AP Physics 1- 4.1 Linear Momentum - Exam Style questions - FRQs- New Syllabus
Linear Momentum AP Physics 1 FRQ
Unit 4: Linear Momentum
Weightage : 10-15%
Question
At time \( t=0 \), Block \( A \) slides along a horizontal surface toward Block \( B \), which is initially at rest, as shown in Figure \( 1 \). The masses of blocks \( A \) and \( B \) are \( 6\ \mathrm{kg} \) and \( 2\ \mathrm{kg} \), respectively. The blocks collide elastically at \( t=1.0\ \mathrm{s} \), and as a result, the magnitude of the change in kinetic energy of Block \( B \) is \( 9\ \mathrm{J} \). All frictional forces are negligible.
Most-appropriate topic codes (AP Physics 1):
• Topic \( 4.1 \) — Linear Momentum (Part \( \mathrm{(b)} \), Part \( \mathrm{(c)} \))
• Topic \( 4.2 \) — Change in Momentum and Impulse (Part \( \mathrm{(c)} \))
• Topic \( 4.3 \) — Conservation of Linear Momentum (Part \( \mathrm{(b)} \), Part \( \mathrm{(c)} \))
• Topic \( 4.4 \) — Elastic and Inelastic Collisions (Part \( \mathrm{(a)} \), Part \( \mathrm{(c)} \))
• Topic \( 1.3 \) — Representing Motion (Part \( \mathrm{(b)} \))
▶️ Answer/Explanation
(a)
Block \( B \) starts from rest, so its initial kinetic energy is \( 0 \). The change in kinetic energy of Block \( B \) is \( 9\ \mathrm{J} \), so its final kinetic energy must be
\( K_B = 9\ \mathrm{J} \)
Use
\( K = \dfrac{1}{2}mv^2 \)
\( 9 = \dfrac{1}{2}(2)v_B^2 \)
\( 9 = v_B^2 \)
\( v_B = 3\ \mathrm{m/s} \)
Therefore, \( \boxed{v_B = 3\ \mathrm{m/s}} \)
Since the mass of Block \( B \) is only \( 2\ \mathrm{kg} \), a \( 9\ \mathrm{J} \) gain in kinetic energy corresponds to a moderate speed of \( 3\ \mathrm{m/s} \).
(b)
At \( t=1.0\ \mathrm{s} \), all three lines meet at \( x=2\ \mathrm{m} \).
After the collision:
\( \bullet \) Block \( B \) moves to the right with speed \( 3\ \mathrm{m/s} \), so its position line is a straight line with a steeper positive slope than before.
\( \bullet \) Block \( A \) continues moving to the right but with a smaller speed than before, so its position line is a straight line with a positive slope smaller than its pre-collision slope.
\( \bullet \) The center-of-mass line remains a straight line with the same slope as before the collision.
A correct continuation from \( t=1.0\ \mathrm{s} \) to \( t=2.0\ \mathrm{s} \) is:
\( \text{Block }A:\ \) line from \( (1.0,\ 2.0) \) to \( (2.0,\ 3.0) \), so slope \( = 1\ \mathrm{m/s} \)
\( \text{Block }B:\ \) line from \( (1.0,\ 2.0) \) to \( (2.0,\ 5.0) \), so slope \( = 3\ \mathrm{m/s} \)
\( \text{Center of Mass}:\ \) straight line continuing with the same pre-collision slope, from \( (1.0,\ 2.0) \) to about \( (2.0,\ 3.5) \)
The center of mass keeps moving uniformly because there is no net external horizontal force on the two-block system.
(c)
The line for the center of mass would not change.
Even if the collision were inelastic and the blocks stuck together, the total momentum of the two-block system would still be conserved because there is no external horizontal force. The motion of the center of mass depends only on the net external force, not on whether the collision is elastic or inelastic.
So the center-of-mass line would remain the same straight line with the same slope as in part \( \mathrm{(b)} \). In the sticking case, the separate lines for Blocks \( A \) and \( B \) after the collision would merge into one common line, but the center-of-mass line itself would be unchanged.