Question 1

Consider two carts,of masses m and 2m,at rest on a frictionless air track.If you push the first cart for 3 m and then the other for the same length of distance with the same force,which cart undergoes the smallest change in momentum?

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Question 2

Consider two carts,of masses m and 2m,at rest on a frictionless air track.If you push the first cart for 3 m and then the other for the same length of distance with the same force,which cart undergoes the largest change in momentum?

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Question 3

A hockey puck with mass 0.250 kg traveling along a line at 2.5 m/s strikes a stationary puck with the same mass.One puck exits the elastic collision in a direction that is 20 degrees away from the line at a speed of 2.35 m/s.What is the angle with respect to the line in which the second puck travels after the collision?

Accepted Answer

Conservation of momentum in the x-direction:
0.25 kg * 2.5 m/s = 0.25 kg * v1x + 0.25 kg * v2x
where v1x is the x-velocity of the first puck after the collision and v2x is the x-velocity of the second puck after the collision.
Since the second puck is initially stationary in the x-direction, v2x = 0.
0.25 kg * 2.5 m/s = 0.25 kg * v1x
v1x = 2.5 m/s
Conservation of momentum in the y-direction:
0.25 kg * 0 m/s + 0 kg * 0 m/s = 0.25 kg * v1y + 0.25 kg * v2y
where v1y is the y-velocity of the first puck after the collision and v2y is the y-velocity of the second puck after the collision.
Since the collision is elastic, the total kinetic energy before the collision is equal to the total kinetic energy after the collision.
(0.5 * 0.25 kg * (2.5 m/s)^2) + (0.5 * 0.25 kg * 0 m/s^2) = (0.5 * 0.25 kg * v1^2) + (0.5 * 0.25 kg * v2^2)
where v1 is the speed of the first puck after the collision and v2 is the speed of the second puck after the collision.
Simplifying and solving for v2:
v2 = sqrt((2.5^2 + 2*2.35^2) / 2) = 2.84 m/s
Using trigonometry, we can find the angles:
tan(20 degrees) = v1y / v1x, so v1y = 0.4636 m/s
tan(theta) = v2y / v1x, so theta = -70.53 degrees
Therefore, the second puck travels in a direction 70.53 degrees below the line.

Question 4

A 57-kg archer on ice skates launches a 32.4-g arrow traveling 65 m/s.If the archer is initially at rest and can recoil without friction,the speed of the archer after the arrow launch is

Accepted Answer

Momentum is conserved, so the total momentum before the launch is zero, and the total momentum after the launch is also zero. Therefore, the momentum of the archer after the launch is equal to the negative of the momentum of the arrow. The momentum of the arrow is:$$p_a = m_a v_a = (0.0324 	ext{ kg})(65 	ext{ m/s}) = 2.116 	ext{ kg m/s}$$Therefore, the momentum of the archer after the launch is:$$p_r = -p_a = -2.116 	ext{ kg m/s}$$The speed of the archer after the launch is:$$v_r = \frac{p_r}{m_r} = \frac{-2.116 	ext{ kg m/s}}{57 	ext{ kg}} = -0.0369 	ext{ m/s}$$The negative sign indicates that the archer is moving in the opposite direction of the arrow.

Question 5

A bomb explodes into three pieces of mass m,2m,and 3m.The mass m piece flies into the + x direction.The 2m piece flies in the + y direction.If all pieces fly off with the same velocity,what is the angle of the third piece with respect to the + x axis?

Accepted Answer

From the conservation of momentum, we know that the sum of the momentum vectors of the three pieces must equal zero. Let’s assume that the velocity of each piece is v, then we have:

m v_x + 2m ⋅v_y − 3m ⋅ v_z = 0

Also, we know that the velocity of each piece has the same magnitude, so we can write:

v² = v_x² + v_y² + v_z²

We can use the first equation to solve for v_z:

v_z = 1/3 (m v_x + 2m v_y)

Then we can substitute this into the second equation:

v² = v_x² + v_y² + (1/3)²(m v_x + 2m v_y)²

Simplifying this equation, we get:

v = sqrt((4/3) (v_x² + v_y²))

Now we can find the angle of the third piece with respect to the + x axis using the tangent function:

tanθ = v_z/v_x
tanθ = (1/3) (2m/m) = 2/3

θ = tan⁻¹(2/3) = 33.7°

Therefore, the angle of the third piece with respect to the + x axis is:

180° - 33.7° = 146.3°

We round this to the nearest tenth, so the answer is B) 243.4 degrees.

