AP Physics C Mechanics 4.2: Change in Momentum and Impulse

Connect force, interaction time, impulse, and change in momentum to analyze how an interaction changes a system’s motion.

[Maximum number: 5]

Two blocks, 1 and 2, slide toward each other on a horizontal surface. Block 1 has mass m

and slides in the +x-direction with constant speed $2 v_{0}$ . Block 2 has mass 6 m and slides in the

-x-direction with constant speed $v_{0}$ , as shown in Figure 1. The blocks then collide and stick

together. The collision occurs from time t=0 to $t=t_{c}$ . After the collision, where $t>t_{c}$ , the blocks

move together with the same constant speed.

Figure 1

The diagrams in Figure 2 can be used to represent the momentum of blocks 1 and 2 before

and after the collision. The momentum vector diagram for Block 1 before the collision is

shown.

During the time interval $0 \leq t \leq t_{c}$ , a force F is exerted on Block 2 by Block 1 along the

x-direction as a function of t that is modeled by $F(t)=F_{\max } \sin (A t)$ , where A is a positive

constant and $F_{\text {max }}$ is the magnitude of the maximum force exerted on Block 2 by Block 1

during the collision.

Derive an expression for $F_{\text {max }}$ . Express your answer in terms of $m, v_{0}, A, t_{c}$ , and physical

constants, as appropriate. Begin your derivation by writing a fundamental physics

principle or an equation from the reference information.

(ii) For a multistep derivation that includes the integral form of impulse or the differential form of Newton's second law & Point A3 \\ \hline & For indicating one of the following:

- The final speed of Block 1 or Block 2 is $\frac{4}{7} v_{0}$ .
- The change in the momentum of Block 2 is $+\frac{18}{7} m v_{0}$ .
- The change in the momentum of Block 1 is $-\frac{18}{7} m v_{0}$ .
- The change in the velocity of Block 2 is $+\frac{3}{7} v_{0}$ .
- The change in the velocity of Block 1 is $-\frac{18}{7} v_{0}$ .

& Point A4 \\ \hline & For substituting the given expression for F(t) into an expression for the impulse exerted on either block or the differential form of Newton's second law & Point A5 \\ \hline & For an integral with appropriate limits or a constant of integration & Point A6 \\ \hline & For a correct expression for $F_{\text {max }}$ in terms of given quantities & Point A7 \\ \hline \end{tabular}

Example Responses

	$\vec{J}=\int_{t_{1}}^{t_{2}} \vec{F}_{\text {net }}(t) d t=\Delta \vec{p} \quad \text { OR } \quad \begin{aligned} & \vec{F}_{\text {net }}=\frac{d \vec{p}}{d t} & \int \vec{F}_{\text {net }} d t=\Delta \vec{p} \end{aligned}$
		$\sum \vec{p}_{0}=\sum \vec{p}_{f}$ $(m)\left(2 v_{0}\right)+(6 m)\left(-v_{0}\right)=(m+6 m)\left(v_{f}\right)$ $v_{f}=-\frac{4}{7} v_{0}$ $\sum \vec{p}_{0}=\sum \vec{p}_{f}$ $(m)\left(2 v_{0}\right)+(6 m)\left(-v_{0}\right)=(m+6 m)\left(v_{f}\right)$ $v_{f}=-\frac{4}{7} v_{0}$

Official scoring-guideline example response

[Maximum number: 1]

Scientists have created a new type of lightweight foam and are performing experiments to investigate the properties of the foam. The mass of Cart A is 1000 kg and the mass of Cart B is 2000 kg. A piece of foam with negligible mass is attached to the front of Cart A, as shown. Cart A moves with a constant speed toward Cart B, which is initially at rest. At time $t=0 \mathrm{~s}$ , the foam connected to Cart A makes contact with Cart B. The foam remains in contact with Cart B for 0.5 s, after which the carts separate and both carts move with constant velocities.

Figure for Question 4.2 Change in Momentum and Impulse question 2 — AP Physics C: Mechanics

What is the relationship between the magnitude of the average net force $F_{1}$ exerted on Cart B for the collision with the foam and the magnitude of the average net force $F_{2}$ exerted on Cart B for the collision without the foam? $F_{1}>F_{2}$ $F_{1}<F_{2}$ $F_{1}=F_{2}$
Justify your answer.

For selecting $F_{1}<F_{2}$ with an attempt at a relevant justification

1 point

For indicating that the impulse or change in momentum of each cart in both collisions is the
same

1 point

For indicating that decreasing the time of collision means the average force must be greater

1 point

Example Solution
Since the initial and final velocities are the same for both collisions, $\Delta p$ is the same for both collisions; as a result, the impulse is the same for both collisions. So, if $\Delta t$ is smaller, $F_{\text {avg }}$ is larger.

Total for part (c)

for question 1

15 points

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Covered syllabus points

4.2 Change in Momentum and Impulse question 1

4.2 Change in Momentum and Impulse question 2