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AP Physics 1 : Impulse and Momentum

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Example Questions

Example Question #21 : Impulse And Momentum

A train of mass $1.8\times10^7 kg$ traveling at $15 \frac{m}{s}$ strikes a car stuck on the tracks of mass 1300 kg . The car becomes stuck on the train.

Determine the final velocity of the train.

Possible Answers:

$15 \frac{m}{s}$

$20 \frac{m}{s}$

$30 \frac{m}{s}$

$10 \frac{m}{s}$

None of these

Correct answer:

$15 \frac{m}{s}$

Explanation:

We will need to use conservation of momentum to solve this problem.

$M_1V_1+M_2V_2=M_{final}V_{final}$

Where M_1 and V_1 refer to the train, and M_2 and V_2 refer to the car. $M_{final}$ and $V_{final}$ refer to the final state of both the train and the car.

Rearranging using algebra......

$\frac{M_1V_1+M_2V_2}{M_{final}}=V_{final}$

$\frac{1.8\times10^7\cdot15+1300\cdot0}{1.8\times10^7}=V_{final}$

Plugging in our values, we get $15 \frac{m}{s}$ .

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Example Question #22 : Impulse And Momentum

A train of mass $1.8\times10^7 kg$ traveling at $15 \frac{m}{s}$ strikes a car stuck on the tracks of mass 1300 kg . The car becomes stuck on the train.

Determine the final velocity of the car.

Possible Answers:

None of these

$15 \frac{m}{s}$

$5 \frac{m}{s}$

$20 \frac{m}{s}$

$10 \frac{m}{s}$

Correct answer:

$15 \frac{m}{s}$

Explanation:

We will need to use conservation of momentum to solve this problem.

$M_1V_1+M_2V_2=M_{final}V_{final}$

Where M_1 and V_1 refer to the train, and M_2 and V_2 refer to the car. $M_{final}$ and $V_{final}$ refer to the final state of both the train and the car.

Rearranging using algebra......

$\frac{M_1V_1+M_2V_2}{M_{final}}=V_{final}$

$\frac{1.8\times10^7\cdot15+1300\cdot0}{1.8\times10^7}=V_{final}$

Plugging in our values, we get $15 \frac{m}{s}$ .

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Example Question #23 : Impulse And Momentum

A train of mass $1.8\times10^7 kg$ traveling at $15 \frac{m}{s}$ strikes a car stuck on the tracks of mass 1300 kg .

Let's assume this collison took .01 seconds to happen. That is, it took the car to accelerate to it's new velocity. Determine the force experienced by the car.

Possible Answers:

$1.95\times10^6 N$

$2.25\times10^6 N$

$3.38\times10^6 N$

None of these

$1.11\times10^6 N$

Correct answer:

$1.95\times10^6 N$

Explanation:

We will need to use conservation of momentum to solve this problem.

$M_1V_1+M_2V_2=M_{final}V_{final}$

Where M_1 and V_1 refer to the train, and M_2 and V_2 refer to the car. $M_{final}$ and $V_{final}$ refer to the final state of both the train and the car.

Rearranging using algebra......

$\frac{M_1V_1+M_2V_2}{M_{final}}=V_{final}$

$\frac{1.8\times10^7\cdot15+1300\cdot0}{1.8\times10^7}=V_{final}$

Plugging in our values, we get $15 \frac{m}{s}$ .

Then, we will need to find out final momentum of the car.

$M_{car}V_{final}=P_{final}$

$1300kg\cdot15\frac{m}{s}=P_{final}$

$M_{car}V_{final}=P_{final}$

$\Delta P=P_{final}-P_{initial}$

Since our intial momentum of the car was , our change in momentum will be equal to the $P_{final}$ .

$\Delta P=P_{final}$

We will use the definition of impulse, which is the change in momentum:

$F\cdot \Delta t= \Delta P$

We will use substitution:

$F\cdot \Delta t= P_{final}$

Plugging in our values, we get $1.95\times10^6 N$

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Example Question #24 : Impulse And Momentum

A car of mass 1110kg is accelerated from $10 \frac{m}{s}$ to $35 \frac{m}{s}$ in 2s.

Determine the average total forces on this car during this time frame.

