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AP Physics 1 : Linear Motion and Momentum

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Example Questions

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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Example Question #491 : Ap Physics 1

A locomotive of mass 167*10^3kg traveling at $4\frac{km}{hr}$ couples to a motionless passenger car without applying the brakes. Assuming no momentum losses, calculate the final velocity.

Possible Answers:

$3.33\frac{km}{hr}$

None of these

$1.05\frac{km}{hr}$

$2.85\frac{km}{hr}$

$5.95\frac{km}{hr}$

Correct answer:

$2.85\frac{km}{hr}$

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 $v_{1,2f}$ :

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

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

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Example Question #31 : Impulse And Momentum

A tennis player hits an incoming tennis ball of mass 50g and speed $40\frac{m}{s}$ right back at his opponent at a speed of $35\frac{m}{s}$ in the exact opposite direction. Suppose the collision between the racket and the tennis ball lasted for 5ms .

What is the magnitude of the average force exerted by the racket on the tennis ball?

Possible Answers:

0.75N

7500N

75N

750000N

750N

Correct answer:

750N

Explanation:

The average force exerted by the racket is related to the impulse by

$F\Delta t=I$

The impulse is related to the change in momentum by

$I=\Delta p=m\Delta v$

Equating the two expressions for impulse gives us

$F\Delta t=m\Delta v$

Simplify to get

$F=\frac{m\Delta v}{\Delta t}$

The change in velocity $\Delta v$ is given as

$\Delta v=v_f-v_i=-35\frac{m}{s}-40\frac{m}{s}=-75\frac{m}{s}$

Plugging in this, along with the given change in time and mass of the tennis ball gives us:

$F=\frac{(-75\frac{m}{s})(0.05kg)}{0.005s}=-750N$

Since the question asks for the magnitude of the average force, we get rid of the negative above to obtain F=750N

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Example Question #32 : Impulse And Momentum

A car takes to accelerate to $100\frac{km}{hr}$ from rest. Determine the average force experience by a driver of mass 100kg .

Possible Answers:

695N

1080N

395N

844N

788N

Correct answer:

695N

Explanation:

Use the equation for momentum:

$F_{net}*\Delta t=P_f-P_i$

$F_{net}*\Delta t=mv_f-mv_i$

Initial velocity was

$F_{net}*\Delta t=mv_f$

$\frac{100km}{hr}*\frac{1000m}{1km}*\frac{1 hr}{3600s}=\frac{27.8m}{s}$

Plug in values:

$F_{net}*4=100*27.8$

$F_{net}=695N$

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Example Question #31 : Impulse And Momentum

A popular topic in early space exploration was how to safely return modules back to the surface of the earth. Early designs contained materials that could only withstand impulses of up to $4500 \frac{kg\cdot m}{s}$ before parts of the module became compromised. One such model of mass 500kg is approaching ocean waters and deploys its shoot, reducing its speed to $8 \frac{m}{s}$ . If the module decelerates to zero velocity in 0.9 seconds upon hitting the water, is the module structurally compromised?

$g=10\frac{m}{s^2}$

Possible Answers:

No; the impulse experienced is $4000\frac{kg\cdot m}{s}$

More information is needed to solve

Yes; the impulse experienced is $5300\frac{kg\cdot m}{s}$

Yes; the impulse experienced is $4700\frac{kg\cdot m}{s}$

No; the impulse experienced is $3400\frac{kg\cdot m}{s}$

Correct answer:

No; the impulse experienced is $4000\frac{kg\cdot m}{s}$

Explanation:

We need to use the equation for impulse to solve this problem. In fact, the time given is completely irrevelevant:

$I =F\cdot t = m \Delta v$

Plugging in our values:

$I = (500kg)(8\frac{m}{s}) = 4000 \frac{kg\cdot m}{s}$

This is less than the threshold, so no, nothing becomes structurally compromised

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Example Question #35 : Impulse And Momentum

An arrow with mass m = 0.2kg is shot straight up in the air with an initial kinetic energy of 10J . What is the momentum of the arrow after . Neglect air resistance.

$g = 10\frac{m}{s^2}$

Possible Answers:

$0\frac{kg\cdot m}{s}$

$4\frac{kg\cdot m}{s}$

$0.2\frac{kg\cdot m}{s}$

$2\frac{kg\cdot m}{s}$

$08\frac{kg\cdot m}{s}$

Correct answer:

$0\frac{kg\cdot m}{s}$

Explanation:

In order to determine the momentum of the arrow, we are going to need its velocity at that time. First, we will convert the initial kinetic energy into initial velocity:

$KE_i = \frac{1}{2}mv_i^2$

Rearranging for velocity:

$v_i = \sqrt{\frac{2KE_i}{m}}$

$v_i = \sqrt{\frac{2(10J)}{0.2kg}}$

$v_i = 10\frac{m}{s}$

Then we can use the following equation to calculate the velocity after 1 second:

v_f = v_i+at

$v_f = 10\frac{m}{s}+\left ( -10\frac{m}{s^2}\right )(1s)$

$v_f = 0\frac{m}{s}$

Then using the expression for momentum:

p = mv

$p = \left ( 0\frac{m}{s}\right )(0.2kg)$

$p= 0 \frac{kg\cdot m}{s}$

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AP Physics 1 : Linear Motion and Momentum

Study concepts, example questions & explanations for AP Physics 1

All AP Physics 1 Resources

Example Questions

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

Example Question #491 : Ap Physics 1

Example Question #31 : Impulse And Momentum

Example Question #32 : Impulse And Momentum

Example Question #31 : Impulse And Momentum

Example Question #35 : Impulse And Momentum

All AP Physics 1 Resources

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