Question 1

A  14,000-kg boxcar is coasting at  1.50 m/s along a horizontal track when it suddenly hits and couples with a stationary  10,000-kg boxcar. What is the speed of the cars just after the collision?

Accepted Answer

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Question 2

A 600-kg car makes a $90 ^ { \circ }$ 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 during the turn?
A) $22.2 \times 10 ^ { 3 } \mathrm {~kg} \cdot \mathrm { m } / \mathrm { s }$
B) $12.0 \times 10 ^ { 3 } \mathrm {~kg} \cdot \mathrm { m } / \mathrm { s }$
C) $19.1 \times 10 ^ { 3 } \mathrm {~kg} \cdot \mathrm { m } / \mathrm { s }$
D) $10.2 \times 10 ^ { 3 } \mathrm {~kg} \cdot \mathrm { m } / \mathrm { s }$
E) $9.55 \times 10 ^ { 3 } \mathrm {~kg} \cdot \mathrm { m } / \mathrm { s }$

Accepted Answer

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Question 3

A 1200-kg cannon suddenly fires a 100-kg cannonball at  35 m/s. What is the recoil speed of the cannon? Assume that frictional forces are negligible and the cannon is fired horizontally.
A)  35 m/s
B)  3.5 m/s
C)  2.9 m/s
D)  3.2 m/s

Accepted Answer

Using the conservation of momentum, the recoil speed of the cannon can be calculated as follows: (mass of cannon) * (recoil speed) = (mass of cannonball) * (speed of cannonball). Solving for the recoil speed gives: 1200 * v = 100 * 35, so v = (100 * 35) / 1200 = 2.9 m/s.

Question 4

A  0.10-kg ball, traveling horizontally at  25 m/s, strikes a wall and rebounds at  19 m/s. What is the magnitude of the change in the momentum of the ball during the rebound?
A) $5.4 \mathrm {~kg} \cdot \mathrm { m } / \mathrm { s }$
B) $1.8 \mathrm {~kg} \cdot \mathrm { m } / \mathrm { s }$
C) $4.4 \mathrm {~kg} \cdot \mathrm { m } / \mathrm { s }$
D) $1.2 \mathrm {~kg} \cdot \mathrm { m } / \mathrm { s }$
E) $72 \mathrm {~kg} \cdot \mathrm { m } / \mathrm { s }$

Accepted Answer

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Question 5

A  1200-kg ore cart is rolling at  10.8 m/s across a flat friction-free surface. A crane suddenly drops  858 kg of ore vertically into the cart. How fast does the cart move just after being loaded with the ore?
A)  4.20 m/s
B)  3.80 m/s
C)  5.70 m/s
D)  6.30 m/s

Accepted Answer

The speed of the cart after being loaded can be found using the principle of conservation of momentum. Initially, the momentum of the system (cart + ore) is the momentum of the cart since the ore is not yet in the cart. After the ore is dropped into the cart, the total mass of the system increases, but the total momentum of the system remains the same (assuming no external forces act on the system). The initial momentum is $p_{initial} = m_{cart} \times v_{cart} = 1200 \, \text{kg} \times 10.8 \, \text{m/s}$. After the ore is added, the total mass is $m_{total} = m_{cart} + m_{ore} = 1200 \, \text{kg} + 858 \, \text{kg}$. Using the conservation of momentum, $p_{initial} = p_{final}$ or $m_{cart} \times v_{cart} = (m_{cart} + m_{ore}) \times v_{final}$. Solving for $v_{final}$, we find $v_{final} = \frac{1200 \times 10.8}{1200 + 858} = 6.3 \, \text{m/s}$.

Question 6

In a police ballistics test,  2.00-g bullet traveling at  700 m/s suddenly hits and becomes embedded in a stationary  5.00-kg wood block. What is the speed of the block immediately after the bullet has stopped moving relative to the block?

Accepted Answer

The answer of In a police ballistics test,  2.00-g...

