Question 1

Astronaut Jennifer's lifeline to her spaceship comes loose and she finds herself stranded, "floating"  100 m from the mothership. She suddenly throws her  2.00-kg wrench at  20 m/s in a direction away from the ship. If she and her spacesuit have a combined mass of  200 kg, how long does it take her to coast back to her spaceship?  
A) 500 s
B) 750 s
C) 1000 s
D) 2.50 min
E) 250 s

Accepted Answer

To solve this problem, we use the conservation of momentum principle. The momentum before and after Jennifer throws the wrench must be equal because there are no external forces acting on the system. The momentum of the wrench is given by its mass times its velocity (2.00 kg * 20 m/s = 40 kg*m/s). Since momentum is conserved, the momentum of Jennifer moving towards the spaceship must also be 40 kg*m/s. To find her velocity, we divide the momentum by her mass (40 kg*m/s / 200 kg = 0.2 m/s). To find the time it takes her to travel 100 m at this speed, we divide the distance by her velocity (100 m / 0.2 m/s = 500 s).

Question 2

A block of mass  m=34 kg and speed  V  is behind a block of mass  M=81 kg and speed of  0.50 m/s, as shown in the figure. The surface is frictionless and the blocks collide and couple. After the collision, the blocks have a common speed of  0.90 m/s. What is the magnitude of the impulse on the 34-kg block due to the collision?

A) $73 \mathrm {~N} \cdot \mathrm { s }$
B) $57 \mathrm {~N} \cdot \mathrm { s }$
C) $32 \mathrm {~N} \cdot \mathrm { s }$
D) $14 \mathrm {~N} \cdot \mathrm { s }$
E) $41 \mathrm {~N} \cdot \mathrm { s }$

Accepted Answer

The answer of A block of mass  m=34 kg...

Question 3

Jennifer hits a stationary 0.20-kg ball, and it leaves her racket at 40 m/s. Time-lapse
photography shows that the ball was in contact with the racket for 40 ms.
(a) What average force did the ball exert on the racket?
(b) What is the ratio of this force to the weight of the ball?

Accepted Answer

The answer of Jennifer hits a stationary 0.20-kg ball, and...

Question 4

A 14-cm diameter champagne bottle rests on its side on top of a frictionless horizontal table. Suddenly, the cork pops and the bottle slides backward, covering a distance of  22.0 cm in  0.44 s. If the mass of the bottle is 500 times the mass of the cork, find the distance from the original position that the cork will land on the table. Neglect air resistance and assume that the cork is very small compared to the bottle.
A)  8.5 cm
B)  3000 cm
C)  85 m
D)  60 m

Accepted Answer

The conservation of momentum principle applies here, stating that the total momentum before the cork pops is equal to the total momentum after the cork pops. Before the cork pops, both the bottle and the cork are at rest, so the total initial momentum is 0. After the cork pops, the momentum of the bottle and the cork must still sum to 0, because there are no external horizontal forces acting on the system.Given that the mass of the bottle is 500 times the mass of the cork, let's denote the mass of the cork as $m$ and the mass of the bottle as $500m$. The velocity of the bottle can be calculated using the distance covered and the time taken, which is $v_b = \frac{0.22\,m}{0.44\,s} = 0.5\,m/s$.Using the conservation of momentum:$$0 = m_{cork}v_{cork} + m_{bottle}v_{bottle}$$$$0 = m \cdot v_{cork} + 500m \cdot (-0.5\,m/s)$$$$v_{cork} = 250\,m/s$$The cork's horizontal velocity is $250\,m/s$. To find the distance it travels, we use the same time duration because in the absence of air resistance and given that the horizontal table is frictionless, the only force acting on both objects after the pop is their mutual interaction at the moment of the pop. Thus, the time of travel for the cork until it lands should be the same as the bottle's, which is $0.44\,s$.Distance traveled by the cork = $v_{cork} \times time = 250\,m/s \times 0.44\,s = 110\,m$.However, none of the provided options match the calculated distance. Given the logic and calculations presented, there seems to be a mistake in the interpretation of the options or in the calculation. The correct approach based on conservation of momentum and the given data should lead to a calculation of the cork's distance, but the discrepancy suggests a reevaluation of the calculation or the assumptions made, such as the time of flight being equal for both cork and bottle, which might not directly apply if considering the cork's significantly higher velocity and potential differences in how air resistance might realistically affect it, despite the instruction to neglect air resistance. The correct approach, adhering strictly to the conservation of momentum and the provided data, indicates a significant oversight in matching the calculated outcome with the provided choices, suggesting a need for clarification or correction in the premise or the interpretation of the results.

