for distance the baseball travels until the two balls meet $\begin{align} d & = v_1 t + \dfrac{1}{2}at^2 \\ \\ x + 2 & = (10)(t) + \dfrac{1}{2}(9.81)t^2 \\ \\ x + 2 & = 10t + 4.905t^2 \\ \\ x & = 10t + 4.905t^2 - 2 \\ \\ \end{align}$ Set ① equal to ② with your distance variable, the x's... $\begin{align} x_{golf} & = x_{baseball} \\ \\ 4.905t^2 & = 10t + 4.905t^2 - 2 \\ \\ 10t & = 2 \\ \\ t & = 0.2\ s \\ \\ \end{align}$ Use the time to find the distance the golf ball travels with ① $\begin{align} x & = 4.905t^2 \\ \\ x & = 4.905(0.2)^2 \\ \\ x & = 0.2\ m \\ \\ \end{align}$ Therefore the balls will meet once the golf ball has travelled 0.2 m in the air.

Kinematic System of Equations II

A soccer ball is kicked into the air and down the field at 20 m/s with an angle of inclination of 35˚ to the horizontal. 2 seconds after the ball was kicked, the receiving player drops a piece of gum out of their mouth, from rest, 1.6 m above the ground. At some point in time, the soccer ball and the gum meet at the same level.

What is true about the point where the soccer ball and the gum meet? Solution

1. The ball and the gum will meet at the same total time, and will meet at the same distance travelled.
2. At the 2 second point the distance above the ground is the same for both, and the time travelled until they meet is the same.
3. The ball and gum will meet after the total time elapsed is the same and the vertical distance travelled is the same.
4. Starting at the 2 second point, the ball and the gum will meet at the same time, and will meet at the same distance above the ground.

At the point where the soccer ball and the gum meet...

• At the point where the ball and gum meet, they will be the same distance above the ground.
• At the point where the ball and gum meet, the same amount of time will have elapsed.
• At the 2 second mark, the soccer ball will have a vertical velocity downward while the gum will have zero initial velocity.

The y-component of the initial velocity of the ball is... Solution

1. 20 × sin(35˚)
2. 20 × cos(35˚)
3. 20 × tan(35˚)
4. None of the above

Make a right angle triangle where the hypotenuse is 20 m/s, the y-component is the opposite side length... $\begin{align} \sin θ˚ & = \dfrac{opp}{hyp} \\ \\ \sin 35˚ & = \dfrac{v_{1 (y)}}{20} \\ \\ v_{1 (y) & = 20\sin 35˚ \\ \\ \end{align}$

Determine the height of the ball above the ground and the y-component of velocity of the ball at the 2 second mark. Solution

Hint Clear Info

d_y = m above
v_y = m/s down

Incorrect Attempts:

CHECK

Hint Unavailable

Givens:
a = -9.81 m/s²
v_{1 (y)} = +20sin30˚ up
t = 2 s $\begin{align} d & = v_1 t + \dfrac{1}{2}at^2 \\ \\ & = (+20sin30˚)(2) + \dfrac{1}{2}(-9.81)(2)^2 \\ \\ & = 3.32\ m \quad \text{above ground} \\ \\ \end{align}$ Find v_{2 (y)}... $\begin{align} v_{2 (y)} & = v_{1 (y)} + at \\ \\ & = +20sin30˚ + (-9.81)(2) \\ \\ & = -8.15\ m/s \quad \text{down} \\ \\ \end{align}$

Determine the height above the ground where the ball and gum meet. Solution

Let x be the distance above the ground where the ball and gum meet. Since we have the height above the ground we can write the distances traveled in terms of the height where they meet, x... TO MAKE IT EASIER FOR YOU, NOW WE WILL DEFINE DOWN WITH A POSITIVE DISPLACEMENT...

• The displacement from the ball to the meeting point is (3.32 - x)
• The displacement from the gum to the meeting point is (1.6 - x)

Redefine Vector Signs of Givens:
(Down = positive, Up = Negative)
a = + 9.81 m/s² down
v_{1 ball} = +8.15 m/s down
v_{1 gum} = 0 m/s
d_ball = +(3.32 - x) down
d_gum = +(1.6 - x) down
① Distance traveled by the ball: $\begin{align} d & = v_1 t + \dfrac{1}{2}at^2 \\ \\ 3.32 - x & = (+8.15)(t) + \dfrac{1}{2}(+9.81)t^2 \\ \\ 3.32 - x & = +8.15t + 4.905t^2 \\ \\ \end{align}$ ② Distance traveled by the gum: $\begin{align} d & = v_1 t + \dfrac{1}{2}at^2 \\ \\ 1.6 - x & = \cancelto{0}{(0)(t)} + \dfrac{1}{2}(+9.81)t^2 \\ \\ 1.6 - x & = + 4.905t^2 \\ \\ x & = 1.6 - 4.905t^2 \\ \\ \end{align}$ Solve the system of equations (2 equations, 2 unknowns) using substitution: ② in ① ... $\begin{align} 3.32 - x & = +8.15t + 4.905t^2 \\ \\ 3.32 - (1.6 - 4.905t^2) & = +8.15t + 4.905t^2 \\ \\ 1.71 + 4.905t^2 & = +8.15t + 4.905t^2 \\ \\ 1.71 & = +8.15t \\ \\ t & = 0.2098 \\ \\ \end{align}$ Calculate height above the ground, x using equation ②... $\begin{align} x & = 1.6 - 4.905t^2 \\ \\ x & = 1.6 - 4.905(0.2098)^2 \\ \\ x & = 1.38\ m \\ \\ \end{align}$

Angle of Inclination, Theta θ

As the angle θ of inclination increases and the object does not move, state whether the following values increase, decrease, or if there is no change: Solution

a) Weight

b) Normal Force

c) Force of Friction

What is Acceleration?

Objects are accelerating when there is a change of direction and/or speed. Solution

1. True
2. False

You know that objects accelerate when the speed changes, but did you also know that an object is technically accelerating when the speed is constant and the direction changes?

For instance, an object traveling at a constant speed while changing direction in a circular path, would be considered as accelerating.

Kinematics and Forces

A vehicle comes to rest over a distance of 100m from an initial velocity of 30 m/s. Calculate the coefficient of kinetic friction (µ_k), neglecting air resistance. Solution

µ_k =

Hint Clear Info

Incorrect Attempts:

CHECK

This question uses a combination of kinematics and forces

FIRST - set up the kinematic equation to solve for acceleration given v₁, v₂, and d... $\begin{align} v_2^2 & = v_1^2 + 2ad \\ \\ a & = \dfrac{v_2^2 - v_1^2}{2d} \\ \\ & = \dfrac{0^2 - 30^2}{2(100)} \\ \\ & = -4.5\ m/s^2 \\ \\ \end{align}$ SECOND - set up a net force equation with the force due to friction...
The only force in the horizontal direction is the force due to friction... this is the net force. Also you know that in the vertical direction: F_N = -F_g ... $\begin{align} F_{NET} & = F_f \\ \\ (m)(a_{system}) & = (µ_k)(F_N) \\ \\ µ_k & = \dfrac{(m)(a_{system})}{F_N} \\ \\ µ_k & = \dfrac{(m)(a_{system})}{-F_g} \quad\quad \\ \\ µ_k & = \dfrac{(m)(a_{system})}{-mg} \\ \\ µ_k & = \dfrac{(\cancel{m})(a_{system})}{-\cancel{m}g} \\ \\ µ_k & = \dfrac{-4.5\ m/s^2}{-9.81\ m/s^2} \\ \\ µ_k & = 0.46 \\ \\ \end{align}$

Net Force with F_A and F_N not always equal to F_g

A parent is pushing their child in a stroller, with a total combined mass of the child and stroller of 30kg, and with an applied force of 100N at an angle of depression of 40˚.

The magnitude of the normal force of the ground on the stroller is equal to the force of gravity on the combined mass of the child and stroller. Solution Video

1. True
2. False

The magnitude of the normal force of the ground on the stroller is equal to the force of gravity on the combined mass of the child and stroller plus the vertical component of the applied force.

Calculate the normal force in this situation, neglecting friction. Solution Video

F_N =

Hint Clear Info

Incorrect Attempts:

CHECK

N

Hint Unavailable

Newton's 2^nd Law: Inclined Planes, with Friction

A crate is stuck to the surface of an inclined plane with the force of friction. If the angle of inclination of the surface from the horizontal is 30˚, find the minimum coefficient of static friction, µ_s required to keep the crate from sliding down. Solution Video

µ_s =

Hint Clear Info

Incorrect Attempts:

CHECK

You don't need the mass, because it will eventually cancel in your equations.

Newton's 2^nd Law: Inclined Planes, with Friction, with Kinematics

Determine the time it takes for a skier to go (from rest) down a ski hill due to gravity if the angle of inclination of the hill is 25˚ from the horizontal, the coefficient of kinetic friction is 0.10, and the distance travelled by the skier on the hill face is 900m. Solution Video

t =

Hint Clear Info

Incorrect Attempts:

CHECK

s

Mass will eventually cancel in the equations

Newton's 2^nd Law: Inclined Planes, Frictionless

Two inclined planes with blocks on them are placed back to back. The inclined plane on the left side has a mass of 3m on it and angle of inclination of 30˚ to the horizontal, while the inclined plane on the right side has a mass of 1m on it, and is at an angle of 60˚ to the horizontal. Find the acceleration of the system of two masses if the inclined planes are frictionless. Solution

a =

Hint Clear Info

Incorrect Attempts:

CHECK

m/s²

Resolve the components that oppose on either side. Mass will eventually cancel in the equations.

Newton's 2^nd Law: Pulley with Two Masses Hanging on Either Side, Frictionless

A mass m₁ hangs on one side of a frictionless pulley while a heavier mass m₂ hangs on the other side. Find the acceleration of the system of masses on this pulley in terms of m₁, m₂, and g. Solution Video

Newton's 2^nd Law: Pulley with Mass on a Table, Frictionless

A hanging 10N weight pulls down on a rope which is attached with a frictionless pulley to a 50N weight resting on the surface of a flat, frictionless table. Find the net acceleration of the system. Solution Video

a_NET =

Hint Clear Info

Incorrect Attempts:

CHECK

m/s²

Net Force = (total mass)(acceleration)

Newton's 2^nd Law: Pulley with Mass on a Table, with Friction

A hanging 12kg mass pulls down on a rope that is attached with a frictionless pulley to a 10kg mass resting on the surface of a flat table with a coefficient of kinetic friction (µ) of 0.40. Find the net acceleration of the system. Solution Video

a_NET =

Hint Clear Info

Incorrect Attempts:

CHECK

m/s²

Net Force = (Force due to Gravity) - (Force due to Friction)

Suspended Masses and Tension

A basic 2000 kg transport helicopter has a maximum thrust of 60 kN, and is carrying a 500 kg crate of survival supplies for an outpost research community with an inextensible, heavy-duty steel cable.

Calculate the tension force in the cable connected to the crate when the helicopter is hovering stationary in the air. Solution

Hint Clear Info

Incorrect Attempts:

CHECK

N

Hint Unavailable

At rest, the tension force is equal and opposite the force due to gravity of the crate. (Define up as positive). $\begin{align} F_{Tension} & = -F_{gravity} \\ \\ F_T & = -(m)(g) \\ \\ & = -(500kg)(-9.81 m/s^2) \\ \\ & = +4905\ N \\ \\ \end{align}$

Determine the acceleration of the helicopter when it takes off vertically with an upward thrust at 60% of maximum thrust. Solution

Hint Clear Info

Incorrect Attempts:

CHECK

m/s²

Hint Unavailable

60% of maximum thrust = (0.6)(60 kN) = 36 kN = 36,000 N

The crate and helicopter are connected with an inextensible cable so they move together and the acceleration of the whole system uses the net force with the total mass of the system of the helicopter plus crate... (Define up as positive)... $\begin{align} F_{NET} & = F_{thrust} + F_{gravity} \\ \\ (m_{total})(\vec{a}_{system}) & = F_{thrust} + (m_{total})(g) \\ \\ \vec{a}_{system} & = \dfrac{F_{thrust} + m_{total})(g)}{m_{total}} \\ \\ & = \dfrac{+36,000\ N + (2000\ kg + 500\ kg)(-9.81\ m/s^2)}{2000\ kg + 500\ kg} \\ \\ & = \dfrac{36,000N - 24,525\ N}{2,500\ kg} \\ \\ & = 4.59\ m/s^2 \\ \\ \end{align}$

Calculate the tension force in the cable connected to the crate when the helicopter takes off. Solution

Hint Clear Info

Incorrect Attempts:

CHECK

N

Net Force = (Tension Force) + (Force due to Gravity)

The cable is weighed down by the force due to gravity of the crate. The upward thrust is adding a tension force due to the acceleration of the system of 4.59 m/s²... (Define up as positive)... $\begin{align} F_{NET\ in\ cable} & = F_{Tension} + F_{gravity} \\ \\ (m_{crate})(a_{system}) & = F_T + (m_{crate})(g) \\ \\ F_T & = (m_{crate})(a_{system}) - (m_{crate})(g) \\ \\ & = m_{crate}(a_{system} - g) \\ \\ & = (500\ kg)(+4.59\ m/s^2 - (-9.81\ m/s^2)) \\ \\ & = (500\ kg)(14.4\ m/s^2) \\ \\ & = 7,200\ N \\ \\ \end{align}$

Circular Motion

An object travelling at constant velocity on a circular track is continually accelerating. Solution

1. True
2. False

Objects moving in circles, whether velocity is constant or not, are always accelerating.