Possible Answers:

F=1.4*10^4N

F=2.4*10^4N

F=4.4*10^4N

F=1.1*10^4N

F=3.9*10^4N

Correct answer:

F=1.4*10^4N

Explanation:

Determine the change in the velocity of the car:

$35\frac{m}{s}-10\frac{m}{s}=25\frac{m}{s}$

Calculate the average acceleration:

$\frac{25\frac{m}{s}}{2 s}=\frac{\Delta v}{\Delta s}=a$

$12.5\frac{m}{s^2}=a$

Use the definition of force:

F=ma

$F=1110kg*12.5\frac{m}{s^2}$

F=1.4*10^4N

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Example Question #25 : Impulse And Momentum

A cart is traveling at $3 \frac{m}{s}$ when it launches a ball straight into the air with initial velocity $15 \frac{m}{s}$ .

You may ignore air resistance.

How much time will the ball take from the moment it is launched to return to it's initial height?

Possible Answers:

None of these

$\Delta t_{airtime}=4.5s$

$\Delta t_{airtime}=30s$

$\Delta t_{airtime}=1.5s$

$\Delta t_{airtime}=3.0s$

Correct answer:

$\Delta t_{airtime}=3.0s$

Explanation:

First, break the airborne time into two pieces, the ascent and descent.

Ascent:

The ball will need to decelerate from $15\frac{m}{s}$ to $0 \frac{m}{s}$ .

Use acceleration due to gravity.

$15\frac{m}{s}-9.8\frac{m}{s^2}*\Delta t=0\frac{m}{s}$

Solve for $\Delta t$

$\Delta t=1.5s$

Descent:

Due to parabolic motion, the ball will have the same magnitude of velocity when it returns to it's height as when it launched, albeit in the opposite direction.

$0\frac{m}{s}-9.8\frac{m}{s^2}*\Delta t=-15\frac{m}{s}$

Solving for $\Delta t$

$\Delta t=1.5s$

Add the times together to get the total airtime.

$\Delta t_{airtime}=3.0s$

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Example Question #26 : Impulse And Momentum

A cart is traveling at $3 \frac{m}{s}$ when it launches a ball straight into the air with initial velocity $15 \frac{m}{s}$ . Ignore air resistance.

How far will the cart have traveled by the time the ball returns to it's initial height?

Possible Answers:

None of these

9m=d

3.5m=d

25m=d

15m=d

Correct answer:

9m=d

Explanation:

First, break the airborne time into two pieces, the ascent and descent.

Ascent:

The ball will need to decelerate from $15\frac{m}{s}$ to $0 \frac{m}{s}$ .

Use acceleration due to gravity.

$15\frac{m}{s}-9.8\frac{m}{s^2}*\Delta t=0\frac{m}{s}$

Solve for $\Delta t$

$\Delta t=1.5s$

Descent:

Due to parabolic motion, the ball will have the same magnitude of velocity when it returns to it's height as when it launched, albeit in the opposite direction.

$0\frac{m}{s}-9.8\frac{m}{s^2}*\Delta t=-15\frac{m}{s}$

Solve for $\Delta t$

$\Delta t=1.5s$

Add the times together.

$\Delta t_{airtime}=3.0s$

Multiply the airtime by the velocity of the cart.

$3.0s*\frac{3m}{s}=d$

9m=d

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Example Question #21 : Impulse And Momentum

An astronaut of mass 100kg throws a 8kg shot put in space. If the shot put has a magnitude of velocity of $3\frac{m}{s}$ in relation to the center of the galaxy, determine the magnitude of velocity of the astronaut in relation to the center of the galaxy.

Possible Answers:

None of these

$v_a=32\frac{m}{s}$

$v_a=4\frac{m}{s}$

$v_a=15\frac{m}{s}$

$v_a=.32\frac{m}{s}$

Correct answer:

$v_a=.32\frac{m}{s}$

Explanation:

Using conservation of momentum:

P_i=P_f

The initial momentum can be taken as zero from our reference frame.

$0\frac{kg*m}{s}=P_{astro}+P_{shotput}$

Plug in values:

$0\frac{kg*m}{s}=100kg*v_a+8kg*-4\frac{m}{s}$

Note: The sign of the velocity for the shot put is negative because it will be in the opposite direction as the astronaut. The question is asking for magnitude though, so we know it will be positive regardless of the direction since magnitude does not take into account the direction of motion.

