Question 1

A plate falls vertically to the floor and breaks up into three pieces, which slide along the floor. Immediately after the impact, a 320-g piece moves along the x-axis with a speed of 2.00 m/s and a 355-g piece moves along the y-axis with a speed of 1.50 m/s. The third piece has a mass of 100 g. In what direction does the third piece move?

Accepted Answer

Conservation of momentum can be used to solve this problem, as the total momentum before the impact is zero (since the plate was falling vertically) and the total momentum after the impact is also zero (since the pieces are moving in different directions). 
Let the velocity of the third piece be (vx, vy). Then we have:
320g(2.00 m/s) + 355g(0 m/s) + 100g(vx, vy) = 0

Solving for vx and vy, we get:
vx ≈ -0.94 m/s
vy ≈ -0.74 m/s

The direction of the third piece's velocity is given by:
θ = tan⁻¹(vy/vx) = tan⁻¹(0.74/0.94) ≈ 39.8° from the x-axis, which is closest to answer choice B.

Question 2

A plate falls vertically to the floor and breaks up into three pieces, which slide along the floor. Immediately after the impact, a 200-g piece moves along the x-axis with a speed of 2.00 m/s, a 235-g piece moves along the y-axis with a speed of 1.50 m/s. The third piece has a mass of 100 g. What is its speed?

Accepted Answer

The answer of A plate falls vertically to the floor...

Question 3

A 50.0-kg person stands on a 200-kg raft, which is at rest. She starts to walk on the raft with a speed of 3.00 m/s with respect to the raft. What is the speed of the person relative to the water?

Accepted Answer

The momentum of the person relative to the water is equal to the momentum of the person relative to the raft plus the momentum of the raft relative to the water. Since the raft is at rest relative to the water, the momentum of the person relative to the water is equal to the momentum of the person relative to the raft. Therefore, the speed of the person relative to the water is equal to the speed of the person relative to the raft, which is 3.00 m/s.

Question 4

A 0.140-kg baseball is dropped from rest from a height of 2.00 m above the ground. What is the magnitude of its momentum just before it hits the ground?

Accepted Answer

We can use the conservation of energy and the fact that at the instant just before hitting the ground, all the potential energy will be converted to kinetic energy. Therefore:

mgh = (1/2)mv^2

where m is the mass of the baseball, g is the acceleration due to gravity, h is the height from which the ball was dropped, and v is the velocity just before hitting the ground.

Solving for v, we get:

v = sqrt(2gh)

Substituting the given values, we get:

v = sqrt(2 * 9.81 m/s^2 * 2.00 m) = 4.43 m/s

The momentum just before hitting the ground is given by:

p = mv = 0.140 kg * 4.43 m/s = 0.620 kg·m/s

Therefore, the best choice is B.

Question 5

A 500-kg cannon fires a 4.0-kg projectile with a velocity of 500 m/s relative to the ground. What is the recoil speed of the cannon?

Accepted Answer

The answer of A 500-kg cannon fires a 4.0-kg projectile...

Question 6

A 0.330-kg volleyball is thrown vertically downward with a speed of 0.150 m/s. It takes it 0.0655 s to reach the ground. What is the magnitude of its momentum just before it hits the ground?

Accepted Answer

The momentum of an object is given by the product of its mass and velocity. The final velocity can be calculated using the initial velocity, acceleration due to gravity, and time. The final velocity is approximately 0.794 m/s, and the momentum is 0.330 kg * 0.794 m/s = 0.262 kg∙m/s.

Question 7

A batter hits a 0.140-kg baseball that was approaching him at 40.0 m/s and, as a result, the ball leaves the bat at 30.0 m/s in the direction of the pitcher. What is the magnitude of the impulse delivered to the baseball?

Accepted Answer

The answer of A batter hits a 0.140-kg baseball that...

Question 8

A 0.140-kg baseball is dropped from rest. It has a speed of 1.20 m/s just before it hits the ground. It rebounds with a speed of 1.00 m/s. The ball is in contact with the ground for 0.0140 s. What is the average force exerted by the ground on the ball during that time?

Accepted Answer

Using the conservation of energy: 
initial potential energy = final kinetic energy + final potential energy + work done by the ground 
mgh = (1/2)mvf^2 + mgh + Favg*d 
where h = 0 (dropped from rest), vf = 1.00 m/s, d = 2h = 0.280 m 
Solving for Favg, we get Favg = 22.0 N upwards.

Question 9

A 0.140-kg baseball is dropped from rest from a height of 2.2 m above the ground. It rebounds to a height of 1.6 m. What change in the ball's momentum occurs when the ball hits the ground?

Accepted Answer

The answer of A 0.140-kg baseball is dropped from rest...

Question 10

A 900-kg car traveling 30.0° south of east at 12.0 m/s collides with a 750-kg car traveling north at 17.0 m/s. The cars stick together. What is the speed of the wreckage just after the collision?

Accepted Answer

Using conservation of momentum, the combined momentum of the cars before the collision equals the momentum of the wreckage after. Calculate the x and y components of momentum separately, then use the Pythagorean theorem to find the resultant speed.

Question 11

A 0.140-kg baseball is dropped and reaches a speed of 1.20 m/s just before it hits the ground. It rebounds with a speed of 1.00 m/s. What is the change of the ball's momentum?