Circular Motion, Centripetal Acceleration, Midpoint of the Circle

A 2.0kg mass swings on a 50cm string around in a vertical circle at 5 m/s. Find the tension in the string at the midpoint on one side of the swing. Solution

Hint Clear Info

Incorrect Attempts:

CHECK

N

Hint Unavailable

Circular Motion, Centripetal Acceleration, Bottom of the Circle

A 1kg mass is swinging around on a string in a vertical loop with a 9cm radius, at a constant velocity of 5m/s, find the force of tension in the string at the bottom of the loop. Solution

Hint Clear Info

Incorrect Attempts:

CHECK

N

Force of tension is negative here

Circular Motion, Centripetal Acceleration, Flat Curve

A car with a mass of 1,400 kg drives around a flat curve at 72 km/h. The curve has a radius of 100m.

Draw a free body diagram and determine the magnitude of the force due to centripetal acceleration. Solution

Hint Clear Info

Incorrect Attempts:

CHECK

N

Hint Unavailable

Find the minimum coefficient of static friction required to keep the car's tires from slipping during the turn. Solution

µ_s =

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

Circular Motion, Centripetal Acceleration, Flat Curve, With Friction

A vehicle travels around a circular, unbanked curve at 12 m/s. If the total mass of the vehicle is 500 kg, and if the force of friction that must be present for the tires not to slip is 1000 N, then determine the smallest radius of the circle. Solution

r =

Hint Clear Info

Incorrect Attempts:

CHECK

m

Hint Unavailable

If the tires do not slip, then the centripetal force and the force of friction are equal magnitude (and opposite direction): $\begin{align} F_c & = F_f \\ \\ \dfrac{mv^2}{r} & = 1000N \\ \\ \dfrac{mv^2}{r} & = 1000N \\ \\ r & = \dfrac{mv^2}{1000\ N} \\ \\ r & = \dfrac{(500\ kg)(12)^2}{1000\ N} \\ \\ r & = 72\ m \\ \\ \end{align}$

Circular Motion, No Friction, Banked Curve

A car with a mass of 1,800 kg drives around a frictionless banked curve with a radius of 120m. The angle of inclination of the banked curve is 12˚ from the horizontal.

Find the speed the car needs to maintain to travel around the curve without sliding up or down the frictionless embankment. Solution

Hint Clear Info

Incorrect Attempts:

CHECK

m/s

Hint Unavailable

Prove that the mass of the vehicle has no effect on the speed required to drive around the curve without slipping. Show your work in your own notes. Solution

Circular Motion, Centripetal Force

Clyde swings a 5.0 kg ball of seaweed around in horizontal circles on the beach. If the center of mass of the seaweed is 1.0m from his hands, and the maximum force that seaweed can withstand before breaking is 85N, then find the maximum rate of revolution in rpm. Solution

Hint Clear Info

Incorrect Attempts:

CHECK

rpm

Hint Unavailable

Circular Motion, No Friction, Vertical Loop (Top)

If the radius of a roller-coaster loop is 30m, find the minimum speed a roller coaster must have in order to complete a vertical loop on a track without stressing the safety rails at the top of the loop, i.e. without coming off the track at the top of the loop. Solution

Hint Clear Info

Incorrect Attempts:

CHECK

m/s

mass cancels

Circular Motion, No Friction, Banked Curve

An engineered highway is being built with a banked curve with a radius of 200m. Find the angle the highway should be banked at for cars to safely drive at 108 km/hr. Solution

θ =

Hint Clear Info

Incorrect Attempts:

CHECK

˚

mass cancels

Conservation of Energy: Theory of Potential Energy (PE) and Kinetic Energy (KE)

Describe the total energy of the system in terms of potential energy and kinetic energy, at the top and bottom of each situation.

A ball is released from rest at a height h, above the ground. Solution

A ball is thrown straight up with an initial velocity v. Solution

Use conservation of energy to solve for the final velocity of a ball thrown horizontally, given an initial velocity and height. Solution

Conservation of Energy: Dropped from Rest

A 33kg mass is dropped from rest, 100m above the ground. Calculate the speed of the object at the instant before it hits the ground. Solution

Hint Clear Info

Incorrect Attempts:

CHECK

m/s

The 'instant' before the object hits the ground, it has an instantaneous maximum velocity. This question can either be solved with energy or kinematics.

Conservation of Energy: Gravities on Different Planets (PE = mgh)

Planet Earth and planet Zuto are being compared in physics class. The gravity on the smaller, imaginary planet Zuto is G_z = -5.00 m/s². A mass m on Earth, and a heavier mass 3m on planet Zuto are thrown upwards and reach the same height. Calculate the factor of speed that the Earth object would have to be thrown compared to the speed of the Zuto object. Solution

Earth speed =

Hint Clear Info

Incorrect Attempts:

CHECK

times the Zuto speed

Mass cancels

Ratios of Kinetic Energy

An object is travelling in space with an initial velocity of 25 m/s.

If the kinetic energy is reduced by 25%, find the new speed. Solution

Hint Clear Info

Incorrect Attempts:

CHECK

m/s

Hint Unavailable

If the object slows down to half its initial speed, calculate the integer ratio of the initial kinetic energy to the final kinetic energy. Solution

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

Energy: Work on an Incline Plane

An object rolls 100m down a frictionless hill that's inclined 30˚ to the horizontal. If the weight is 120N, find:

The work due to gravity. Enter your answer as a positive value. Solution

Hint Clear Info

Incorrect Attempts:

CHECK

J

Hint Unavailable

The work due to the normal force. Solution

Hint Clear Info

Incorrect Attempts:

CHECK

J

Hint Unavailable

Conservation of Energy with Potential Energy of Spring Converted to Kinetic Energy

A ball is compressed onto a spring vertically. Given that mass is 2.0 kg, the spring is compressed 1.0 m and the spring constant, k is 200 N/m, calculate the velocity v_r of the ball at the moment it was released from the spring. Solution

Hint Clear Info

Incorrect Attempts:

CHECK

m/s

Hint Unavailable

Conservation of Energy with Potential Energy of Spring Converted to Kinetic Energy, with Friction

A new 1.0 kg wind-up toy car developed by NASA is a marvel in engineering because it has near-zero friction in the internal mechanism. Winding the toy car to the max, stretches a spring 5.0 cm. The spring has a spring constant of 100 N/m. The car is released and travels a distance of 900cm over a surface with friction. Determine the maximum velocity of the car, and the coefficient of kinetic friction (µ_k). Solution

Hint Clear Info

v_max = m/s
µ_k =

Incorrect Attempts:

CHECK

Hint Unavailable

Conservation of Energy: Kinetic Energy converted to Potential Energy of a Spring or Work

A safety mechanism in a train station is built at the end of a track to catch a runaway train - the mechanism basically mimics one extremely gigantic spring. A 90,000kg crash-test train is tested on this spring. Determine the spring constant of the spring, required to stop a train with an initial velocity of 25m/s, with a maximum deceleration of 40m/s². Solution

k =

Hint Clear Info

Incorrect Attempts:

CHECK

N/m

First calculate the compression of the spring, then calculate the spring constant.

Conservation of Energy (PE, KE) and Centripetal Force

A rollercoaster car with mass m, slides along a frictionless track from a height h, down an incline into a loop with a radius r. Find the minimum height h (in terms of r) the car must start from in order to remain on the track at the top of the loop without falling. (Hint: At the top the loop, F_N = 0) Solution

Hint Clear Info

h = r

Incorrect Attempts:

CHECK

Hint Unavailable

Momentum: Inelastic Collision, Objects Stuck Together

At a car derby, two of the cars crash and stick together. Car 'A' had an initial speed of 20 m/s and a mass of 2000kg, while Car 'B' was at rest with a mass of 1000kg. Find the speed of the mass of stuck-together cars after the crash. Solution Video

Hint Clear Info

Incorrect Attempts:

CHECK

m/s

Momentum is conserved: (total initial momentum) = (total final momentum)

Momentum: Inelastic Collision, Same Directions

A 50kg skateboarder pushes a 5kg skateboard out in-front of herself with a velocity of 2.0 m/s [N]. If she runs with a velocity of 3.0 m/s [N] and jumps on the skateboard, then what is the velocity of the skateboarder just after she jumps on the skateboard? Solution Video

Hint Clear Info

Incorrect Attempts:

CHECK

m/s

Hint Unavailable

$\begin{align} m_1v_1 + m_2v_2 & = m_T v_T \\ \\ (50)(3) + (5)(2) & = (50 + 5)(v_T) \\ \\ 150 + 10 & = (55)(v_T) \\ \\\ v_T & = \dfrac{160}{55} \\ \\ v_T & = 2.9\ m/s \\ \\ \end{align}$

Momentum: Elastic Explosion

An explosion sends two objects flying outwards in opposite directions from the center of the explosion. The velocities of the objects are considered to be zero before the explosions. The mass of object 'A' is 5m, and the mass of object 'B' is 2m. Find the final velocity of one object in terms of the velocity of the other. Solution

Hint Clear Info

━━ V_fB

Incorrect Attempts:

CHECK

Both objects start from rest: initial velocity = 0 m/s

Momentum: Inelastic Collision, No Stick

An artillery round with a mass 100g and an initial speed of 300m/s pierces a stationary block of test wood with mass 10kg. The artillery round exits the block of wood with a considerable speed of 90 m/s. Find the speed with which the test block of wood moves. Solution

Hint Clear Info

Incorrect Attempts:

CHECK

m/s

Hint Unavailable

Momentum: Impulse and Average Force

A 1.5kg object gets slugged (impacted) by a rather powerful force at a speed of 50 m/s. If the contact lasts for 0.03s, find the average force exerted by this slugging. Solution

Hint Clear Info

Incorrect Attempts:

CHECK

N

Hint Unavailable

Momentum: Elastic Collision, Combined with Conservation of Energy

A 5 kg ball rolls down a frictionless hill, which is inclined at an angle of 20˚ to the horizontal. The ball starts rolling from the top of the hill - a vertical height of 10 m. The ball collides elastically with a 10kg block resting at the base of the hill causing the block to move at 9.33 m/s.

Find the speed of the ball at the base of the hill before the collision, and the final speed of the ball after the collision. Solution

Hint Clear Info

V_{base of hill} = m/s
V_final = m/s

Incorrect Attempts:

CHECK

To solve, use conservation of, then momentum

Determine the (hypotenuse) distance the ball bounces back up the hill coming to rest. Solution

Hint Clear Info

Incorrect Attempts:

CHECK

m

Hint Unavailable

Momentum in 2-Dimensions: Elastic Collision

A pool ball (ball'A') with a mass m, and an initial horizontal velocity of 5m/s strikes another pool ball with the same mass that is initially at rest (ball 'B'). Ball 'A' strikes ball 'B' and deflects at an angle 30˚ to the horizontal, while ball 'B' deflects at 60˚ to the horizontal below the horizontal. Find the final velocities of both pool balls. Solution

Hint Clear Info

v_fA =
v_fB =

Incorrect Attempts:

CHECK

m/s

Split the velocities into their 'x' and 'y' components

Momentum in 2-Dimensions: Elastic 'Explosion'

A 1000kg satellite travelling in space at a blistering 300m/s must alter it's path 8˚ from the direction of its travel. In order to change its path slightly, the rocket blasts a shot of gas 2000m/s at an angle perpendicular to the initial plane on which it is travelling. Determine the mass of gas that must be blasted outwards, assuming the mass of the satellite remains constant before and after the blast. Solution

m =

Hint Clear Info

Incorrect Attempts:

CHECK

kg

Use 'x' and 'y' components

Simple Harmonic Motion (SHM): Spring

What factors affect the period of a spring: (check all that apply) Solution

1. Mass, m
2. Gravity, g
3. Spring constant, k
4. Length of spring, l
5. Displacement, x

Simple Harmonic Motion (SHM): Pendulum

What factors affect the period of a pendulum: (check all that apply) Solution

1. Mass, m
2. Gravity, g
3. Displacement, x
4. Length of pendulum, l
5. Height, h

Simple Harmonic Motion (SHM): Frequency and Period

An object oscillating in simple harmonic motion completes 30 cycles in 2.75s. Determine the frequency f, and period T. Solution

Hint Clear Info

f = Hz
T = s

Incorrect Attempts:

CHECK

Hint Unavailable

Simple Harmonic Motion (SHM): Springs

A spring with a 100N/m spring constant is loaded with a 3kg mass and the period is measured. Then another unknown mass is added to the 3kg mass and the period is measured to be 1.0 second longer. Determine the unknown mass. Solution

m =

Hint Clear Info

Incorrect Attempts:

CHECK

kg

Hint Unavailable

Spring before $\begin{align} T_1 & = 2\pi\sqrt{\dfrac{m}{k}} \\ \\ & = 2\pi\sqrt{\dfrac{3}{100}} \\ \\ & = 1.088\ s \\ \\ \end{align}$ Spring after, (let x represent the added mass to solve for x): $\begin{align} T_2 & = 2\pi\sqrt{\dfrac{3 + x}{100}} \\ \\ T_1 + 1.0 & = 2\pi\sqrt{\dfrac{3 + x}{100}} \\ \\ 2.088 & = 2\pi\sqrt{\dfrac{3 + x}{100}} \\ \\ \sqrt{\dfrac{3 + x}{100}} & = \dfrac{2.088}{2\pi} \\ \\ \left(\sqrt{\dfrac{3 + x}{100}}\right)^2 & = \left(\dfrac{2.088}{2\pi}\right)^2 \\ \\ \dfrac{3 + x}{100} & = 0.11 \\ \\ 3 + x & = 11 \\ \\ x & = 8\ kg \\ \\ \end{align}$

Simple Harmonic Motion (SHM): Period (T) and Angular Frequency (ω)

A simple pendulum has a period of 2.0s and an initial displacement of 0.04m. At a later time, the amplitude is reduced to 0.03m.