$v_a=.32\frac{m}{s}$

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Example Question #28 : Impulse And Momentum

A batter makes contact with a baseball for 0.1 seconds, exerting a constant force of 100 Newtons during the contact.

What is the ball's change in momentum as a result of the contact?

Possible Answers:

100Ns

5Ns

1Ns

10 Ns

50Ns

Correct answer:

10 Ns

Explanation:

We use the equation:

$\Delta p = F \Delta t$

Where the change in momentum is $\Delta p$ .

The force, , is 100N and t is the time during which the force is exerted ( $\Delta t=0.1s$ ) as given by the problem statement, we plug in to the formula:

$\Delta p=(100N)*(0.1s)=10Ns$

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Example Question #29 : Impulse And Momentum

A locomotive of mass 167*10^3kg traveling at $4\frac{km}{hr}$ couples to a motionless coach without applying the brakes. Calculate the momentum change (the impulse experienced) of the coach.

Possible Answers:

$85.1*10^3\frac{kg*m}{s}$

$47.8*10^3\frac{kg*m}{s}$

$91.1*10^3\frac{kg*m}{s}$

$66.7*10^3\frac{kg*m}{s}$

$52.3*10^3\frac{kg*m}{s}$

Correct answer:

$52.3*10^3\frac{kg*m}{s}$

Explanation:

Use conservation of momentum

P_i=P_f

$m_{1i}v_{1i}+m_{2i}v_{2i}=m_{1,2f}v_{1,2f}$

Plug in values:

$(167*10^3)*4+(67*10^3)*0=(234*10^3)v_{1,2f}$

Solve for $v_{1,2f}$ :

$v_{1,2f}=\frac{167*10^3*4}{234*10^3}$

$v_{1,2f}=2.85\frac{km}{hr}$

Definition of impulse:

$J=\Delta P=m(v_f-v_i)$

Convert $\frac{km}{hr}$ to $\frac{m}{s}$

$2.85\frac{km}{hr}*\frac{1hr}{3600s}*\frac{1000m}{1km}=.79\frac{m}{s}$

Plug in values:

$J=\Delta P=67*10^3(.79-0)$

$J=52.3*10^3\frac{kg*m}{s}$

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Example Question #30 : Impulse And Momentum

A locomotive of mass 167*10^3kg traveling at $4\frac{km}{hr}$ couples to a motionless coach without applying the brakes. This took .5s . Determine the average force experienced by the coach.

Possible Answers:

105kN

207kN

302kN

382kN

507kN

Correct answer:

105kN

Explanation:

Use conservation of momentum

P_i=P_f

$m_{1i}v_{1i}+m_{2i}v_{2i}=m_{1,2f}v_{1,2f}$

Plug in values:

$(167*10^3)*4+(67*10^3)*0=(234*10^3)v_{1,2f}$

Solve for $v_{1,2f}$ :

$v_{1,2f}=\frac{167*10^3*4}{234*10^3}$

$v_{1,2f}=2.85\frac{km}{hr}$

Definition of impulse:

$J=\Delta P=m(v_f-v_i)$

Convert $\frac{km}{hr}$ to $\frac{m}{s}$

$2.85\frac{km}{hr}*\frac{1hr}{3600s}*\frac{1000m}{1km}=.79\frac{m}{s}$

Plug in values:

$J=\Delta P=67*10^3(.79-0)$

$J=52.3*10^3\frac{kg*m}{s}$

$J=F*\Delta t$

Plug in values

$52.3*10^3\frac{kg*m}{s}=F*.5seconds$

Solve for :

F=105kN

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AP Physics 1 : Impulse and Momentum

Study concepts, example questions & explanations for AP Physics 1

All AP Physics 1 Resources

Example Questions

Example Question #21 : Impulse And Momentum

Example Question #22 : Impulse And Momentum

Example Question #23 : Impulse And Momentum

Example Question #24 : Impulse And Momentum

Example Question #25 : Impulse And Momentum

Example Question #26 : Impulse And Momentum

Example Question #21 : Impulse And Momentum

Example Question #28 : Impulse And Momentum

Example Question #29 : Impulse And Momentum

Example Question #30 : Impulse And Momentum

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