Question 7

A dinner plate falls vertically to the floor and breaks up into three pieces, which slide horizontally 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 the speed of the third piece?
A)  2.57 m/s
B)  6.83 m/s
C)  5.33 m/s
D)  3.50 m/s
E)  2.51 m/s

Accepted Answer

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Question 8

A firecracker breaks up into two pieces, one of which has a mass of  200 g and flies off along the  +x -axis with a speed of  82.0 m/s. The second piece has a mass of  300 g and flies off along the  +y -axis with a speed of  45.0 m/s. What is the total momentum of the two pieces?
A) $21.2 \mathrm {~kg} \cdot \mathrm { m } / \mathrm { s } \text { at } 56.3 ^ { \circ } \text { from the } + x \text {-axis }$
B) $21.2 \mathrm {~kg} \cdot \mathrm { m } / \mathrm { s } \text { at } 39.5 ^ { \circ } \text { from the } + x \text {-axis }$
C) $361 \mathrm {~kg} \cdot \mathrm { m } / \mathrm { s } \text { at } 56.3 ^ { \circ } \text { from the } + x \text {-axis }$
D) $361 \mathrm {~kg} \cdot \mathrm { m } / \mathrm { s } \text { at } 0.983 ^ { \circ } \text { from the } + x \text {-axis }$
E) $93.5 \mathrm {~kg} \cdot \mathrm { m } / \mathrm { s } \text { at } 28.8 ^ { \circ } \text { from the } + x \text {-axis }$

Accepted Answer

The answer of A firecracker breaks up into two pieces,...

Question 9

Two ice skaters suddenly push off against one another starting from a stationary position. The 45-kg skater acquires a speed of  0.375 m/s relative to the ice. What speed does the  60-kg skater acquire relative to the ice?
A)  0.50 m/s
B)  0.38 m/s
C)  0.00 m/s
D)  0.28 m/s
E)  0.75 m/s

Accepted Answer

The law of conservation of momentum states that the total momentum of a closed system is constant if no external forces are acting on it. Since the skaters push off against each other from a stationary position, their initial total momentum is 0. For the 45-kg skater with a speed of 0.375 m/s, their momentum is $45 \times 0.375 = 16.875$ kg·m/s. To conserve momentum, the 60-kg skater must have the same magnitude of momentum but in the opposite direction, so their speed $v$ is found by $60v = 16.875$, giving $v = 16.875 / 60 = 0.28125$ m/s, which rounds to 0.28 m/s.

Question 10

A dinner plate falls vertically to the floor and breaks up into three pieces, which slide horizontally 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 relative to the  +x -axis does the third piece move?
A) $39.9 ^ { \circ } \text { from the } + x \text {-axis }$
B) $216.9 ^ { \circ } \text { from the } + x \text {-axis }$
C) $219.8 ^ { \circ } \text { from the } + x \text {-axis }$
D) $36.9 ^ { \circ } \text { from the } + x \text {-axis }$
E) $39.8 ^ { \circ } \text { from the } + x \text {-axis }$

Accepted Answer

The conservation of momentum in the x and y directions must be applied. The total momentum before the break is zero, so the sum of the momenta of the pieces must also be zero. Solving for the third piece's velocity components and using trigonometry gives the direction as $219.8 ^{\circ}$ from the +x-axis.

Question 11

An empty train car of mass $2.0 \times 10 ^ { 4 }$ kg coasts along at  10 m/s. A  3000-kg boulder is suddenly dropped vertically into the car. Find the speed of the car immediately after the boulder is dropped in.

Accepted Answer

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Question 12

Find the magnitude and direction of the net momentum of the system shown in the figure. Express the direction by giving the angle the net momentum makes with the  +x -axis.

Accepted Answer

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Question 13

Two astronauts, of masses  60 kg and  80 kg, are initially right next to each other and at rest in outer space. They suddenly push each other apart. How far has the heavier astronaut moved when the lighter astronaut has moved  12 m?
A)  7.0 m
B)  16 m
C)  21 m 
D)  9.0 m
E)  12 m

Accepted Answer

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Question 14

Two astronauts, of masses  60 kg and  80 kg, are initially right next to each other and at rest in outer space. They suddenly push each other apart. What is their separation after the heavier astronaut has moved  12 m?
A)  28 m
B)  24 m
C)  9.0 m
D)  16 m
E)  21 m

Accepted Answer

The principle of conservation of momentum states that the total momentum of a closed system is constant if no external forces are acting on it. Since both astronauts push each other away with equal and opposite forces and start from rest, their momenta are equal in magnitude but opposite in direction. Let $d$ be the distance the lighter astronaut moves. The ratio of their distances moved is inversely proportional to their masses, so $\frac{d}{12} = \frac{80}{60}$ or $d = 16$. Their separation is the sum of the distances they both moved, which is $12 + 16 = 28$ meters.