Calculate the length of the pendulum. Solution Video

Hint Clear Info

Incorrect Attempts:

CHECK

m

Hint Unavailable

Calculate the angular velocity, ω. Solution

ω =

Hint Clear Info

Incorrect Attempts:

CHECK

rad/s

Hint Unavailable

$\begin{align} ω & = 2 \pi f \\ \\ & = 2 \pi \dfrac{1}{T} \\ \\ & = 2 \pi \dfrac{1}{2.0s} \\ \\ & = 3.14\ rad/s \\ \\ \end{align}$

Calculate the angular acceleration, α. Solution

α =

Hint Clear Info

Incorrect Attempts:

CHECK

rad/s²

Hint Unavailable

$\begin{align} α & = \dfrac{ω}{T} \\ \\ & = 2 \pi f \dfrac{1}{T} \\ \\ & = 2 \pi \dfrac{1}{T} \dfrac{1}{T} \\ \\ & = \dfrac{2 \pi}{T^2} \\ \\ & = \dfrac{2 \pi}{(2.0s)^2} \\ \\ & = 1.57\ rad/s^2 \\ \\ \end{align}$

Determine the tangential velocity, v_tangential. Solution

Hint Clear Info

Incorrect Attempts:

CHECK

m/s

Hint Unavailable

$\begin{align} v_{tangential} & = ω r \\ \\ & = 2 \pi \dfrac{1}{T} (0.99m) \\ \\ & = 2 \pi \dfrac{1}{2.0s} (0.99m) \\ \\ & = 3.1\ m/s \\ \\ \end{align}$

Find the percent reduction in energy from the initial to the later time. Solution

Hint Clear Info

Incorrect Attempts:

CHECK

%

Hint Unavailable

Simple Harmonic Motion (SHM): a_max

A spring is stretched 10cm from its equilibrium position and released. The period T, of the oscillation is 0.75s. Find the:

Maximum acceleration Solution

Hint Clear Info

Incorrect Attempts:

CHECK

m/s²

Hint Unavailable

Displacement over 0.5s Solution

Hint Clear Info

Incorrect Attempts:

CHECK

m

Hint Unavailable

Acceleration units

The following is a correct determination of units for acceleration. Solution $\begin{align} & = ω^2 A \\ \\ & = \left( \dfrac{rad}{s}\right)^2 (m) \\ \\ & = \dfrac{\cancel{rad^2}\,m}{s^2} \\ \\ & = \dfrac{m}{s^2} \\ \\ \end{align}$

1. True
2. False

Simple Harmonic Motion (SHM): Amplitude and Displacement at t_x

The distance between the points of maximum acceleration on an oscillating object is 54 cm. The time it takes the object to go from the point of maximum acceleration to the nearest point of maximum speed is 0.05 seconds. Determine the amplitude in order to calculate displacement at t = 0.75 seconds. Solution

Hint Clear Info

Incorrect Attempts:

CHECK

m

Hint Unavailable

Simple Harmonic Motion (SHM): Springs, Spring Constant, and Angular Frequency

A giant spring supports a 2000 kg mass and oscillates at a frequency of 1 Hz. Find the spring constant of this giant spring. Solution

k =

Hint Clear Info

Incorrect Attempts:

CHECK

N/m

Hint Unavailable

Simple Harmonic Motion (SHM)

A 1500 kg car is distributed evenly over 4 identical springs (one at each wheel). If the average displacement of each spring is 10cm, determine the frequency. Solution

f =

Hint Clear Info

Incorrect Attempts:

CHECK

Hz

Hint Unavailable

First determine the spring constant (k) for each spring. The mass on each spring is 1500kg/4 = 375kg $\begin{align} F_{spring} = -kx & = F_g \\ \\ kx & = mg \\ \\ k & = \dfrac{mg}{x} \\ \\ k & = \dfrac{(375kg)(10m/s^2)}{0.1m} \\ \\ k & = 37,500 N/m \\ \\ \end{align}$ Then, calculate the frequency with the equation: $\begin{align} \dfrac{1}{f} & = 2\pi\sqrt{\dfrac{m}{k}} \\ \\ \dfrac{1}{f} & = 2\pi\sqrt{\dfrac{375}{37,500}} \\ \\ f & = \dfrac{1}{2\pi\sqrt{\dfrac{375}{37,500}}} \\ \\ f & = \frac{5}{\pi} \\ \\ f & = 1.59\ Hz \\ \\ \end{align}$

Gravitation: Units Intro

The Mega or M symbol is equivalent to Solution

1. 1.0 × 10⁹
2. 1.0 × 10⁶
3. 1.0 × 10³
4. 1.0 × 10¹²

Mega (1.0 × 10⁶) is the next factor of 1,000 greater than kilo (1.0 × 10³).

Mega = 1,000,000

kilo = 1,000

Gravitational Potential Energy

The two equations for gravitational potential energy are interchangeable at all altitudes. Solution $U = mgh \quad\quad\quad\quad\quad U = \dfrac{GMm}{r}$

1. True
2. False

U = mgh is an approximation that can be used only near the surface of the Earth.

For high Earth altitudes and in space, the equation $U = \dfrac{GMm}{r} \ $ must be used.

Centripetal Force, F_c

An object thrown straight up into the air experiences a centripetal force around the Earth. Solution

1. True
2. False

The object only has F_c when the it is rotating in an orbit, tangential to the F_c. Since an object thrown straight up is not moving tangential to the Earth, then it does not have a centripetal force.

Forces

The centripetal and gravitational forces are equal in magnitude for an object in constant orbit. Solution

1. True
2. False

Furthermore, this gives (derives) the equation for orbital velocity: $\begin{align} F_c & = F_g \\ \\ \dfrac{mv^2}{r} & = \dfrac{GMm}{r^2} \\ \\ \dfrac{v^2}{r} & = \dfrac{GM}{r^2} \\ \\ \dfrac{v^2}{1} & = \dfrac{GM}{r} \\ \\ v^2 & = \dfrac{GM}{r} \\ \\ v & = \sqrt{\dfrac{GM}{r}} \\ \\ \end{align}$

Gravitational Constant, G

The gravitational constant, G is equal in all parts of the universe. Solution G = 6.67 × 10^-11 Nm²/kg²

1. True
2. False

The gravitational constant is the same anywhere in the universe. It is used to convert units of mass and distance to a force.

Gravitation: Kepler's Laws

Which of the following is Kepler's first law? Solution

1. A line joining a planet and the Sun sweeps out equal areas during equal intervals of time.
2. The orbit of every planet is an ellipse with the Sun at one of the two foci.
3. The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit.
4. The acceleration of an object is proportional to the force acting upon it.

Most people think the orbit is a perfect circle, but this is not the case. Extreme examples of elliptical orbits can be observed in the orbit shape of comets.

Kepler's second law means that a planet moves faster when it's closer to the Sun. Solution

1. True
2. False

Kepler's second law: A line joining a planet and the Sun sweeps out equal areas during equal intervals of time. Since the orbit is elliptical, the speed of the planet changes when it is in different parts of its orbit.

Derive Kepler's Third Law. Solution

Set the force of gravity (F_g) = centripetal force (F_c): $\begin{align} F_g & = F_c \\ \\ \dfrac{GMm}{r^2} & = \dfrac{mv^2}{r} \\ \\ \dfrac{GM}{r^2} & = \dfrac{v^2}{r} \\ \\ \dfrac{GM}{r} & = \dfrac{v^2}{1} \\ \\ ① \ \ \ \dfrac{GM}{r} & = v^2 \\ \\ \end{align}$ Write an equation for the speed of a circle (approximation: $\begin{align} v_{circular} & = \dfrac{d}{t} \\ \\ ② \ \ \ v & = \dfrac{2 \pi r}{T} \\ \\ \end{align}$ Substitute this velocity (v) into equation ① $\begin{align} \dfrac{GM}{r} & = \left( \dfrac{2 \pi r}{T} \right)^2 \\ \\ \dfrac{GM}{r} & = \left( \dfrac{4 \pi^2 r^2}{T^2} \right) \\ \\ \dfrac{GM}{1} & = \dfrac{4 \pi^2 r^3}{T^2} \\ \\ \dfrac{GM}{4 \pi^2} & = \dfrac{r^3}{T^2} \quad\quad\quad\quad\quad\quad \dfrac{GM}{4 \pi^2} \ is \ constant \\ \\ \end{align}$ Kepler's third law: The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit.

Gravitation: Radius and Escape Velocity

Increasing the radius of a planet, would decrease the escape velocity. Solution

1. True
2. False

$\begin{align} V_{esc} & = \sqrt{\dfrac{2GM}{R_{planet}}} \\ \\ \end{align}$ Increasing 'r' will decrease v_esc

Gravitation: v_escape

Which of the following would double the escape velocity required to escape a planet's orbit? Solution

1. Double the mass of the projected object
2. Double the radius from the planet's center
3. Increase the mass of the planet by a factor of four
4. Increase the radius from the planet's center by a factor of four

Note that the only mass in the equation is for the central gravitational body. The mass of the projected object is irrelevant to orbital velocity.
$\begin{align} v_{escape} & = \sqrt{\dfrac{2GM}{r}} \\ \\ & = \sqrt{\dfrac{2G(4M)}{r}} \\ \\ & = 2\sqrt{\dfrac{2G(M)}{r}} \\ \\ \end{align}$ We pull out the square root of 4 as 2.

(Or to double the escape velocity, we could also decrease the radius by a factor of four.)

Gravitation: v_escape

A rocket ship must have a velocity greater than the 'escape velocity' of a planet in order to overcome the planet's gravity. If a new rocket ship has a maximum velocity of 1.0 × 10⁴m/s, determine which of the following planets the rocket could not escape and would get stuck on, and state the escape velocity of that planet. (G = 6.67 × 10^-11 Nm²/kg²) Solution

Planet Name	Mass (kg)	Radius (m)
Mars	6.42 × 1023	3.39 × 106
Earth	5.97 × 1024	6.38 × 106
Moon (of Earth)	7.35 × 1022	1.74 × 106
Zygote	4.35 × 1025	6.12 × 107

1. Mars
2. Earth
3. Moon
4. Zygote

v_escape =

Hint Clear Info

Incorrect Attempts:

CHECK

m/s

Hint Unavailable

Compare maximum mass/radius ratio of a planet that the rocket can leave and escape. Can escape when planet's M/R ratio is less than 7.493 × 10¹⁷.