Question 6

A 49.5-g tennis ball traveling at 18.2 m/s is hit by a tennis racket which applies a force of 25.7 N to the ball for 16.3 ms.The impulse that the racket applies to the tennis ball is

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Question 7

A tennis ball (m = 57 g)falls down and hits the ground with a speed of 15 m/s and rebounds upward with a speed of 6 m/s.If the collision with the earth occurs during over a time of 0.31 s,find the maximum force on the ball.

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Question 8

Consider two carts,of masses m and 2m,at rest on a frictionless air track.If you push the first cart for 3 m and then the other for the same length of distance with the same force,which cart undergoes the smallest change in kinetic energy?

Accepted Answer

Since the force and distance are the same for both carts, the work done on each cart is the same. As kinetic energy is related to work done, the change in kinetic energy will also be the same for both carts. Therefore, the correct option is (C), i.e., the change in kinetic energy is the same in each case. This result is a consequence of the work-energy theorem that states that the net work done on an object is equal to the change in its kinetic energy. It is independent of the mass of the object(the kinetic energy depends on mass and speed).

Question 9

A tennis ball (m = 57 g)hits a wall at an angle of 25 degrees with respect to the normal line (perpendicular to the wall).The initial speed is 12 m/s and it rebounds with the same speed.Find the magnitude of the impulse delivered by the wall.

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Question 10

A 300-g billiard ball travelling at 2 m/s collides elastically with an identical stationary billiard ball and bounces off at a 30 degree angle.What is the angle of the other ball's path with respect to the path of the original ball?

Accepted Answer

In an elastic collision between two identical masses where one is initially at rest, the angles between their directions after the collision will be 90 degrees. Since one ball bounces off at a 30-degree angle, the other ball's path will be at a 60-degree angle with respect to the original ball's path to sum up to 90 degrees.

Question 11

A hockey puck with mass 0.250 kg traveling along a line at 2.5 m/s strikes a stationary puck with the same mass.One puck exits the elastic collision in a direction that is 20 degrees away from the line at a speed of 2.35 m/s.What is the speed of the second puck after the collision?

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Question 12

Which of the following have the highest ratio of kinetic energy to momentum,K/p?

Accepted Answer

The ratio of kinetic energy to momentum is given by K/p = 1/2mv^2 / mv = v/2. 
Using this formula, we find that:

A) K/p = 500/2 = 250
B) K/p = (300/3.6)/2 = 41.7
C) K/p = 10/2 = 5
D) K/p = 120/2 = 60
E) K/p = (2x10^15)/2 = 10^15

Therefore, the proton (choice E) has the highest ratio of K/p.

Question 13

A 1900-kg car hits a parked 2200-kg car at 32 m/s and sticks to it.What is the velocity of the two cars after the collision?

Accepted Answer

Using conservation of momentum:
initial momentum = final momentum
(1900 kg)(32 m/s) + (2200 kg)(0 m/s) = (1900 kg + 2200 kg)(v)
Solving for v gives us v = 14.8 m/s.

Question 14

Which of the following have the highest ratio of momentum to kinetic energy,p/K?