Question 5

A very small  51-g steel ball is released from rest and falls vertically onto a steel plate. The ball strikes the plate and is in contact with it for  0.50 ms. The ball rebounds elastically and returns to its original height. The total time interval for a round trip is  3.00 s. What is the magnitude of the average force exerted on the ball by the plate during contact with the plate?
A)  2490 N
B)  1500 N
C)  3000 N
D)  2000 N
E)  3500 N

Accepted Answer

The answer of A very small  51-g steel ball...

Question 6

A 0.140-kg baseball is thrown with a velocity of 27.1 m/s.It is struck by the bat with an average force of 5000 N, which results in a velocity of 37.0 m/s in the opposite direction from the original
Velocity. How long were the bat and ball in contact? 
A) $3.07 \times 10 ^ { - 2 } \mathrm {~s}$
B) $1.79 \times 10 ^ { - 3 } \mathrm {~s}$
C) $4.30 \times 10 ^ { - 3 } \mathrm {~s}$
D) $1.28 \times 10 ^ { - 2 } \mathrm {~s}$

Accepted Answer

To find the time of contact between the bat and the ball, we use the impulse-momentum theorem, which states that the impulse on an object is equal to the change in its momentum. The impulse can be calculated as the force applied times the time it was applied, and the change in momentum is the mass of the ball times the change in velocity. The initial velocity of the ball is 27.1 m/s, and the final velocity is -37.0 m/s (negative because it's in the opposite direction). The change in velocity is therefore -37.0 m/s - 27.1 m/s = -64.1 m/s. The change in momentum is then $0.140 \, \text{kg} \times -64.1 \, \text{m/s} = -8.974 \, \text{kg} \cdot \text{m/s}$. Given the force of 5000 N, the time of contact can be found by rearranging the impulse formula: $time = \frac{Impulse}{Force} = \frac{-8.974 \, \text{kg} \cdot \text{m/s}}{5000 \, \text{N}} = -1.79 \times 10^{-3} \, \text{s}$. The negative sign indicates the direction of the impulse opposite to the initial direction of the ball, but for the duration, we consider the absolute value.

Question 7

A 475-gram ball is traveling horizontally at  12.0 m/s to the left when it is suddenly struck horizontally by a bat, causing it to reverse direction and initially travel at  8.50 m/s to the right. If the bat produced an average force of  1275 N on the ball, for how long (in milliseconds) was it in contact with the ball?

Accepted Answer

The answer of A 475-gram ball is traveling horizontally at...

Question 8

A 0.17-kg baseball is thrown with a speed of 38 m/s and it is hit straight back toward the
pitcher with a speed of 62 m/s. What is the magnitude of the impulse exerted upon the ball
by the bat?

Accepted Answer

The answer of A 0.17-kg baseball is thrown with a...

Question 9

A girl of mass  55 kg throws a ball of mass  0.80 kg against a wall. The ball strikes the wall horizontally with a speed of  25 m/s, and it bounces back with this same speed. The ball is in contact with the wall  0.050 s. What is the magnitude of the average force exerted on the wall by the ball?
A)  400 N
B)  13,750 N
C)  55,000 N 
D)  800 N
E)  27,500 N

Accepted Answer

The answer of A girl of mass  55 kg...

Question 10

A super dart of mass  20 g, traveling at  350 m/s, strikes a steel plate at an angle of $30 ^ { \circ }$ with the plane of the plate, as shown in the figure. It bounces off the plate at the same angle but at a speed of 320  m/s. What is the magnitude of the impulse that the plate gives to the bullet?

A) $4.3 \mathrm {~N} \cdot \mathrm { s }$
B) $300 \mathrm {~N} \cdot \mathrm { s }$
C) $0.30 \mathrm {~N} \cdot \mathrm { s }$
D) $6.7 \mathrm {~N} \cdot \mathrm { s }$
E) $0.52 \mathrm {~N} \cdot \mathrm { s }$

Accepted Answer

The answer of A super dart of mass  20...

Question 11

A forklift pushes a  100-kg  crate, starting from rest, with a horizontal force of magnitude  F . The graph in the figure shows the  x  component of this force as a function of time. What is the instantaneous velocity of the crate at time  t=10 s?
 
A)  25 cm/s
B)  0.00 cm/s
C)  75 cm/s
D)  -75 cm/s
E)  -25 cm/s

Accepted Answer

The answer of A forklift pushes a  100-kg ...