The rocket ship cannot leave Earth because the escape velocity for Earth is 11,155 m/s, which is higher than 1.0 × 10⁴ m/s. $\begin{align} v_{escape} & = \sqrt{\dfrac{2GM}{r}} \\ \\ & = \sqrt{\dfrac{2(6.67 x 10^{-11})(5.97 x 10^{24})}{6.4 x 10^6}} \\ \\ & = 11,155\ m/s \\ \\ \end{align}$

Force of Gravity

A very skilled soccer player has managed to balance two identical 435 g soccer balls with an 11 cm radius, on top of one another. Determine the force of gravity of one ball on the other. (G = 6.67 × 10^-11 Nm²/kg²) Solution

Hint Clear Info

× 10

Incorrect Attempts:

CHECK

N

Hint Unavailable

Make sure to use 'r' (separation distance) between the centers of masses of the balls:

The separation distance of the centers of mass equals 2(0.11m) = 0.22m $\begin{align} F_g & = \dfrac{Gm_1m_2}{r^2} \\ \\ & = \dfrac{(6.67 × 10^{-11})(0.435)(0.435)}{(0.22)^2} \\ \\ & = 2.6 × 10^{-10} N \\ \\ \end{align}$

Gravitational: Force, Potential, and Potential Energy

Gravitational potential $\ V = \dfrac{GM}{r}$ can be calculated using which of the following equivalent methods? Solution

1. Gravitational potential energy divided by orbit radius
2. Gravitational field strength divided by the orbit radius
3. Gravitational force times orbit radius
4. Gravitational field strength times orbit radius

$\begin{align} g & = \dfrac{GM}{r^2} \\ \\ V & = \dfrac{GM}{r^2} × r \\ \\ V & = \dfrac{GM}{r} \\ \\ \end{align}$ That's it!

Gravitational Field Strength

Determine the gravitational field strength of a 2.0 × 10⁵ kg space station located 200 km above the surface of the earth. (Given: Radius_Earth = 6.4 × 10⁶ m, G = 6.67 × 10^-11 Nm²/kg², Mass_Earth = 5.98 × 10²⁴ kg) Solution

Hint Clear Info

Incorrect Attempts:

CHECK

N/kg

Hint Unavailable

The mass used is always the central body, in this case it's the Earth.

Mass of the object in orbit is irrelevant in determining gravitational field strength.
200km = 200 × 10³ m
$\begin{align} g & = \dfrac{GM_{Earth}}{r^2} \\ \\ & = \dfrac{(6.67 × 10^{-11})(5.98 × 10^{24})}{\left((6.4 × 10^6) + (200 × 10^3)\right)^2} \\ \\ & = 9.16\ m/s^2 \\ \\ \end{align}$ Note the units of 'g' can also be stated as (N/kg).

Gravitational Potential

Rounding the distance between the center of masses of the Sun and Earth gives 1.496 × 10¹¹ m. If a 1kg meteoroid is located some distance between the Sun and the Earth, determine the distance from Earth that this meteoroid would experience a net gravitational potential, V equal to zero. (Given: Mass_Earth = 5.98 × 10²⁴ kg, Mass_Sun = 2.0 × 10³⁰ kg) Solution

Hint Clear Info

× 10

Incorrect Attempts:

CHECK

m

Hint Unavailable

V at the meteoroid equals zero when V_Sun = V_Earth.
Let d be the distance from the Sun to the Earth.
Let x be the distance from the meteoroid to the Earth.
So d - x is the distance from the Sun to the meteoroid. Let $\begin{align} V_{Sun} & = V_{Earth} \\ \\ \dfrac{GM}{r} & = \dfrac{GM}{r} \\ \\ \dfrac{GM_{Sun}}{r} & = \dfrac{GM_{Earth}}{r} \\ \\ \dfrac{\cancel{G}M_{Sun}}{d-x} & = \dfrac{\cancel{G}M_{Earth}}{x} \\ \\ \dfrac{M_{Sun}}{d-x} & = \dfrac{M_{Earth}}{x} \\ \\ (M_{Sun})(x) & = (M_{Earth})(d-x) \\ \\ (2.0 × 10^{30})(x) & = (5.98 × 10^{24})(d-x) \\ \\ (2.0 × 10^{30})(x) & = (5.98 × 10^{24})\left((1.496 × 10^{11})-x\right) \\ \\ 2.0 × 10^{30}x & = 8.946 × 10^{35} - 5.98 × 10^{24}x \\ \\ 2.0 × 10^{30}x & = 8.946 × 10^{35} \\ \\ x & = 4.47 × 10^{5}\ m \quad\quad\quad\quad\quad or \ \ 447\ km \\ \\ \end{align}$

Theory: Kinetic Energy and Radius

A satellite in high Earth orbit has altered its orbit and is slowly drifting towards Earth. As the orbital radius of the satellite decreases, the kinetic energy increases. Solution

1. True
2. False

As the satellite moves closer to the center of mass of the Earth, the radial distance (r) gets smaller, and the force of gravity (F_g) gets stronger.

The stronger force accelerates the satellite towards Earth, so velocity (v) increases.

Therefore kinetic energy increases because: $KE = \dfrac{1}{2}mv^2$

Schwarzschild Radius

The Schwarzschild radius is the radius of an object so dense that the escape velocity of this object would equal the speed of light. This is common of black holes. Determine the Schwarzschild radius for a black hole with the mass of the Earth. (Given: Mass_Earth = 5.98 × 10²⁴ kg, G = 6.67 × 10^-11 Nm²/kg², c = 3.0 × 10⁸ m/s) Solution

r =

Hint Clear Info

Incorrect Attempts:

CHECK

m

Hint Unavailable

$\begin{align} v_{escape} & = \sqrt{\dfrac{2GM}{r}} \\ \\ c & = \sqrt{\dfrac{2GM}{r}} \\ \\ c^2 & = \dfrac{2GM}{r} \\ \\ r & = \dfrac{2GM}{c^2} \\ \\ & = \dfrac{2(6.67 × 10^{-11})(5.98 × 10^{24})}{(3.0 × 10^{8})^2} \\ \\ & = 0.0089 m \\ \\ \end{align}$

Conservation of Gravitational Potential Energy

A 100g meteoroid is travelling directly towards Earth at 1,000m/s. If the meteoroid is 1.0 × 10¹⁰ m from the surface of the Earth, calculate an estimation of the speed of impact of the meteoroid at the Earth's surface. (Given: G = 6.67 × 10^-11 Nm²/kg², Mass_Earth = 5.98 × 10²⁴ kg, Radius_Earth = 6.38 × 10⁶ m) Solution

Hint Clear Info

Incorrect Attempts:

CHECK

m/s

Hint Unavailable

Conservation of Energy $\begin{align} \Delta KE & = \Delta U \\ \\ KE_2 - KE_1 & = U_2 - U_1 \\ \\ \dfrac{1}{2}\cancel{m}v_2^2 - \dfrac{1}{2}\cancel{m}v_1^2 & = \dfrac{GM\cancel{m}}{r_2} - \dfrac{GM\cancel{m}}{r_1} \\ \\ \dfrac{1}{2}v_2^2 - \dfrac{1}{2}v_1^2 & = \dfrac{GM}{r_2} - \dfrac{GM}{r_1} \\ \\ v_2^2 - v_1^2 & = 2\dfrac{GM}{r_2} - 2\dfrac{GM}{r_1} \\ \\ v_2^2 & = 2\dfrac{GM}{r_2} - 2\dfrac{GM}{r_1} + v_1^2 \\ \\ v_2 & = \sqrt{2\dfrac{GM}{r_2} - 2\dfrac{GM}{r_1} + v_1^2} \\ \\ v_2 & = \sqrt{2\dfrac{(6.67 × 10^{-11})( 5.98 × 10^{24})}{6.38 × 10^6} - 2\dfrac{(6.67 × 10^{-11})( 5.98 × 10^{24})}{6.38 × 10^6 + 1.0 × 10^{10}} + (1000)^2} \\ \\ & = 11,223 m/s \\ \\ \end{align}$

Orbit Velocity

Derive the equation for orbit velocity from conservation of energy. Solution

$\begin{align} KE_{orbit} & = KE \\ \\ \frac{GMm}{2r} & = \frac{1}{2}mv^2 \\ \\ \frac{GM\cancel{m}}{\cancel{2}r} & = \frac{1}{\cancel{2}}\cancel{m}v^2 \\ \\ \frac{GM}{r} & = v^2 \\ \\ v^2 & = \frac{GM}{r} \\ \\ v & = \sqrt{\frac{GM}{r}} \\ \\ \end{align}$

Orbit Velocity

A 1000kg GPS satellite has been designed for medium Earth orbit between 2,000 and 35,000km above the surface of the Earth. The maximum velocity of the satellite is 3000m/s. If the minimum gravitational field strength that satellite can have is 1.2 N/kg, determine if the satellite can make it to medium Earth orbit, hence calculate the height. Solution

Hint Clear Info

Incorrect Attempts:

CHECK

km

Hint Unavailable

$\begin{align} v_{orbit} & = \sqrt{\dfrac{GM}{r}} \\ \\ v_{orbit}^2 & = \dfrac{GM}{r} \\ \\ M & = \dfrac{v^2 r}{G} \\ \\ \\ \\ \\ \\ \\ \\ g & = \dfrac{GM}{r^2} \\ \\ g & = \dfrac{G\left(\dfrac{v^2 r}{G}\right)}{r^2} \\ \\ g & = \dfrac{v^2}{r} \\ \\ r & = \dfrac{v^2}{g} \\ \\ & = \dfrac{(3000)^2}{1.2} \\ \\ & = 7.5 × 10^6 \ m \\ \\ & = 7.5 × 10^6 - 6.38 × 10^6 \quad\quad \text{subtract radius of Earth} \\ \\ & = 1.12 × 10^6 \ m \quad\quad or \ 1120 \ km\\ \\ \end{align}$ The satellite will not make it high enough to medium Earth orbit.

Orbit Velocity

A satellite orbiting the Earth with a speed of 0.75km/s. Given the radius of the Earth is 6.38 × 10⁶ m, find the hight above the earth. (Givens: Mass_Earth = 5.98 × 10²⁴ kg, G = 6.67 × 10^-11 Nm²/kg²) Solution

Hint Clear Info

× 10

Incorrect Attempts:

CHECK

m

Hint Unavailable

$\begin{align} v_{orbit} & = \sqrt{\dfrac{GM}{r}} \\ \\ v & = \sqrt{\dfrac{GM}{(height \ + \ r_E)}} \\ \\ v^2 & = \dfrac{GM}{(height \ + \ r_E)} \\ \\ (height \ + \ r_E) & = \dfrac{GM}{v^2} \\ \\ height & = \dfrac{GM}{v^2} - r_E \\ \\ & = \dfrac{(6.67 × 10^{-11})(5.98 × 10^{24})}{(750)^2} - 6.38 × 10^6 \\ \\ & = 7.03 × 10^8 \ m \\ \\ \end{align}$

Electric Intro

The mu or micro sign, µ is equivalent to Solution

Hint Clear Info

× 10

Incorrect Attempts:

CHECK

Hint Unavailable

milli (m) is 1.0 × 10^-3

micro (µ) is 1.0 × 10^-6

(See that these are different by a factor of 10³ or 1,000)

Electric Force

Changing the separation distance of two 10µC point charges by a factor of 3/4 will affect the electric force by what factor? Solution

1. $\dfrac{16}{9}$
2. $\dfrac{4}{3}$
3. $\dfrac{3}{4}$
4. $\dfrac{9}{16}$

Sub the 3/4 into the equation. Notice that it gets squared in the denominator. Then, the reciprocal (flipped form) is factored out: $\begin{align} F & = \dfrac{kq_1q_2}{r^2} \\ \\ & = \dfrac{kq_1q_2}{\left(\dfrac{3}{4} r\right)^2} \\ \\ & = \dfrac{kq_1q_2}{\dfrac{9}{16} r^2} \\ \\ & = \dfrac{16 kq_1q_2}{9 r^2} \\ \\ & = \dfrac{16}{9} \cdot \dfrac{kq_1q_2}{r^2} \\ \\ \end{align}$

Electric Force

Determine the electric force between the nucleus of a hydrogen atom and its valence electron if the separation distance is 1.07 × 10^-14 m. (Given: k = 8.99 × 10⁹ Nm²/C², q_e^- = 1.6 × 10^-19 C, q_p⁺ = 1.6 × 10^-19 C) Solution

Hint Clear Info

Incorrect Attempts:

CHECK

N

Hint Unavailable

The charge on a proton or electron is ±1.6 × 10^-19 C $\begin{align} F & = \dfrac{kq_1q_2}{r^2} \\ \\ & = \dfrac{(8.99 × 10^9)(1.6 × 10^{-19})(-1.6 × 10^{-19})}{(1.07 × 10^{-14})^2} \\ \\ & = \dfrac{(8.99 × 10^9)(1.6 × 10^{-19})(-1.6 × 10^{-19})}{(1.145 × 10^{-28})} \\ \\ & = -2.0\ N \\ \\ \end{align}$

Electric Charges

A positive test charge will move in which direction when placed 1.0 m [N 40˚ W] from a stationary –0.58 µC charge? Solution

1. [N 40˚ W]
2. [E 40˚ S]
3. [S 40˚ E]
4. [N 60˚ E]

The positive test charge will move towards the stationary -0.58µC charge.

So if the charge is placed [N 40˚ W] then it will move towards directly (parallel) in the opposite, [S 40˚ E] direction.