Question 15

A  328-kg car moving at  19.1 m/s in the  +x  direction hits from behind a second car moving at  13.0 m/s in the same direction. If the second car has a mass of  790 kg and a speed of  15.1 m/s right after the collision, what is the velocity of the first car after this sudden collision?
A)  14.0 m/s
B)  18.2 m/s
C)  -14.0 m/s
D)  24.2 m/s

Accepted Answer

Using the conservation of momentum: (328 kg * 19.1 m/s) + (790 kg * 13.0 m/s) = (328 kg * v) + (790 kg * 15.1 m/s). Solving for v gives approximately 14.0 m/s.

Question 16

Two air track carts move along an air track towards each other. Cart A has a mass of  450 g and moves toward the right with a speed of  0.850 m/s. Cart B has a mass of  300 g and moves toward the left with a speed of  1.12 m/s. What is the total momentum of the two-cart system?
A)  0.047 kg . m/s toward the right
B)  0.750 kg . m/s toward the right
C)  0.750 kg . m/s toward the left
D)  0.719 kg . m/s toward the right
E)  0.719 kg . m/s toward the left

Accepted Answer

The total momentum of the system can be calculated by adding the momentum of each cart, taking into account their directions. For Cart A, momentum = mass * velocity = 0.450 kg * 0.850 m/s = 0.3825 kg*m/s toward the right. For Cart B, momentum = mass * velocity = 0.300 kg * (-1.12 m/s) = -0.336 kg*m/s (toward the left, hence the negative sign). Adding these, the total momentum = 0.3825 kg*m/s - 0.336 kg*m/s = 0.0465 kg*m/s, which rounds to 0.047 kg*m/s toward the right.

Question 17

A  0.140-kg baseball is dropped and reaches a speed of  1.20 m/s  just before it hits the ground and bounces. It rebounds with an upward velocity of  1.00 m/s. What is the change of the ball's momentum during the bounce?
A)  0.000 kg . m/s
B)  0.0280 kg . m/s  upwards
C)  0.0280 kg . m/s  downwards
D)  0.308 kg . m/s downwards
E)  0.308 kg . m/s upwards

Accepted Answer

The change in momentum is calculated using the formula Δp = m(v_final - v_initial). Before the bounce, the velocity is downwards at 1.20 m/s (taking down as negative), and after the bounce, it is upwards at 1.00 m/s (up is positive). Thus, Δp = 0.140 kg * (1.00 - (-1.20)) m/s = 0.140 kg * 2.20 m/s = 0.308 kg*m/s upwards.

Question 18

A  100-g  ball falls from a window that is  12 m above ground level and experiences no significant air resistance as it falls. What is its momentum when it strikes the ground?
A)  3.3 kg . m/s
B)  4.8 kg . m/s
C)  1.5 kg . m/s
D)  2.4 kg . m/s
E)  1.8 kg . m/s

Accepted Answer

The answer of A  100-g  ball falls from...

Question 19

In a police ballistics test, a 10.0-g bullet moving at  300 m/s  is fired into a 1.00-kg block at rest. The bullet goes through the block almost instantaneously and emerges with  50.0%  of its original speed. What is the speed of the block just after the bullet emerges?
A)  1.50 m/s
B)  273 m/s
C)  3.00 m/s
D)  2.97 m/s

Accepted Answer

The conservation of momentum principle applies here. Before the collision, the total momentum is the momentum of the bullet since the block is at rest. After the collision, the total momentum is the sum of the momentums of the bullet and the block. The momentum of an object is given by its mass times its velocity (p = mv). Let's denote the mass of the bullet as $m_b = 0.01 \, \text{kg}$, the initial velocity of the bullet as $v_{bi} = 300 \, \text{m/s}$, the final velocity of the bullet as $v_{bf} = 0.5 \times 300 = 150 \, \text{m/s}$, the mass of the block as $m_{block} = 1.00 \, \text{kg}$, and the final velocity of the block as $v_{block}$.Using the conservation of momentum:$$m_b \cdot v_{bi} + 0 = m_b \cdot v_{bf} + m_{block} \cdot v_{block}$$Plugging in the values:$$0.01 \cdot 300 = 0.01 \cdot 150 + 1.00 \cdot v_{block}$$Solving for $v_{block}$:$$3 = 1.5 + v_{block}$$$$v_{block} = 1.5 \, \text{m/s}$$

Question 20

A 60-kg swimmer suddenly dives horizontally from a 150-kg raft with a speed of  1.5 m/s . The raft is initially at rest. What is the speed of the raft immediately after the diver jumps if the water has negligible effect on the raft?

