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AP Physics 1 : Impulse and Momentum

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Example Question #11 : Impulse And Momentum

When catching an object, an average person can stand a maximum impact force of 20000N. Forces greater than this would most likely break bones in the person's hand. If a person catches a 500g baseball that moves at $-50\frac{m}{s}$ , what is the minimum time the person should take to stop it in order to avoid seriously hurting his hand?

Possible Answers:

$1.25*10^{-4}s$

$1.25 *10^{-3} s$

$2.50*10^{-3}s$

$2.50*10^{-4}s$

1.00s

Correct answer:

$1.25 *10^{-3} s$

Explanation:

In order to solve this problem we need to use the relationship between force and impulse:

$F=\frac{\Delta P}{t}$

Since the ball is moving with a Velocity of $-50\frac{m}{s}$ , we have that

$\Delta P=P_f-P_i=mV_f-mV_i$

$\Delta P=(0.5kg)(0 \frac{m}{s})-(0.5kg)(-50\frac{m}{s})=\frac{25kg\cdot m}{s}$

Note that the final velocity of the ball is zero since it comes to a stop. We want the force, , experienced by the person's hand to be less than or equal to the maximum impact force

F_i_m_p=20,000N

Mathematically:

$F\leq F_i_m_p$

$\frac{1}{F}\geq \frac{1}{F_i_m_p}$

Use the impulse equation and solve for time:

$t=\frac{\Delta P}{F}\geq \frac{\Delta P}{F_i_m_p}$

$t\geq \frac{\frac{25kg\cdot m}{s}}{20000N}=1.25*10^{-3}s$

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Example Question #12 : Impulse And Momentum

What is the impulse of a bowling ball that has mass 10kg that hits bowling pins going at $2\frac{m}{s}$ , and slows to $1\frac{m}{s}$ after striking the bowling pins?

Possible Answers:

$0N\cdot s$

$10N\cdot s$

$20N\cdot s$

$5N\cdot s$

Correct answer:

$10N\cdot s$

Explanation:

Impulse is defined as the change in momentum.

$J=\mid p_{final}-p_{initial}\mid$

Where is the impulse, $p_{final}$ is final momentum and $p_{initial}$ is the initial momentum. Recall:

p=mv

Where is mass and is velocity.

Our bowling ball has a mass of 10kg, with initial velocity of $2 \frac{m}{s}$ and final velocity of $1 \frac{m}{s}$ .

$p_{final}=m*v_{final}=10kg*1\frac{m}{s}=10 N\cdot s$

$p_{initial}=m*v_{initial}=10kg*2\frac{m}{s}= 20 N\cdot s$

Plug in these two values into the equation for impulse and solve.

$J=\mid 20-10\mid N\cdot s= 10N\cdot s$

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Example Question #13 : Impulse And Momentum

Imagine a baseball player hitting a home run. If the 1 kg ball is thrown at $40 \: \textup{m/s}$

and it leaves the bat at $60 \:\textup{m/s}$ . What is the impluse applied by the bat to the ball?

Assume the collision lasts $\textup{1/10}$ of a second and the ball leaves at the same angle it entered.

Possible Answers:

$300 \: N\cdot s$

$150 \: N\cdot s$

$100 \: N\cdot s$

$120 \: N\cdot s$

$220 \: N\cdot s$

Correct answer:

$100 \: N\cdot s$

Explanation:

Impulse is the change in momentum. So all that is needed for this problem is to solve for the change in momentum.

$\Delta P = (mv)_{final}-(mv)_{initial}$

Note!!!! momentum is a vector quantity. This means that we must accound for the change in the ball's direction. This can be done by defining one of the directions negative.

For convience, I'll define the initial velocity as negative. Plugging in numbers we get

$100 \: N\cdot s$ .

We never accounted for the time of the collision. We would have needed to do this if the problem asked for the force the bat applied.

$F = \Delta P/\Delta t$

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Example Question #11 : Impulse And Momentum

A 70 g bullet is fired at a block of lead resting on a frictionless surface. The bullet has an initial speed of $400 \frac{m}{s}$ , while the block is initially at rest. After hitting the block, the bullet rebounds with a speed of $300 \frac{m}{s}$ . How fast is the lead block moving after the bullet rebounds off of it?

Possible Answers:

$-7 \frac{m}{s}$

$0 \frac{m}{s}$

$1 \frac{m}{s}$

$-1 \frac{m}{s}$

$7 \frac{m}{s}$

Correct answer:

$7 \frac{m}{s}$

Explanation:

To solve this problem, we will use conservation of momentum. The initial momentum of the system must be equal to the final momentum of the system if no external forces act on it. It is important to note the directions and signs of the velocities. From this information, we may write:

$P_i = P_f\Rightarrow \frac{m_{bullet}\left( v_{bullet,i} -v_{bullet,f}\right )}{m_{block}}=v_{block,f}$

$v_{block,f}=\frac{.07kg}{7kg} \left( 400 \frac{m}{s}- (-300 \frac{m}{s})\right) = 7 \frac{m}{s}$

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Example Question #12 : Impulse And Momentum

A 6000 kg truck travelling at $25 \frac{m}{s}$ to the right has a head-on collision with a 3000 kg car moving at $50 \frac{m}{s}$ to the left. During the collision, the two vehicles become stuck together. With what speed does the two-car pair move after the collision?

Possible Answers:

$25 \frac{m}{s}$

$-33.3 \frac{m}{s}$

$-25 \frac{m}{s}$

$0 \frac{m}{s}$

$33.3 \frac{m}{s}$

Correct answer:

$0 \frac{m}{s}$

Explanation:

To solve this problem, we must first note that this is a collision problem. Secondly, we must be careful with the signs of the velocities associated with each vehicle.

$P_i=P_f\Rightarrow m_1v_1+m_2v_2=\left ( m_1+m_2 \right ) v_f$

Because of the parameters of the problem, it is easy to see that the total momentum of the system initially is equal to zero. Therefore, the final momentum of the system must also be zero.

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Example Question #16 : 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 left

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

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

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

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

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

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 #17 : 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:

$v_{1f}=0$

$v_{2f}=3v$

$v_{1f}=-v$

$v_{2f}=2v$

$v_{1f}=-2v$

$v_{2f}=v$

$v_{1f}=-3v$

$v_{2f}=0$

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 #18 : Impulse And Momentum

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:

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

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

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

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

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

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 #11 : 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:

None of these

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

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

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

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

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 #20 : 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:

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

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

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

None

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

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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AP Physics 1 : Impulse and Momentum

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