Accepted Answer

The change in momentum is equal to the final momentum minus the initial momentum.

Initial momentum = 0, since the ball is initially at rest.

Final momentum = (mass of the ball) x (final velocity)

Final momentum = (0.14 kg) x (1.00 m/s)

Final momentum = 0.14 kg∙m/s

Therefore, the change in momentum is:

Change in momentum = Final momentum - Initial momentum

Change in momentum = 0.14 kg∙m/s - 0

Change in momentum = 0.14 kg∙m/s

Since the ball rebounds with a speed of 1.00 m/s, the change in momentum is in the upward direction (opposite to the direction of gravity), so the correct answer is C) 0.308 kg∙m/s upwards.

Question 12

Two ice skaters, one with a mass of 45 kg and the other with a mass of 60 kg, push off against one another starting from a stationary position. After they have pushed off, the relative speed at which they separate is 0.600 m/s. What is the speed of the 60-kg skater?

Accepted Answer

The answer of Two ice skaters, one with a mass...

Question 13

A 900-kg car traveling east at 15.0 m/s collides with a 750-kg car traveling north at 20.0 m/s. The cars stick together. In what direction does the wreckage move just after the collision?

Accepted Answer

The answer of A 900-kg car traveling east at 15.0...

Question 14

A 900-kg car traveling east at 15.0 m/s collides with a 750-kg car traveling north at 20.0 m/s. The cars stick together. What is the speed of the wreckage just after the collision?

Accepted Answer

The answer of A 900-kg car traveling east at 15.0...

Question 15

A batter hits a foul ball. The 0.140-kg baseball that was approaching him at 40.0 m/s leaves the bat at 30.0 m/s in a direction perpendicular to the line between the batter and the pitcher. What is the magnitude of the impulse delivered to the baseball?

Accepted Answer

Impulse is the change in momentum. The initial momentum is 0.140 kg * 40.0 m/s = 5.6 kg*m/s. The final momentum is 0.140 kg * 30.0 m/s = 4.2 kg*m/s. Since the directions are perpendicular, use the Pythagorean theorem: √(5.6² + 4.2²) = 7.00 Ns.

Question 16

A 600-kg car makes a 90° turn. Its speed before the turn is 21.0 m/s and after the turn it is 24.0 m/s. What is the magnitude of the change in the car's momentum?

Accepted Answer

The change in momentum is equal to the final momentum minus the initial momentum. The final momentum is equal to the mass times the final velocity, and the initial momentum is equal to the mass times the initial velocity. Therefore, the change in momentum is equal to:$$\Delta p = p_f - p_i = mv_f - mv_i$$$$= (600 	ext{ kg})(24.0 	ext{ m/s}) - (600 	ext{ kg})(21.0 	ext{ m/s})$$$$= 19.1 	imes 10^3 	ext{ kg∙m/s}$$

Question 17

A 0.330-kg volleyball is dropped from rest. It takes it 1.30 s to reach the ground. What is the magnitude of its momentum just before it hits the ground?

Accepted Answer

Momentum is mass times velocity. The mass of the volleyball is 0.330 kg. The velocity of the volleyball just before it hits the ground can be found using the equation v = gt, where g is the acceleration due to gravity (9.8 m/s^2) and t is the time it takes for the volleyball to reach the ground (1.30 s). Plugging in these values, we get v = 9.8 m/s^2 * 1.30 s = 12.74 m/s. Therefore, the momentum of the volleyball just before it hits the ground is 0.330 kg * 12.74 m/s = 4.21 kg∙m/s.

Question 18

Two ice skaters push off against one another starting from a stationary position. The 45-kg skater acquires a speed of 0.375 m/s. What speed does the 60-kg skater acquire?

Accepted Answer

From conservation of momentum, we know that the total momentum before the push-off is zero. After the push-off, the momentum is still zero. Let the velocity of the 60-kg skater be v. Therefore, we have:
(45 kg)(0.375 m/s) + (60 kg)(v) = 0
v = - (45/60)(0.375) = -0.28125 m/s
Since the 60-kg skater acquires a velocity in the opposite direction, the magnitude of the velocity is 0.281 m/s. Therefore, the best choice is B.

Question 19

A 3.0-kg object moves to the right at 4.0 m/s. It collides head-on with a 6.0-kg object moving to the left at 2.0 m/s. Which statement is correct?

Accepted Answer

The total momentum before the collision is zero because the momentum of the 3.0-kg object (3.0 kg * 4.0 m/s = 12 kg∙m/s) is canceled out by the momentum of the 6.0-kg object (6.0 kg * -2.0 m/s = -12 kg∙m/s). Since momentum is conserved, the total momentum after the collision is also zero.

Question 20

A 2.00-g bullet hits and becomes embedded in a 5.00-kg wood block which is hanging from a 1.20-m long string. This causes the block to swing through an arc of 3.50°. What was the speed of the bullet before it hit the block?

Accepted Answer

The speed of the bullet can be found using conservation of energy and momentum. First, calculate the height the block rises using the arc angle and string length. Then, use potential energy to find the velocity of the block and bullet system after collision. Finally, apply conservation of momentum to find the initial speed of the bullet.

Quiz 9: Linear Momentum and Collisions