Accepted Answer

The answer of A 60-kg swimmer suddenly dives horizontally from...

Quiz 7: Linear Momentum

A 14,000-kg boxcar is coasting at 1.50 m/s along a horizontal track when it suddenly hits and couples with a stationary 10,000-kg boxcar. What is the speed of the cars just after the collision?

A 600-kg car makes a 90∘90 ^ { \circ }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 during the turn?

A 1200-kg cannon suddenly fires a 100-kg cannonball at 35 m/s. What is the recoil speed of the cannon? Assume that frictional forces are negligible and the cannon is fired horizontally.

A 0.10-kg ball, traveling horizontally at 25 m/s, strikes a wall and rebounds at 19 m/s. What is the magnitude of the change in the momentum of the ball during the rebound?

A 1200-kg ore cart is rolling at 10.8 m/s across a flat friction-free surface. A crane suddenly drops 858 kg of ore vertically into the cart. How fast does the cart move just after being loaded with the ore?

In a police ballistics test, 2.00-g bullet traveling at 700 m/s suddenly hits and becomes embedded in a stationary 5.00-kg wood block. What is the speed of the block immediately after the bullet has stopped moving relative to the block?

A firecracker breaks up into two pieces, one of which has a mass of 200 g and flies off along the +x -axis with a speed of 82.0 m/s. The second piece has a mass of 300 g and flies off along the +y -axis with a speed of 45.0 m/s. What is the total momentum of the two pieces?

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

An empty train car of mass 2.0×1042.0 \times 10 ^ { 4 }2.0×104 kg coasts along at 10 m/s. A 3000-kg boulder is suddenly dropped vertically into the car. Find the speed of the car immediately after the boulder is dropped in.

Find the magnitude and direction of the net momentum of the system shown in the figure. Express the direction by giving the angle the net momentum makes with the +x -axis.

Two astronauts, of masses 60 kg and 80 kg, are initially right next to each other and at rest in outer space. They suddenly push each other apart. How far has the heavier astronaut moved when the lighter astronaut has moved 12 m?

Two astronauts, of masses 60 kg and 80 kg, are initially right next to each other and at rest in outer space. They suddenly push each other apart. What is their separation after the heavier astronaut has moved 12 m?

A 328-kg car moving at 19.1 m/s in the +x direction hits from behind a second car moving at 13.0 m/s in the same direction. If the second car has a mass of 790 kg and a speed of 15.1 m/s right after the collision, what is the velocity of the first car after this sudden collision?

Two air track carts move along an air track towards each other. Cart A has a mass of 450 g and moves toward the right with a speed of 0.850 m/s. Cart B has a mass of 300 g and moves toward the left with a speed of 1.12 m/s. What is the total momentum of the two-cart system?

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

A 100-g ball falls from a window that is 12 m above ground level and experiences no significant air resistance as it falls. What is its momentum when it strikes the ground?

In a police ballistics test, a 10.0-g bullet moving at 300 m/s is fired into a 1.00-kg block at rest. The bullet goes through the block almost instantaneously and emerges with 50.0% of its original speed. What is the speed of the block just after the bullet emerges?

A 60-kg swimmer suddenly dives horizontally from a 150-kg raft with a speed of 1.5 m/s . The raft is initially at rest. What is the speed of the raft immediately after the diver jumps if the water has negligible effect on the raft?

A 600-kg car makes a $90 ^ { \circ }$ 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 during the turn?

An empty train car of mass $2.0 \times 10 ^ { 4 }$ kg coasts along at 10 m/s. A 3000-kg boulder is suddenly dropped vertically into the car. Find the speed of the car immediately after the boulder is dropped in.