Electric Potential Energy

Determine the change in electric potential energy when a 2.5 µC charge and a 1.5 mC start 1.0 m apart, and are moved to within 0.25 m away. Solution

Hint Clear Info

Incorrect Attempts:

CHECK

J

Hint Unavailable

$\begin{align} F & = \dfrac{kq_1q_2}{r^2} \\ \\ \dfrac{U}{r} & = \dfrac{kq_1q_2}{r^2} \\ \\ \dfrac{U}{1} & = \dfrac{kq_1q_2}{r} \\ \\ U & = \dfrac{kq_1q_2}{r} \\ \\ \Delta U & = U_2 - U_1 \\ \\ & = \dfrac{kq_1q_2}{r_2} - \dfrac{kq_1q_2}{r_1} \\ \\ & = kq_1q_2 \left(\dfrac{1}{r_2} - \dfrac{1}{r_1} \right) \\ \\ & = (8.99 × 10^9)( 2.5 × 10^{-6})(1.5 × 10^{-3}) \left(\dfrac{1}{0.25} - \dfrac{1}{1.0} \right) \\ \\ & = 101 \ J \\ \\ \end{align}$

Separation Distance and Energy

The magnitude of the Total Energy between two charges is always lower than the magnitudes of the individual, Kinetic and Potential Energies. Solution

1. True
2. False

Net Electrostatic Force

A -6µC charge is placed between two stationary charges in the diagram below. Determine the direction the -6µC charge will move. (Given: k = 8.99 × 10⁹ Nm²/C²) Solution

1. To the right
2. To the left
3. It will not move
4. There is not enough information to determine

$\begin{align} F & = \dfrac{kq_1q_2}{r^2} \\ \\ F_{left} & = \dfrac{(8.99 × 10^9)(12 × 10^{-6})(-6 × 10^{-6})}{(0.04)^2} \\ \\ & = -404.55 \ N \\ \\ \\ \\ \\ \\ \\ \\ F_{right} & = \dfrac{(8.99 × 10^9)(30 × 10^{-6})(-6 × 10^{-6})}{(0.08)^2} \\ \\ & = -252.84 \ N \\ \\ \\ \\ \\ \\ \\ \\ \end{align}$ The force to the left is stronger. Therefore the -6µC charge will move to the left.

A +1C test charge is placed at point 'B'. Determine the net electrostatic force of the stationary charges 'A' 'C' and 'D' on point 'B'. (Given: k = 8.99 × 10⁹ Nm²/C²) Solution

Hint Clear Info

× 10

Incorrect Attempts:

CHECK

N

Hint Unavailable

Determine the directions the charges will push q_B.
F_A push right, F_C pull right, F_D push left. $\begin{align} F_{\text{A on B}} & = \dfrac{kq_1q_2}{r^2} \\ \\ & = \dfrac{(8.99 × 10^9)(5)(1)}{(0.1)^2} \\ \\ & = 4.5 × 10^{12} \ N \ \text{[right]} \\ \\ F_{\text{C on B}} & = \dfrac{kq_1q_2}{r^2} \\ \\ & = \dfrac{(8.99 × 10^9)(3)(1)}{(0.08)^2} \quad\quad\quad\quad \text{Use absolute value of 3C} \\ \\ & = 4.2 × 10^{12} \ N \ \text{[right]} \\ \\ F_{\text{D on B}} & = \dfrac{kq_1q_2}{r^2} \\ \\ & = \dfrac{(8.99 × 10^9)(10)(1)}{(0.12)^2} \\ \\ & = 6.2 × 10^{12} \ N \ \text{[left]} \\ \\ \\ \\ \\ \\ F_{\text{NET}} & = 4.5 × 10^{12} \ N \ + \ 4.2 × 10^{12} \ N \ - \ 6.2 × 10^{12} \ N \\ \\ & = 2.5 × 10^{12} \ N \ \text{[right]} \\ \\ \end{align}$

Electric Fields

The S.I. units of electric field are: Solution

1. Vm
2. kgm/s
3. kg²/m³
4. N/C

$\begin{align} E & = \dfrac{F}{q} \\ \\ & = \dfrac{N}{C} \\ \\ \end{align}$ (Also acceptable as V/m)

Electric Field and Electric Potential

Electric potential (V) is a scalar, and electric field (E) is a vector. Solution

1. True
2. False

Vectors depend on direction, while scalars do not.

Increasing the separation distance of two charged plates by a factor of 1.5, and decreasing the electric field by one third would not change the magnitude of electric potential. Solution

1. True
2. False

$\begin{align} V & = E \cdot d \\ \\ & = \left( \dfrac{2}{3}E \right) \cdot \left( \dfrac{3}{2}d \right) \\ \\ & = \dfrac{1}{1} E \cdot d \\ \\ & = E \cdot d \\ \\ & \text{Result Unchanged} \\ \\ \end{align}$

Electric Field

A stationary +30 µC charge is located 44 cm from another stationary -10 µC charge. Determine the magnitude of the electric field at the midpoint between the two charges. Solution

Hint Clear Info

× 10

Incorrect Attempts:

CHECK

N/C

Hint Unavailable

Electric field is a vector, so use proper vector addition. The electric field points away from the positive charge, and towards the negative charge. At the midpoint between a +30µC and -10µC charge, the electric field is pointing in the same direction, therefore the absolute value (magnitudes only) of the fields add. E field is a vector. $\begin{align} E & = \dfrac{k q}{r^2} \\ \\ E_{NET} & = E_1 - E_2 \\ \\ & = \dfrac{k q_1}{r^2} - \dfrac{k q_2}{r^2} \\ \\ & = \dfrac{(8.99 × 10^9) (30 × 10^{-6})}{(0.22)^2} - \dfrac{(8.99 × 10^9) (-10 × 10^{-6})}{(0.22)^2} \\ \\ & = \dfrac{(8.99 × 10^9) (30 × 10^{-6})}{(0.22)^2} + \dfrac{(8.99 × 10^9) (10 × 10^{-6})}{(0.22)^2} \\ \\ & = 7.43 × 10^{6} \ N/C \\ \\ \end{align}$

Energy of a Charge in an Electric Field

If the energy to move a +5 C charge through a 20 V potential difference is equal to the work to move another +2 C charge through an electric field with plates separated by a distance of 40 cm, determine the magnitude of the electric field. Solution

Hint Clear Info

Incorrect Attempts:

CHECK

N/C

Hint Unavailable

$\begin{align} U & = qV \\ \\ \text{Energy of charge 2} & = \text{Work of charge 1} \\ \\ U_2 & = U_1 \\ \\ q_2(V_2) & = q_1(V_1) \\ \\ q_2(V_2) & = q_1(E\cdot d) \quad\quad\quad\quad V = E\cdot d \\ \\ E & = \dfrac{q_2(V_2)}{q_1 d} \quad\quad\quad\quad \text{Rearrange, isolate of 'E'} \\ \\ & = \dfrac{(5)(20)}{(2)(0.4)} \\ \\ & = 125 \ N/C \\ \\ \end{align}$

Electric Fields due to Point Charges and Plates

An electric field can be generated by a point charge and is uniform at a constant radial distance from the center of this charge. Alternatively, a uniform electric field can be generated by the separation of charged plates. The electric field 20cm from a point charge is equal to the electric field generated by a potential difference of 100V across plates separated by 20cm. Determine the magnitude of the point charge. (Given: k = 8.99 × 10⁹ Nm²/C²) Solution

Hint Clear Info

× 10

Incorrect Attempts:

CHECK

C

Hint Unavailable

$\begin{align} \text{Electric field from charge} & = \text{Electric field from plates} \\ \\ \dfrac{kq}{r^2} & = \dfrac{V}{d} \\ \\ q & = \dfrac{V \cdot r^2}{d\cdot k} \\ \\ & = \dfrac{100 \cdot (0.2)^2}{(0.2)\cdot (8.99 × 10^9)} \\ \\ & = 2.2 × 10^{-9} \ C \\ \\ \end{align}$

Electric Force

A +1 C charge is located 1.0m West of a stationary +1 C charge at Point 'B' and another +2 C charge is located 1.0 m South of Point 'B'. Determine the magnitude of the force felt at Point 'B'. Solution

Hint Clear Info

× 10

Incorrect Attempts:

CHECK

N

Hint Unavailable

Electric force is a vector quantity so we must use vector addition. $\begin{align} F_{west} & = \dfrac{kq_1q_2}{r^2} \\ \\ F_{west} & = \dfrac{(8.99 × 10^9)(1)(1)}{(1)^2} \\ \\ \\ \\ \\ \\ \\ \\ F_{south} & = \dfrac{kq_1q_2}{r^2} \\ \\ F_{south} & = \dfrac{(8.99 × 10^9)(2)(1)}{(1)^2} \\ \\ \\ \\ \\ \\ \\ \\ \left( F_{NET} \right)^2 & = \left( F_{west} \right)^2 + \left( F_{south} \right)^2 \\ \\ F_{NET} & = \sqrt { \left( \dfrac{(8.99 × 10^9)(1)(1)}{(1)^2} \right)^2 + \left( \dfrac{(8.99 × 10^9)(2)(1)}{(1)^2} \right)^2 } \\ \\ & = 2.0 × 10^{10} \ N \ \\ \\ \end{align}$

Electric Potential

A 2.0 × 10^-6 C charge is placed at point 'A' 40 cm from a -1.0 × 10^-6 C charge at point 'B'. Determine the electric potential at point 'P' 30 cm perpendicularly above point 'B'. (Given: k = 8.99 × 10⁹ Nm²/C²) Solution

Hint Clear Info

Incorrect Attempts:

CHECK

kV

Hint Unavailable

Electric potential (V) is a scalar quantity, so we can simply add and subtract without vector addition.

Make sure to find and use the hypotenuse distance between 'AP' ... = 0.5m $\begin{align} V_{\text{Total}} & = V_1 + V_2 \\ \\ & = \dfrac{kq_1}{r_1} + \dfrac{kq_2}{r_2} \\ \\ & = \dfrac{(8.99 × 10^9)(2.0 × 10^{-6})}{0.5} + \dfrac{(8.99 × 10^9)(-1.0 × 10^{-6})}{0.3} \quad\quad\quad\quad (0.3)^2 + (0.4)^2 = (0.5)^2 \\ \\ & = 5993 \ V \\ \\ & = 6 \ kV \\ \\ \end{align}$

Electric Potential Energy is Conserved

Determine the electric potential necessary to accelerate an electron to the speed of light, 3.0 × 10⁸ m/s. (Given: q_e^- = 1.6 × 10^-19 C, m_e^- = 9.1 × 10^-31 kg) Solution

Hint Clear Info

× 10

Incorrect Attempts:

CHECK

V

Energy is conserved, think about which two equations to set equal.

$\begin{align} \text{Initial Energy} & = \text{Final Energy} \\ \\ \text{Electric Potential Energy} & = \text{Kinetic Energy} \\ \\ U & = KE \\ \\ qV & = \dfrac{1}{2}mv^2 \\ \\ V & = \dfrac{1}{2q}mv^2 \\ \\ & = \dfrac{1}{2(1.6 × 10^{-19})}(9.1 × 10^{-31})(3.0 × 10^8)^2 \\ \\ & = 2.56 × 10^5 \ V \\ \\ \end{align}$

A -2.0 × 10^-6 C charge with a mass of 1.0 × 10^-8 kg is located 2.0 m from a stationary +1.0 × 10^-6 C charge. If the negative charge accelerates directly towards the central charge, determine the final velocity of the negative charge. (k = 8.99 × 10⁹ Nm²/C²) Solution

Hint Clear Info

Incorrect Attempts:

CHECK

m/s

Energy is conserved, think about which two equations to set equal.

$\begin{align} \text{Initial Energy} & = \text{Final Energy} \\ \\ \text{Electric Potential Energy} & = \text{Kinetic Energy} \\ \\ U & = KE \\ \\ \dfrac{kq_1q_2}{r} & = \dfrac{1}{2}mv^2 \\ \\ v^2 & = \dfrac{2kq_1q_2}{mr} \\ \\ v & = \sqrt{\dfrac{2kq_1q_2}{mr}} \\ \\ & = \sqrt{\dfrac{\left|2(8.99 × 10^9)(2.0 × 10^{-6})(1.0 × 10^{-6})\right|}{(1.0 × 10^{-8})(2)}} \\ \\ & = 1341 \ m/s \\ \\ \end{align}$

Conventional Current

Conventional current is loosely defined as the flow of positive charges. Solution

1. True
2. False

While it is the delocalized (free) electrons in a metal that carries a current, by convention the flow of positive charges in the opposite direction is used.