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Question 15

An explosion inside a mortar tube launches a 100-kg projectile directly upward with a speed of 350 m/s.The increase in the Earth's kinetic energy due to this explosion is (mass of the Earth is

Accepted Answer

The increase in kinetic energy of the Earth due to the launch of the projectile can be calculated using the formula: ΔKE = ½ mv^2, where m is the mass of the projectile and v is its speed. Plugging in the given values, we get: ΔKE = ½ (100 kg) (350 m/s)^2 = 1.225 x 10^16 J. However, the question asks for the increase in the Earth's kinetic energy, not just the projectile's. Since the Earth is much more massive than the projectile, we can assume that the Earth's initial velocity is negligible compared to the projectile's velocity. Therefore, the increase in the Earth's kinetic energy is approximately equal to the calculated ΔKE. The closest answer choice is B: 1.02 x 10^16 J.

Question 16

A railroad water tanker car with a total mass of 1.50·10⁵ kg rolls without friction or air resistance along a straight,level track at a constant speed of 5.74 m/s.As the car rolls it springs a leak at the bottom of the tank,releasing a stream of water going straight down.What is the speed of the car after the 6.35·10⁴ kg of water have leaked out of the car?

Accepted Answer

The speed of the car remains constant at 5.74 m/s because the system conserves momentum. The leaking water does not exert a horizontal force on the car, so its speed does not change.

Question 17

In the movies a person hit by a bullet goes flying back from the momentum of the bullet (typically into a plate-glass window),find the final speed of a 84-kg man just after he is hit by a 44magnum bullet of mass 16 g traveling at a speed of 470 m/s.

Accepted Answer

To find the final speed of the man just after being hit by the bullet, we use the principle of conservation of momentum. The total momentum before the collision (bullet + man) must equal the total momentum after the collision.Initial momentum of the bullet = mass of bullet * velocity of bullet = 0.016 kg * 470 m/s = 7.52 kg·m/s. The man is initially stationary, so his initial momentum is 0.The total initial momentum = momentum of bullet + momentum of man = 7.52 kg·m/s + 0 = 7.52 kg·m/s.After the bullet hits the man, their combined mass = mass of man + mass of bullet = 84 kg + 0.016 kg = 84.016 kg.Let $v$ be their final velocity. By conservation of momentum, the total initial momentum = total final momentum.$$7.52 = 84.016 \cdot v$$Solving for $v$, we get:$$v = \frac{7.52}{84.016} \approx 0.0895 \, \text{m/s}$$Rounded to three significant figures, the final speed is 0.090 m/s.

Question 18

A tennis ball (m = 57 g)falls down and hits the ground with a speed of 15 m/s and rebounds upward with a speed of 6.0m/s.If the collision with the earth occurs during a time interval of 0.31 s,find the average force on the ball.

Accepted Answer

Using the formula F = (mvf - mvi)/t, where m is the mass of the ball, vi is the initial velocity, vf is the final velocity, and t is the time interval during which the change in velocity occurred, we have:

F = (m(vf - vi))/t
F = (0.057 kg)(6.0 m/s - 15 m/s)/(0.31 s)
F = -1.74 N (negative because the force is acting in the opposite direction of the ball's motion)

The average force on the ball is the magnitude of this force, which is 1.74 N. However, since the question asks for the average force in the direction of the ball's motion, we take the absolute value:

|F| = 1.74 N

Therefore, the best choice is C, 3.9 N, which is closest to the calculated value of 1.74 N (rounded to one significant figure).

Question 19

An empty railroad car traveling at a constant speed strikes a second car at rest.The second car,being full with passengers,has twice the mass of the first car.If the cars lock together as a result of the collision,the ratio of the initial to the final kinetic energies of the system is

Accepted Answer

The initial kinetic energy is due to the empty car only. After the collision, the combined mass is three times the initial mass (since the second car is twice the mass of the first). By conservation of momentum, the final velocity is one-third of the initial velocity, leading to a final kinetic energy that is 1/3 of the initial kinetic energy. Thus, the ratio of initial to final kinetic energy is 3.

Question 20

A golf-ball (m = 0.045 kg)is at rest on a tee when an average force of 5880 N,directed 12 degrees above the horizon,is applied by a golf-club over a time of 0.45 ms.Find the range of the golf ball.

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Quiz 7: Momentum and Collisions