Question 12

A steady horizontal force lasting for  2.1 s gives a  1.25 kg object an acceleration of $3.2 \mathrm {~m} / \mathrm { s } ^ { 2 }$ on a frictionless table. What impulse does this force give to the object?
A) $2.6 \mathrm {~kg} \cdot \mathrm { m } / \mathrm { s }$
B) $26 \mathrm {~kg} \cdot \mathrm { m } / \mathrm { s }$
C) $8.4 \mathrm {~kg} \cdot \mathrm { m } / \mathrm { s }$
D) $11 \mathrm {~kg} \cdot \mathrm { m } / \mathrm { s }$

Accepted Answer

Impulse is calculated as the product of force and time. The force can be found using $ F = ma $, where $ m = 1.25 \, \text{kg} $ and $ a = 3.2 \, \text{m/s}^2 $. Thus, $ F = 1.25 \times 3.2 = 4 \, \text{N} $. The impulse is then $ 4 \, \text{N} \times 2.1 \, \text{s} = 8.4 \, \text{kg} \cdot \text{m/s} $.

Question 13

A golf ball of mass  0.050 kg is at rest on the tee. Just after being struck, it has a velocity of  102 m/s. If the club and ball were in contact for  0.81 ms, what is the average force exerted on the ball by the club?
A)  5.5 kN
B)  4.9 kN
C)  7.1 kN
D)  6.3 kN

Accepted Answer

The average force exerted on the ball can be calculated using the formula $F = \frac{\Delta p}{\Delta t}$, where $\Delta p$ is the change in momentum and $\Delta t$ is the time interval. The change in momentum $\Delta p = m \Delta v = 0.050 \times 102 = 5.1 \, \text{kg m/s}$. The time interval $\Delta t = 0.81 \, \text{ms} = 0.81 \times 10^{-3} \, \text{s}$. Thus, $F = \frac{5.1}{0.81 \times 10^{-3}} = 6296.3 \, \text{N}$, which rounds to $6.3 \, \text{kN}$.

Question 14

A batter applies an average force of 8000 N to a baseball for 1.1 ms. What is the magnitude
of the impulse delivered to the baseball by the bat?

Accepted Answer

The answer of A batter applies an average force of...

Question 15

A 0.24 kg blob of clay is thrown at a wall with an initial horizontal velocity of 16 m/s. If the clay comes to a stop in 91 ms, what is the average horizontal force on the clay due to the wall?
A) 35 N
B) 42 N
C) 51 N
D) 26 N

Accepted Answer

The average force can be calculated using the formula $F = \frac{\Delta p}{\Delta t}$, where $\Delta p$ is the change in momentum and $\Delta t$ is the time interval. The change in momentum, $\Delta p$, is given by $m \cdot \Delta v$, where $m$ is the mass of the clay (0.24 kg) and $\Delta v$ is the change in velocity (from 16 m/s to 0 m/s, so $\Delta v = -16$ m/s). Thus, $\Delta p = 0.24 \cdot -16 = -3.84$ kg·m/s. The time interval, $\Delta t$, is 91 ms, which is $0.091$ s. Plugging these values into the formula gives $F = \frac{-3.84}{0.091} \approx -42.2$ N. The negative sign indicates the direction of the force is opposite to the initial motion, but since we're asked for the magnitude, the answer is 42 N.

Question 16

A golf club exerts an average horizontal force of 1000 N on a 0.045-kg golf ball that is initially at rest on the tee. The club is in contact with the ball for 1.8 ms. What is the speed of the golf ball just
As it leaves the tee?
A) 40 m/s
B) 45 m/s
C) 30 m/s
D) 35 m/s
E) 50 m/s

Accepted Answer

Impulse = change in momentumImpulse = F * tChange in momentum = m * vF * t = m * vv = F * t / mv = 1000 N * 1.8 ms / 0.045 kgv = 40 m/s

Question 17

A  0.140 kg baseball is thrown horizontally with a velocity of  28.9 m/s. It is struck with a constant horizontal force that lasts for  1.85 ms, which results in a velocity of  37.0 m/s in the opposite direction. What was the magnitude of the horizontal force?
A)  4.99 kN
B)  0.613 kN
C)  2.80 kN
D)  2.19 kN

Accepted Answer

The magnitude of the horizontal force can be calculated using the change in momentum over time, which is the impulse. The change in velocity is from 28.9 m/s to -37.0 m/s (since it's in the opposite direction, we consider it negative), so the total change in velocity is 65.9 m/s. The change in momentum ($\Delta p$) is mass ($m$) times change in velocity ($\Delta v$), so $\Delta p = 0.140 \times 65.9$. The time ($t$) is 1.85 ms, which is $1.85 \times 10^{-3}$ s. Force ($F$) is the change in momentum over time, so $F = \frac{\Delta p}{t}$. Plugging in the numbers, we find the force to be approximately 4.99 kN.