Magnetic Field Units: Tesla

The Tesla unit is equivalent to which S.I. units? Solution

1. $\dfrac{N}{A \cdot m}$
2. $\dfrac{N}{s \cdot C \cdot m}$
3. $\dfrac{N \cdot m}{C \cdot s}$
4. $\dfrac{A}{N \cdot m}$

$\begin{align} F & = qvBsin\theta \\ \\ F & = qvBsin90 \\ \\ F & = qvB \\ \\ (N) & = (C)\left(\dfrac{m}{s}\right)(T) \\ \\ T & = \dfrac{N \cdot s}{C \cdot m} \\ \\ & = \dfrac{N \cdot s}{C \cdot m} \\ \\ & = \dfrac{N}{m} \cdot \dfrac{s}{C} \quad\quad\quad\quad\quad\quad Note: \ A = \dfrac{C}{s} \\ \\ & = \dfrac{N}{m} \cdot \dfrac{1}{A} \\ \\ & = \dfrac{N}{A \cdot m} \\ \\ \end{align}$

Magnetic Fields: Intro

Select the false statement regarding magnetic field lines. Solution

1. Magnetic field lines from one object never cross
2. Magnetic field lines exit a material directed outwards from the center of mass
3. The density of magnetic field lines is directly proportional to the strength of the magnetic field.
4. Magnetic field lines are a form of light

Although light is an electro-magnetic wave, the magnetic component of light is not to be confused with magnetic field lines.

Magnetic Fields: Grip Rule

The Right Hand Grip Rule, for the flow of charges in a current-carrying wire, points the thumb in the direction of: Solution

1. The flow of positive charges
2. The flow of negative charges
3. The magnetic field
4. The electric field

The right hand grip rule is used for the flow of positive charges (conventional current) in a current carrying wire.

The fingers point in the circular direction of the magnetic field.

If the conventional current in a current-carrying wire is directed into the page, at a location above the wire, the magnetic field will point in what direction? Solution

1. To the right
2. To the left
3. Into the page
4. Out of the page

Conventional current is the flow of positive charges, so use the right hand grip rule:
The thumb points in the same direction as current, and the fingers indicate the direction of the magnetic field.

Point the thumb into the page, and notice the fingers wrap around in a clockwise direction. Above the wire, the magnetic field will point to the right.

The Left Hand Rule

The Left Hand Rule points the thumb in the direction of force, the index finger in the direction of magnetic field, and the second finger in the direction of positive charges. Solution

1. True
2. False

Note that the Right Hand Rule would be used for the direction of movement of a negative charge.

The Right Hand Rule

The Right Hand Rule points the thumb in the direction of force, the index finger in the direction of magnetic field, and the second finger in the direction of negative charges (electrons). Solution

1. True
2. False

The Right Hand Rule would be used for the direction of movement of a negative charge.

(Note that this is not the same as the Right Hand Grip Rule).

Charges Moving in a Magnetic Field

In which of the following examples is the force directed into the page? Solution

1. I
2. II
3. III
4. IV

Force is out of the page (use left hand rule for negative charges)

III) Force is downwards (use right hand rule for negative charges)

IV) Force is upwards (use right hand rule for negative charges)

Determine the direction of the force, if any, on the positive charge (+q) in the uniform magnetic field below. Solution

1. Into the page
2. Up
3. Down
4. The force equals zero, so no direction

Force equals zero if the angle between charge velocity (v) and magnetic field (B) is zero.

Draw the path of the negative charge (-q) in the uniform magnetic field below. Solution

The maximum force on a current-carrying wire in a magnetic field is produced when: Solution

1. The wire is perpendicular to the magnetic field
2. The wire is 45˚ to the magnetic field
3. The wire is -180˚ to the magnetic field
4. The wire is parallel to the magnetic field

As theta increases from 0˚ to 90˚, the value of $sin\theta$ increases. Therefore the maximum force is experienced when the angle between the wire and magnetic field is 90˚. $\begin{align} F & = ILBsin\theta \\ \\ \end{align}$

Magnetic Force on a Current-Carrying Wire

A current-carrying wire in a uniform magnetic field generates a force. The direction of this force can be reversed by reversing the direction of the current or the magnetic field. Solution

1. True
2. False

The uniform magnetic field (B) is represented with the large arrows below.

Two current-carrying wires in different directions are represented by the gray circles.

You can see the resulting forces are shown in blue.

Changing the direction of the current in a uniform magnetic field changes the direction of the force (in blue) changes.

Charges Moving in a Magnetic Field: Mass Spectrometer

Which of the following would decrease the radius of curvature of a charge moving through a uniform magnetic field in a mass spectrometer? Solution

1. Increasing the charge
2. Increasing the mass of the charge
3. Decreasing the magnetic field
4. Increasing gravity

Things that (directly) increase radius of curvature: m
- Increasing mass → increases the radius
- Decreasing mass → decreases the radius


Things that (inversely) decrease radius of curvature: q, B
- Increasing charge → decreases the radius
- Decreasing charge → increases the radius
- Increasing magnetic field → decreases the radius
- Decreasing charge → increases the radius

Solenoid

A solenoid with an iron core is an electromagnet in which the strength is directly proportional to the amount of current. Solution

1. True
2. False

An electromagnet is created by an electric field (B).

For a solenoid with n turns, i current, and L length: $B = \dfrac{u_0 n i}{L}$ As you can see, the magnitude of the magnetic field (B) is directly proportional to current (i).

Determining Radius of Curvature of a Charge Moving in a Magnetic Field

A 1.67 × 10^-27kg proton is accelerated from rest through a potential difference of 12V. The charge was initially moving from the top of the page to the bottom of the page. The proton exits the electric field and enters a uniform 10mT magnetic field directed out of the page.

Determine the speed of the proton as it exits the electric field. (Given: q = 1.602 × 10^-19C) Solution

Hint Clear Info

× 10

Incorrect Attempts:

CHECK

m/s

Hint Unavailable

The energy is conserved as electric potential energy (U) is converted into kinetic energy (KE). $\begin{align} KE & = U \\ \\ \dfrac{1}{2}mv^2 & = qV \\ \\ v & = \sqrt{\dfrac{2qV}{m}} \\ \\ & = \sqrt{\dfrac{2(1.602 × 10^{-19})(12)}{1.67 × 10^{-27}}} \\ \\ & = 4.80 × 10^4 \ m/s \\ \\ \end{align}$

Determine the radius of curvature of the charge in the magnetic field. Solution

Hint Clear Info

Incorrect Attempts:

CHECK

m

Hint Unavailable

The magnetic force (F_B) balances the centripetal force (F_C). $\begin{align} F_B & = F_C \\ \\ qvBsin\theta & = \dfrac{mv^2}{r} \\ \\ qBsin\theta & = \dfrac{mv}{r} \\ \\ qBsin90 & = \dfrac{mv}{r} \\ \\ qB & = \dfrac{mv}{r} \\ \\ r & = \dfrac{mv}{qB} \\ \\ & = \dfrac{(1.67 × 10^{-27})(4.80 × 10^4)}{(1.602 × 10^{-19})(10 × 10^{-3})} \\ \\ & = 0.05 \ m \\ \\ \end{align}$

Determine whether the charge will rotate clockwise or counterclockwise in the magnetic field. [1] Solution

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

Use the left hand rule for positive charges. The index finger points out of the page, and the second finger points down. The force is directed to the left. This will cause the charge to rotate clockwise.

Magnetic Field in a Wire Loop

A wire loop with a 40 cm radius is carrying a 20 A (conventional) current counterclockwise. Determine the maximum magnitude of the magnetic field and the direction. (Given: µ₀ = 4$\pi$ × 10^-7 Tm/A) Solution

1. 3.14 × 10^-5 T, Out of the page
2. 3.14 × 10^-5 T, Into the page
3. 1.00 × 10^-5 T, Into the page
4. 1.00 × 10^-5 T, Out of the page

$\begin{align} B & = \dfrac{\mu_0 i}{2r} \\ \\ & = \dfrac{(4\pi × 10^{-7}) (20)}{2(0.4)} \\ \\ & = 3.14 × 10^{-5} \ T \\ \\ \end{align}$ Using the right hand grip rule, the magnetic field points out of the page at the center of the loop.

Magnetic Field from Magnetic Poles

A magnetic field generated by North and South poles is always directed from North to South.

1. True
2. False

Force on a Current-Carrying Wire

Determine the current necessary to levitate a 10 g, 10 cm wire set perpendicular to a 740 mT magnetic field. Solution

Hint Clear Info

Incorrect Attempts:

CHECK

A

Hint Unavailable

Think what levitation is. Levitation is the point where the magnetic force is equal and opposite to the force due to gravity: $\begin{align} F_B & = F_g \\ \\ iLBsin\theta & = mg \\ \\ i & = \dfrac{mg}{LBsin\theta} \\ \\ & = \dfrac{(0.01)(9.81)}{(0.1)(740 × 10^{-3})sin90} \\ \\ & = 1.3 \ A \\ \\ \end{align}$

Force on a Current-Carrying Wire

A current carrying wire is placed in a magnetic field. Determine the direction of force. Solution

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

Conventional current points into the page. Magnetic field is always directed from N to S, which in this case is to the left. Using the left hand rule for conventional current, the force is directed upwards.

Light Sources

Waves from light sources that are coherent have the same frequency and are in phase with each other. Solution

1. True
2. False

This is what coherent means -- that the light sources have the same frequency and are in phase with each other. This is significant when it comes to constructive and destructive interference of waves.

Light Speeds

Different colors of light like red, violet, or green, have different speeds (v). Solution Video

1. True
2. False

The speed doesn't change, but wavelength and frequency do change.

Principle of Superposition (Review)

For total destructive interference to occur between two waves, what must be the same? Solution Video

1. Amplitude
2. Frequency
3. Wavelength
4. Path Difference Multiple of ½λ

1. II and IV
2. I, III, and IV
3. I and III
4. I, II, III, and IV

Amplitude must be the same so that total destructive interference occurs.

Frequency and wavelength must be the same so that the crests and troughs can align.

The path difference must be n multiple of ±(n)½λ.

Interference of Two Sources

The following diagram depicts two coherent sources, where solid lines represent wavefront crests and dotted lines indicate troughs. At which point(s) does destructive interference occur? Solution Video

1. A only
2. B only
3. C only
4. C or A

Destructive interference occurs where a crest overlaps a trough.

(Constructive interference occurs where two crests overlap, or where two troughs overlap).

Diffraction

What is diffraction? [2] Solution Video

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

Diffraction is the curvature of a wavefront (of any wave like light, sound, water) that occurs when the wave passes through a narrow opening, or aperture.

The diffraction pattern looks like 'ripples in a pond' and can create interference patterns if more than one source is present.

Young's Double Slit Experiment

Thomas Young's experiment involved a monochromatic light source, projected through a single slit, then projected through a double slit.

Young used slits that act as new light sources according to Huygen's principle. Solution Video

1. True
2. False

Huygen's principle suggests that a wavefront is capable of acting as a 'source' for new waves.

As the light passes through a narrow opening, it diffracts in wave-like patterns. Each narrow opening, or slit is considered as a light 'source'.

The main significance of Young's slit experiment is that: Solution

1. Wavelengths are proportional to path difference
2. The slits used act as light sources
3. It demonstrates that light is not just particle photons, but that it also has properties of waves.
4. It shows the diffraction of light from two different sources

This is a bit tricky because all of these factors area a little bit significant.

However, there is one thing is more significant than the rest. Light has properties of particles (Einstein and photoelectric effect of photons) and waves.


Young's experiment is the experiment that proves light has properties of waves!

Calculate the path difference pd, in the two waves originating from double slits if the angle from the horizontal is 30˚ and the distance, d between the slits is 0.5 mm. Solution Video

pd =

Hint Clear Info

Incorrect Attempts:

CHECK

mm

Hint Unavailable

Distance (d) and path difference (pd) are same units.
They don't need to be in any particular unit, as long as they are the same. $\begin{align} sin \theta & = \dfrac{opposite}{hypotenuse} \\ \\ sin \theta & = \dfrac{pd}{d} \\ \\ pd & = d · sin \theta \\ \\ pd & = (0.5\ mm)sin 30˚ \\ \\ pd & = 0.25\ mm \\ \\ \end{align}$

Young's Double Slit

The following diagram just shows two out of the many waves from diffraction in a double slit experiment. In reality, what will actually occur on a screen? Solution

1. Bright fringes only
2. Dark fringes only
3. The waves do not interfere constructively nor destructively
4. Both bright fringes and dark fringes, at different locations on the screen

Light waves with a path difference equal to an integer number of wavelengths (n wavelengths) will constructively interfere and create bright fringes on the screen.

In the example given, the path difference is equal to 1 wavelength (1 λ).

The following path differences will also produce bright fringes: 0 λ, 1 λ, 2 λ, 3 λ, 4 λ, ...

Calculations with Young's Double Slit

Fringes are only bright regions where constructive interference occurs. Solution Video

1. True
2. False

Fringes are not only bright regions.

Dark fringes appear where destructive interference occurs between two waves that are a multiple of ½λ apart in phase (path).

Dark fringes are between the bright fringes on the screen.

