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AP Physics 1 : Linear Motion and Momentum

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Example Questions

Example Question #11 : Impulse And Momentum

Some students are investigating momentum using carts and a spring. They have a way to release the spring while the carts are in motion without disturbing the motion. They put the spring and release mechanism between two carts whose masses are $m=3\,kg$ and $m=1\,kg$ with the lighter cart in the front. They put the entire system in motion to the right with a velocity of $v_0=2\frac{m}{s}$ . They then trigger the spring to release. As a result, the front, less massive cart has a new velocity of $v=5\frac{m}{s}$ to the right. What was the speed and direction of the more massive cart after the "explosion"?

Possible Answers:

$v=2\frac{m}{s}$ to the right

$v=1\frac{m}{s}$ to the left

$v=1\frac{m}{s}$ to the right

$v=0\frac{m}{s}$ (the cart is motionless)

$v=2\frac{m}{s}$ to the left

Correct answer:

$v=1\frac{m}{s}$ to the right

Explanation:

The key is that both carts will receive the same impulse ( $F\,\Delta t$ ) from the spring, but in opposite directions. Since impulse is change in momentum, find the change in light cart's momentum:

$\Delta p=p_{final}-p_{initial}=mv_f-mv_i =1kg\times 5\frac{m}{s}-1kg\times2\frac{m}{s}=3\frac{kg\cdot m}{s}$

This is directed to the right. So the change in the momentum of the heavier cart is:

$\Delta p = -3\frac{kg\cdot m}{s}$ , to the left. Solve for its final momentum:

$p_f=p_i+\Delta p=3kg\times2\frac{m}{s}-3\frac{kg \cdot m}{s}=3\frac{kg\cdot m}{s}$

Use this to find its velocity:

$p=mv\rightarrow 3\frac{kg\cdot m}{s}=3kg\times v$

$v=1\frac{m}{s}$

The positive sign means that it is moving to the right.

We could also use momentum conservation. Find total momentum before:

$p=mv=4kg\times 2\frac{m}{s}=8\frac{kg\cdot m}{s}$

The less massive cart has:

$p=mv=1kg\times5\frac{m}{s}=5\frac{kg\cdot m}{s}$ after the explosion, so the more massive cart has the other $3\frac{kg\cdot m}{s}$

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Example Question #14 : Impulse And Momentum

A billiard ball travels at $v_{1}=v$ toward another billiard ball traveling at $v_{2}=-2v$ . They collide elastically. Which option correctly describes the final velocities of the billiard balls? Assume they have the same mass.

Possible Answers:

$v_{1f}=-v$

$v_{2f}=2v$

$v_{1f}=0$

$v_{2f}=3v$

$v_{1f}=-3v$

$v_{2f}=0$

$v_{1f}=-2v$

$v_{2f}=v$

Correct answer:

$v_{1f}=-2v$

$v_{2f}=v$

Explanation:

Due to conservation of momentum, the initial momentum must equal the final momentum of the system. Both billiard balls are of equal mass, and since the collision is elastic than the billiard balls will simply exchange momentum. This is a problem that is best to think about before starting to solve any equations because sometimes the correct answer is one you can deduce without any calculations. Therefore:

$v_{1f}=-2v$

$v_{2f}=v$

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Example Question #436 : Newtonian Mechanics

A 30 kg object is moving along with the velocity given below. Calculate the magnitude of the momentum vector $\vec{p}$ .

$\vec{v}=5\hat{i}-2\hat{j}$

Possible Answers:

$162 kg\frac{m}{s}$

$90kg\frac{m}{s}$

$210kg\frac{m}{s}$

$58kg\frac{m}{s}$

$300kg\frac{m}{s}$

Correct answer:

$162 kg\frac{m}{s}$

Explanation:

We begin by writing down the definition of an object's linear momentum $\vec{p}$

$\vec{p}=m\vec{v}$

$\vec p=30kg\left(5\hat{i}-2\hat{j} \right )\frac{m}{s}$

$\vec p=\left( 150\hat{i}-60\hat{j}\right )kg\frac{m}{s}$

We then find the magnitude of the momentum by taking the square root of the sum of squares of its components.

$\left| \vec{p}\right| = \sqrt{p_x^2+p_y^2}$

$\left|\vec p\right|=\sqrt{150^2+60^2}kg\frac{m}{s}=162kg\frac{m}{s}$

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Example Question #15 : Impulse And Momentum

An asteroid of mass 560 kg is traveling with the velocity $<391, 498, 121> \frac{m}{s}$ .

What is the magnitude of the momentum of the asteriod?

Possible Answers:

$7.3\times10^5 kg\frac{m}{s}$

$13.1\times10^5 kg\frac{m}{s}$

None of these

$3.6\times10^5 kg\frac{m}{s}$

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

Correct answer:

$3.6\times10^5 kg\frac{m}{s}$

Explanation:

First, we will need to find the magnitude of the velocity vector.

$\left | \vec{v} \right |=\sqrt{v_x^2+v_y^2+v_z^2}$

Plugging in our values

$\left | \vec{v} \right |=\sqrt{391^2+498^2+121^2}$

$645 \frac{m}{s}=|\vec{v}|$

Momemtum is defined as

$\vec{P}=m\vec{v}$

Thus,

$\left | \vec{P} \right | =m \left | \vec{v} \right |$

We plug in our values

$\left | \vec{P} \right | =560kg\cdot \left | 645\frac{m}{s}\right |$

$3.6\times10^5 kg\frac{m}{s}=|\vec{P}|$

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Example Question #16 : 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 .

Determine the initial momentum of the system.

Possible Answers:

$2.7\times10^8 kg\frac{m}{s}$

$2.4\times10^7 kg\frac{m}{s}$

$5.7\times10^8 kg\frac{m}{s}$

$1.7\times10^8 kg\frac{m}{s}$

None

Correct answer:

$2.7\times10^8 kg\frac{m}{s}$

Explanation:

$P_{total}=P_1+P_2.......$

The train and car are our only two objects in the system.

The initial momentum of the car is zero.

So the only momentum that will contribute is that of the train.

p=mv

$p=1.8\times10^7\cdot15$

Plugging in our values, we get $2.7\times10^8 kg\frac{m}{s}$

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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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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 #11 : Impulse And Momentum

Example Question #14 : Impulse And Momentum

Example Question #436 : Newtonian Mechanics

Example Question #15 : Impulse And Momentum

Example Question #16 : Impulse And Momentum

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

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