Question 18

Calculate the impulse due to a force of 4.5 N that lasts for 1.4 s.
A) 6.9 kg· m/s
B) 5.0 kg· m/s
C) 6.3 kg· m/s
D) 5.7 kg· m/s

Accepted Answer

The answer of Calculate the impulse due to a force...

Question 19

The graph in the figure shows the  x  component  F  of the net force that acts for  10 s on a  100-kg crate. What is the change in the momentum of the crate during the  10 s that this force acts?

A) $25 \mathrm {~kg} \cdot \mathrm { m } / \mathrm { s }$
B) $- 75 \mathrm {~kg} \cdot \mathrm { m } / \mathrm { s }$
C) $- 100 \mathrm {~kg} \cdot \mathrm { m } / \mathrm { s }$
D) $75 \mathrm {~kg} \cdot \mathrm { m } / \mathrm { s }$
E) $- 25 \mathrm {~kg} \cdot \mathrm { m } / \mathrm { s }$

Accepted Answer

The answer of The graph in the figure shows the...

Question 20

A batter hits a 0.140-kg baseball that was approaching him at 30 m/s and, as a result, the
ball leaves the bat at 40 m/s in the reverse of its original direction. The ball remains in
contact with the bat for 2.0 ms. What is the magnitude of the average force exerted by the
bat on the ball?

Accepted Answer

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

Quiz 7: Linear Momentum

Jennifer hits a stationary 0.20-kg ball, and it leaves her racket at 40 m/s. Time-lapse photography shows that the ball was in contact with the racket for 40 ms. (a) What average force did the ball exert on the racket? (b) What is the ratio of this force to the weight of the ball?

A 0.140-kg baseball is thrown with a velocity of 27.1 m/s.It is struck by the bat with an average force of 5000 N, which results in a velocity of 37.0 m/s in the opposite direction from the original Velocity. How long were the bat and ball in contact?

A 0.17-kg baseball is thrown with a speed of 38 m/s and it is hit straight back toward the pitcher with a speed of 62 m/s. What is the magnitude of the impulse exerted upon the ball by the bat?

A girl of mass 55 kg throws a ball of mass 0.80 kg against a wall. The ball strikes the wall horizontally with a speed of 25 m/s, and it bounces back with this same speed. The ball is in contact with the wall 0.050 s. What is the magnitude of the average force exerted on the wall by the ball?

A forklift pushes a 100-kg crate, starting from rest, with a horizontal force of magnitude F . The graph in the figure shows the x component of this force as a function of time. What is the instantaneous velocity of the crate at time t=10 s?

A steady horizontal force lasting for 2.1 s gives a 1.25 kg object an acceleration of 3.2 m/s23.2 \mathrm {~m} / \mathrm { s } ^ { 2 }3.2 m/s2 on a frictionless table. What impulse does this force give to the object?

A golf ball of mass 0.050 kg is at rest on the tee. Just after being struck, it has a velocity of 102 m/s. If the club and ball were in contact for 0.81 ms, what is the average force exerted on the ball by the club?

A batter applies an average force of 8000 N to a baseball for 1.1 ms. What is the magnitude of the impulse delivered to the baseball by the bat?

A 0.24 kg blob of clay is thrown at a wall with an initial horizontal velocity of 16 m/s. If the clay comes to a stop in 91 ms, what is the average horizontal force on the clay due to the wall?

A golf club exerts an average horizontal force of 1000 N on a 0.045-kg golf ball that is initially at rest on the tee. The club is in contact with the ball for 1.8 ms. What is the speed of the golf ball just As it leaves the tee?

A 0.140 kg baseball is thrown horizontally with a velocity of 28.9 m/s. It is struck with a constant horizontal force that lasts for 1.85 ms, which results in a velocity of 37.0 m/s in the opposite direction. What was the magnitude of the horizontal force?

Calculate the impulse due to a force of 4.5 N that lasts for 1.4 s.

The graph in the figure shows the x component F of the net force that acts for 10 s on a 100-kg crate. What is the change in the momentum of the crate during the 10 s that this force acts?

A batter hits a 0.140-kg baseball that was approaching him at 30 m/s and, as a result, the ball leaves the bat at 40 m/s in the reverse of its original direction. The ball remains in contact with the bat for 2.0 ms. What is the magnitude of the average force exerted by the bat on the ball?

A steady horizontal force lasting for 2.1 s gives a 1.25 kg object an acceleration of $3.2 \mathrm {~m} / \mathrm { s } ^ { 2 }$ on a frictionless table. What impulse does this force give to the object?