Monochromatic light originates from two slits with a separation distance of 0.2 mm. If two adjacent (1^st order) bright fringes are 15˚ apart, determine the wavelength of the wave. Solution Video

λ =

Hint Clear Info

Incorrect Attempts:

CHECK

mm

Hint Unavailable

n (or sometimes m) is the number of spaces between fringes:

"adjacent (1^st order) bright fringes" means n = 1
d = 0.2 mm
$\begin{align} n \lambda & = d · sin \theta \\ \\ \lambda & = \dfrac{d · sin \theta}{n} \\ \\ \lambda & = \dfrac{0.2 sin 15˚}{1} \\ \\ \lambda & = 0.05\ mm \\ \\ \end{align}$

Determine which of the following equations could be used for dark fringes (total destructive interference), where n is the integer number of wavelengths (1, 2, 3...), Solution Video

1. $n + \lambda = d ·\text{sin} \theta$
2. $\left(n + \dfrac{1}{2}\right)\lambda = d· \text{sin} \theta$
3. $n·\lambda = d ·\text{cos} \theta$
4. $\left(\dfrac{1}{2}·n\right)\lambda = d ·\text{sin} \theta$

Destructive interference occurs when the path difference is a multiple of ½λ.

Double Slit Small Angle Approximation

For light passing between two slits that are close together, y represents the distance from the central bright fringe to other fringes on a screen. The horizontal distance from the double slits to the screen is represented by x.

For very small angles, the following relationship is true. Solution Video $\text{sin}\,\theta ≈ \text{tan}\,\theta$

1. True
2. False

This approximation is true for small angles. For example:

tan(10˚) = 0.176
sin(10˚) = 0.174

These values are close enough to say tan(z˚) ≈ sin(z˚)

Assuming a small angle and given the relationship above, with the diagram and equation below, which of the following formulas is correct? (Use the small angle approximation and then use the information in the diagram to make an equation, then isolate for wavelength) Solution $n \lambda = d · sin \theta$

d is the distance between the slits
x or D is the distance from the slits to the screen
m or n is order, the number of spaces between fringes
y is the distance between fringes on the screen.
(Interchangeable: n is m, x is D).

Hey, it's the small angle approximation!
$\begin{align} tan \theta & = \dfrac{opposite}{adjacent} \\ \\ tan \theta & = \left(\dfrac{y}{x}\right) \\ \\ \end{align}$ $\begin{align} n \lambda & = d · sin \theta \\ \\ \\ \\ n \lambda & ≈ d · tan \theta \\ \\ n \lambda & ≈ d · \left(\dfrac{y}{x}\right) \\ \\ n \lambda & ≈ \dfrac{d · y}{x} \\ \\ \lambda & ≈ \dfrac{d · y}{n · x} \\ \\ \end{align}$
(remember, n is the number spaces between the bright fringes.)

1. $\lambda ≈ \dfrac{d · y}{n · x}$
2. $\lambda ≈ \dfrac{n · x}{d · y}$
3. $\lambda ≈ \dfrac{x · y}{n · d}$
4. $\lambda ≈ \dfrac{x + y}{n - d}$

A monochromatic light source shines between two thin slits spread 0.1 mm apart, onto a screen 10 cm from the slits. The central, brightest fringe is 15cm from an adjacent bright fringe. The angle between the light waves is very small. Determine the wavelength of the light, in centimeters. Solution

λ =

Hint Clear Info

Incorrect Attempts:

CHECK

cm

Hint Unavailable

Use the small angle approximation
See how some of the units of distance are different, so convert all units to the same thing:
d = 0.1 mm = 0.01 cm
x = 10 cm
y = 15 cm
Adjacent bright fringes means n = 1 $\begin{align} \lambda & ≈ \dfrac{d · y}{n · x} \\ \\ \lambda & ≈ \dfrac{(0.01) · (15)}{(1) · (10)} \\ \\ \lambda & ≈ \dfrac{0.15}{10} \\ \\ \lambda & ≈ 0.015\ cm \\ \\ \end{align}$ (Also accept answer: 0.15 mm)

Double Slit Small Angle Approximation

The following equation only works for small angles between a fringe and the central maximum. Solution $\lambda ≈ \dfrac{d · y}{n · x}$

1. True
2. False

Because it is based on the small angle approximation where sin ≈ tan $\begin{align} n \lambda & = d · sin \theta \\ \\ \\ \\ n \lambda & ≈ d · tan \theta \\ \\ n \lambda & ≈ d · \left(\dfrac{y}{x}\right) \\ \\ n \lambda & ≈ \dfrac{d · y}{x} \\ \\ \lambda & ≈ \dfrac{d · y}{n · x} \\ \\ \end{align}$

Destructive Interference

It is still possible to hear sound when destructive interference affects sound waves. Solution

1. True
2. False

Destructive interference does not have to be total destructive interference.

If the path difference is not exactly $½ \lambda$, then there will be some destructive interference but it will not be complete and some sound will still be heard.

Applications of Two Wave Sources

Similar to a double slit experiment, 250 Hz tuning forks emit sound waves from two sources spaced 2.0 m apart. At two places 3.5 m horizontally away from the sources, there is no sound heard. If the speed of sound in air at sea level is 340 m/s, determine the distance between the adjacent (1^st order) quiet spots. Solution

Hint Clear Info

Incorrect Attempts:

CHECK

m

Hint Unavailable

The distance between the sources, d = 2.0m
The horizontal distance between the source and the interference pattern, x = 3.5m
This distance between adjacent quiet spots is the separation distance like on a screen, y

Adjacent spots means n = 1

Sub f and v for wavelength: $\begin{align} v & = f · \lambda \\ \\ \lambda & = \dfrac{v}{f} \\ \\ \end{align}$ $\begin{align} \lambda & = \dfrac{d · y}{n · x} \\ \\ \dfrac{v}{f} & = \dfrac{d · y}{n · x} \\ \\ y & = \dfrac{n · x · v}{f · d} \\ \\ & = \dfrac{(1) · (3.5) · (340)}{(250) · (2)} \\ \\ & = \dfrac{(1) · (3.5)}{(2)} × 1.36 \\ \\ & = 2.38\ m \\ \\ \end{align}$

Single Slit

Modern experiments of single slit diffraction are still based on Huygen's principle of wavefronts, but use parallel light rays rather than a non-uniform light source (like a candle or the Sun). Solution

1. True
2. False

Huygen's principle suggests that a wavefront is capable of acting as a 'source' for new waves. As the light passes through a narrow opening, it diffracts in wave-like patterns. Each narrow opening, or slit is considered as a light 'source'.

The light rays must be parallel to make calculations more accurate. This means controllable light sources (like lasers) are used rather than non-uniform sources (like the Sun).

Which of the following formulas is used for constructive interference? Solution

1. $y = \dfrac{(m + \tfrac{1}{2}) \lambda x}{w}$
2. $y = \dfrac{m \lambda x}{w}$
3. None of the above
4. Any (all) of the above

    y   is the distance from the central fringe to the n^th order fringe.
    m   is the fringe order number, or spaces between the central and n^th fringe, (any non-zero integer).
    $\lambda$   is wavelength.
    x   is the distance from the slit to the screen.
    w   is the width of the slit.

Waves that destructively interfere are out of phase by an whole number multiple of +½λ or -½λ.
Waves that constructively interfere are out of phase by a whole number multiple of +λ or -λ.
The other formula presented here is used for destructive interference: $\begin{align} d· sin\theta & = (m + \tfrac{1}{2}\lambda) \\ \\ d· tan\theta & = (m + \tfrac{1}{2}\lambda) \\ \\ d· (\tfrac{y}{x}) & = (m + \tfrac{1}{2}\lambda) \\ \\ y & = \dfrac{(m + \tfrac{1}{2}\lambda) x }{d} \\ \\ \end{align}$

A wave passes through a single slit that is 0.2 mm wide and 3 cm from a screen. If the third order bright fringe is measured 2.1 cm from the central bright fringe, determine the wavelength, in millimeters. Solution

Hint Clear Info

Incorrect Attempts:

CHECK

mm

Hint Unavailable

'Bright fringes' means use constructive interference formula.

Givens:
y = 2.1 cm = 21 mm
m = 3
x = 3 cm = 30 mm
w = 0.2 mm
Find $\lambda$ $\begin{align} y & = \dfrac{m \lambda x}{w} \\ \\ \lambda & = \dfrac{w y}{m x} \\ \\ & = \dfrac{(.2)(21)}{(3)(30)} \\ \\ & = \dfrac{(.2)(21)}{(3)(30)} \\ \\ & = 0.05\ mm \\ \\ \end{align}$

Diffraction Gratings

Diffraction gratings provide evidence for Huygen's principle. Solution

1. True
2. False

Huygen's principle suggests that a wavefront is capable of acting as a 'source' for new waves (wavelets).

As the light passes through a narrow opening, it diffracts in wave-like patterns. Each narrow opening, or slit is considered as a light 'source'. Each slit or ridge in a diffraction grating acts as a new source.

The distances between bright fringes from diffraction gratings that have 100's or 1000's of slits are easier to measure on a screen. Solution

1. True
2. False

A greater number of slits allows for more destructive interference between bright fringes, resulting in bright fringes that are easier to see and tell apart (resolve) from one another (bottom of image).

The diffraction grating behaves like 100's or 1000's of pairs of double slits.

Creative Commons Attribution: © B. Crowell, 2002

Which of the following does not have diffraction gratings? Solution

1. A holograph image
2. A DVD
3. A prism
4. A CD

A prism is actually just refraction of light, rather than diffraction (with gratings).

Calculations with bright fringes created by diffraction gratings use the same formula as double slits: $n \lambda = d · sin \theta$

1. True
2. False

    n   is the fringe order number, or spaces between the central and n^th fringe, (any non-zero integer).
    $\lambda$   is wavelength.
    d   is the distance between the slits.
    $\lambda$   is the angle between the central axis and the wave axis.

Thin Film Interference

Thin films can create rainbow-like shimmer on some surfaces, like oil on water, or a surfactant like soap, on air.

In thin film interference a phase shift, or path difference occurs when light enters a medium with a lower index of refraction. Solution

1. True
2. False

The thin film must have a higher index of refraction (n) so the phase shift, or path difference occurs when light enters a medium with a higher index of refraction, for instance from air into oil, or air into soap.

The phase shift only occurs at the when the light goes from air→oil, or air→soap--not when light goes from oil→air or soap→air. Therefore only one of the reflected waves has this path difference.

Thin film interference would not occur without which of the following? Solution

1. The frequency is much lower than the period
2. The wavelength is far greater than the film width
3. The refractive index of the first medium is far lower than the refractive index of the second medium
4. The width of the film is less than the wavelength of light

This is required for the different colors to be seen due to the different interference patterns.

In thin film interference, why do different colors (wavelengths) appear from the presence of oil on water? Solution

1. The film has different indexes of refraction
2. The light diffracts at the medium-film boundary
3. The width of the film varies
4. The destructive interference due to path differences

The reason is that thickness of the film varies. As the light refracts through the film and undergoes total internal reflection, the phase difference depends on the thickness of the film.

Different thicknesses will cause different phase differences causing different wavelengths of light to interfere destructively.

(note that destructive interference does not make colors, instead the colors are observed due to constructive interference).

In thin film interference, some of the light will reflect off the surface of the medium, while the rest of the light gets refracted and internally reflected on the far side of the thin film layer. This creates a phase difference that causes constructive or destructive interference patterns.

1. True
2. False

If the film is 0.02 mm thin, then what will be the path (phase) difference between the reflected and the refracted wave? Solution

Hint Clear Info

Incorrect Attempts:

CHECK

mm

Hint Unavailable

The phase change difference is 2× the width of the thin film... = 0.04 mm

Describe how an antireflective coating could be chosen to reduce certain amounts of light (colors). [1] Solution

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

Antireflective properties are based on destructive interference. The thickness of the thin film determines the path difference and the wavelength of light that gets destroyed.

Make a certain width or combinations of widths that can make certain light colors (wavelengths) have an amplitude of zero through destructive interference.

In thin film interference, if the thickness of the thin film was exactly the same everywhere, we would see an even distribution of the same color. Solution

1. True
2. False

Because the different colors we see are based on the thickness of the thin film.

Polarization

Sunlight has an infinite amount of 'polarization' orientations. Solution

1. True
2. False

Sunlight is perfectly un-polarized.

Polarized Light Intensity

Polarizing materials reduce the intensity of the incoming light.

Polarizer in front of computer screen that polarizes light 45˚ (Credit: Rogilbert, Waugsberg, France)

Light with an initial intensity (I₀) enters a polarizing material, resulting in what intensity? Solution

1. ½ I₀
2. ¼ I₀
3. ¾ I₀
4. ⅛ I₀

Only the first polarizer has the intensity reduced by half, ½ I₀. All the next polarizers (2^nd, 3^rd... ) use the equation I = I₀cos²θ.

Light that passes through 3 different polarizers each at 45˚, will have what intensity? Solution

1. ¼ I₀
2. ⅜ I₀
3. ⅛ I₀
4. ¾ I₀

Light that passes through the first polarizer always is ½I₀.

Light that passes through the second polarizer has ½I₀cos²45˚

Light that passes through the third polarizer has:
$\begin{align} & (\tfrac{1}{2}I_0cos^245˚)cos^245˚ \\ \\ & = \tfrac{1}{2}I_0cos^445˚ \\ \\ & = \tfrac{1}{2}I_0(0.25) \\ \\ & = \tfrac{1}{8}I_0 \\ \\ \end{align}$
(also I = I₀/2ⁿ, where n = 3)

Multiple Polarizations

Light passes through three polarizers rotated at different angles. The first polarizer is vertical, the second is 30˚ from the first, and the third is 45˚ from the second. Calculate the ratio of the light at the end to the incident light (I₀). Solution

Hint Clear Info

━━

Incorrect Attempts:

CHECK

Hint Unavailable

After the first polarizer: $\begin{align} I_1 & = \dfrac{1}{2}I_0 \\ \\ \end{align}$ After the second polarizer: $\begin{align} I_2 & = \left(\dfrac{1}{2}I_0\right)cos^2 30˚ \\ \\ \end{align}$ After the third polarizer: $\begin{align} I_3 & = \left(\left(\dfrac{1}{2}I_0\right)cos^2 30˚\right)cos^2 45˚ \\ \\ I_3 & = \left(\left(\dfrac{1}{2}I_0\right)0.75 \right)(0.5) \\ \\ I_3 & = \dfrac{3}{16}I_0 \\ \\ \end{align}$ The ratio of final to initial is (3/16):1, or the ratio of output to incident light in fraction form is... $\begin{align} & = \dfrac{I_0}{I_3} \\ \\ & = \dfrac{\cancel{I_0}}{\dfrac{3}{16}\cancel{I_0}} \\ \\ \\ & = \dfrac{16}{3} \\ \\ \end{align}$

Polarization

A polarized light source with an intensity of 2 W/m² is polarized by 30˚. Calculate the intensity after polarization. Solution

I =

Hint Clear Info

Incorrect Attempts:

CHECK

W/m²

Hint Unavailable

Use the following equation for polarized light $\begin{align} I & = I_0 \cos^2\theta \\ \\ & = (2)\cos^2 30˚ \\ \\ & = 1.5\ W/m^2 \\ \\ \end{align}$

Brewster's Angle

Brewster's angle is which two statements? (More than one answer: check all that apply) Solution

1. The angle of reflection that polarized light transmits through a surface with no reflection
2. The angle of refraction that polarized light reflects on a surface and (the reflection) becomes unpolarized
3. The angle of incidence that certain polarized light transmits through a surface with no reflection
4. The angle of incidence that unpolarized light reflects on a surface and (the reflection) becomes polarized

For polarized light: The angle of incidence that certain polarized light transmits through a surface with no reflection.

For unpolarized light (like sunlight): The angle of incidence that unpolarized light reflects on a surface and (the reflection) becomes polarized

(Note that the surface must be a dielectric (e.g. glass, not shiny metal).

Brewster's Angle

Calculate Brewster's angle for light from going from air to glass (n_air = 1.0, n_glass = 1.5). Solution

θ =

Hint Clear Info

Incorrect Attempts:

CHECK

˚

Hint Unavailable

n₁ is the initial medium, air

n₂ is the second medium, glass. $\begin{align} tan\theta & = \dfrac{n_2}{n_1} \\ \\ \theta & = \tan^{-1}\left(\dfrac{n_2}{n_1}\right) \\ \\ \theta & = \tan^{-1}\left(\dfrac{1.5}{1.0}\right) \\ \\ \theta & = 56.3˚ \\ \\ \end{align}$

Polarization

Determine if it is possible to use one polarizer to rotate light by 90˚. [1] Solution

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

$\begin{align} I & = I_0 \cos^2 \theta \\ \\ I & = I_0 \cancel{\cos^2 90˚}^0 \\ \\ I & = 0 \\ \\ \end{align}$ No it is not possible because the light would have a final intensity (I) of zero.

Polarization

Using two sequential (one after the next) polarizers to rotate light 45˚ each time, would effectively rotate light 90˚. Solution

1. True
2. False

Through first polarizer: $\begin{align} I & = I_0 \cos^2 \theta \\ \\ I & = I_0 \cos^2 45˚ \\ \\ \end{align}$ Through second polarizer: $\begin{align} I & = (I_0) \cos^2 \theta \\ \\ I & = (I_0 \cos^2 45˚) cos^2 45˚ \\ \\ I & = I_0 (\cos^4 45˚) \\ \\ \end{align}$

Reminder of Equations and Angle Units

Which of the following equations must use radians? Solution

1. $I = I_0 \cos^2 \theta$
2. $\theta_{min} = \dfrac{1.22 · \lambda}{A}$
3. $n \lambda = d · \sin \theta$
4. $\tan\theta = \dfrac{n_2}{n_1}$

The Raleigh resolution equation must be in radians. $\theta_{min} = \dfrac{1.22 · \lambda}{A}$ (Fringe, Brewster's, and Polarization are all in degrees).

Atomic Particles

Determine the correct masses of the subatomic particles below. Solution

Electron

Neutron and Proton

1.7 × 10^-27 kg

9.3 × 10^-31 kg

A proton and neutron have the same mass.

Electrons are lighter!

Given the correct values below, state the charge on a neutron, proton, and electron. Solution

Neutron

Proton

Electron

+ 1.6 × 10^-19 C

- 1.6 × 10^-19 C

0.0 C

Protons and electrons have the same magnitude of charge, with opposite signs!

Neutrons are neutral (0.0 C of charge).

Planck's Constant

The units for the Planck Constant (h = 6.63 × 10^-34) is/are: Solution

1. J
2. kg·ms^-1
3. J·s
4. eV·s

1. I only
2. II only
3. III only
4. II or III
5. III or IV

More precisely, Planck's constant is 6.62606957 × 10⁻³⁴, but you don't need to know that much detail.

It's not easy to wrap your head around. Basically Planck's constant is the amount of Energy per Frequency for a photon (of light). Yup.

(FYI: another form of Planck's constant is 4.135667516 × 10⁻¹⁵ eV·s)

Substitute the units in the following equation, and isolate for h: $\begin{align} E & = hf \\ \\ J & = (h)(s^{-1}) \\ \\ h & = Js \\ \\ \end{align}$

Mass Energy Equivalence and electronVolts

The following unit analysis is correct. Solution $\begin{align} & \dfrac{eV}{c^2} \\ \\ & = \dfrac{eV}{\left(\dfrac{m}{s}\right)^2} \\ \\ & = eV × \dfrac{s^2}{m^2} \\ \\ & = J × \dfrac{s^2}{m^2} \\ \\ & = \left(\dfrac{kg·m^2}{s^2}\right) × \dfrac{s^2}{m^2} \\ \\ & = kg \\ \\ \end{align}$

1. True
2. False

c = speed of light

eV = electron volt = Joules

Quantum Photons

Calculate the energy of a photon with a 500 nm wavelength. (c = 3.0 × 10⁸ m/s) Solution

Hint Clear Info

× 10

Incorrect Attempts:

CHECK

J

Hint Unavailable

500 nm = 500 × 10^-9 m $\begin{align} E & = hf \\ \\ & = \dfrac{hc}{λ} \\ \\ & = \dfrac{(6.63 × 10^{-34})(3.0 × 10^8)}{ 500 × 10^{-9} } \\ \\ & = 3.98 × 10^{-19} J \\ \\ \end{align}$

Planck Formula Variations

Which one of the following formulas is wrong? Solution

1. $p = \dfrac{E}{c}$
2. $p = \dfrac{E}{fλ}$
3. $λ = \dfrac{hc}{Ef}$
4. $λ = \dfrac{h}{mv}$
5. None of the above

It should be: $\begin{align} E & = hf \\ \\ & = h \dfrac{c}{λ} \\ \\ λ & = \dfrac{hc}{E} \\ \\ \end{align}$

Energy, Joules and Electronvolts

There are 1 × 10^-11 J in 500 MeV. (1 eV = 1.6 × 10^-19 J) Solution

1. True
2. False

Check to see if this is true 1 × 10^-11 J = 500 MeV...

The 'M' in 500 MeV stands for Mega, which is 10⁶.

1 eV = 1.6 × 10^-19 J $\begin{align} & 500\ MeV \\ \\ & = 500 × 10^6 \ eV\\ \\ & = (500 × 10^6 \cancel{eV})(1.6 × 10^{-19}\ \dfrac{J}{\cancel{eV}}) \\ \\ & = 8 × 10^{-11}\ J \\ \\ \end{align}$

Quantum

Calculate the wavelength of photons with 1 eV. (c = 3.0 × 10⁸ m/s, h = 6.63 × 10^-34Js, 1 eV = 1.6 × 10^-19 J) Solution

λ =

Hint Clear Info

× 10

Incorrect Attempts:

CHECK

m

Hint Unavailable

$\begin{align} E & = hf \\ \\ E & = h \dfrac{c}{λ} \\ \\ λ & = \dfrac{c\ h}{E} \\ \\ λ & = \dfrac{(3.0 × 10^8)(6.63 × 10^{-34})}{1.6 × 10^{-19}} \\ \\ λ & = 1.2 × 10^{-6} \ m \\ \\ \end{align}$

Quantum

Photons with a total wattage of 10.0 W over a period of 10.0 s have what wavelength? (c = 3.0 × 10⁸ m/s, h = 6.63 × 10^-34Js) Solution

λ =

Hint Clear Info

× 10

Incorrect Attempts:

CHECK

m

Hint Unavailable

Calculate the energy from watts $\begin{align} power & = \dfrac{energy}{time} \\ \\ energy & = power × time \\ \\ & = 10.0 × 10.0 \\ \\ & = 100\ J \\ \\ \end{align}$ $\begin{align} E & = hf \\ \\ E & = \dfrac{hc}{λ} \\ \\ λ & = \dfrac{hc}{E} \\ \\ & = \dfrac{(3.0 × 10^8)(6.63 × 10^{-34})}{100\ J} \\ \\ & = 2.0 × 10^{-27} m \\ \\ \end{align}$

Quantum Levels

For an electron that falls down to a lower energy level, energy is absorbed. Solution

1. True
2. False

When the electron falls down to a lower level, energy is released proportional to the distance in energy levels, in electronvolts (eV).

Electron Diffraction

Particles like electrons can diffract when passed through either a single or double slit. Solution

1. True
2. False

Beams or electron particles were directed through single and double slits, and each time, diffraction patterns were observed.

This shows that matter can have wave-like properties!

Louis de Broglie (1924)

Louis de Broglie correctly hypothesized what about matter? Solution

1. Matter has a mass
2. Matter has wave-like properties
3. Matter is made of subatomic particles
4. Matter cannot be created or destroyed

For instance, the diffraction of electrons (particles!) creates fringe patterns with constructive and destructive interference.

This is interesting because it's typically thought waves that can do this, and not particles.

Schrödinger's Model of the Atom

The shape of the electron position in an atom is a: Solution

1. Sine wave
2. Sphere
3. Circle
4. Ring

The electrons moving very fast form a 3-dimensional cloud in the shape of a sphere.

Photoelectric

For a given light beam with an initial intensity I₀, increasing the intensity of the light source will increase the energy of excited electrons. Solution

1. True
2. False

Increasing intensity increases the number of photons of light, and this can excite a greater number of electrons, but the energy of each individual excited electron will remain the same, there is no difference in kinetic energy.

(Note - increasing the frequency of the light source will increase the kinetic energy of the emitted electrons, aka photoelectrons.)

Photoelectric

Hey what's threshold frequency? [1] Solution

Hint Clear Info

Incorrect Attempts:

CHECK

Hint Unavailable

For a certain kind of metal, there is a minimum frequency for electrons to be emitted from the material via the photoelectric effect.

As you can imagine, photons with a lower frequency will not be able to excite electrons in the metal.

Photoelectric Work Function

What part is the "work function" in the following equation? Solution $KE_{max} = hf - φ$

1. KE_max
2. hf
3. φ
4. h

The work function (φ) is the minimum, threshold work (energy) required for an electron to be released from a metal material.

The frequency of incoming light, hf ultimately gives the photoelectron its net energy.

The whole equation is sometimes seen as $KE_{max} = hf - hf_0$ A metal that releases electrons more easily in general, has a lower work function, φ.

Photoelectric Work Function

if the work function is greater than the energy of the incident (incoming) photons, then no electrons (photoelectrons) will be emitted from the metal material. Solution

1. True
2. False

If the work function is higher than the energy of the incident light, then no electrons will be emitted.