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AP Physics C Mechanics - 4.2 Change in Momentum and Impulse- Exam Style questions- FRQs

Change in Momentum and Impulse AP  Physics C Mechanics FRQ

Unit 4: Linear Momentum

Weightage : 15-25%

AP Physics C Mechanics Exam Style Questions – All Topics

Question

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 \(2v_{0}\). Block 2 has mass \(6m\) 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.
A. 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.
i. Draw arrows on the grids to represent the momentum vectors of Block 2 before the collision and the two-block system before and after the collision.
• Arrows should start at the zero-momentum line.
• The length of the arrows should be proportional to the relative magnitudes of the vectors.
• Represent an arrow of zero length by drawing a dot at zero.
ii. During the time interval \(0\le t\le t_{c}\) a force \(F\) is exerted on Block 2 by Block 1 along the \(x\)-direction as a function of \(t\) that is modeled by \(F(t)=F_{\text{max}}\sin(At)\), 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.
B. Consider a new scenario where Block 1 initially slides in the \(+x\)-direction with a new constant speed \(v_{1}\) and Block 2 again initially slides in the \(-x\)-direction with constant speed \(v_{0}\). The blocks collide and stick together. In this new scenario, the two-block system has constant speed \(v_{0}\) after the collision.
Derive an expression for \(v_{1}\) in terms of \(v_{0}\). Begin your derivation by writing a fundamental physics principle or an equation from the reference information.

Most-appropriate topic codes (AP Physics C: Mechanics):

• Topic \(4.2\) — Change in Momentum and Impulse (Part \( \mathrm{A(ii)} \))
• Topic \(4.3\) — Conservation of Linear Momentum (Parts \( \mathrm{A(i)} \), \( \mathrm{B} \))
• Topic \(4.4\) — Elastic and Inelastic Collisions (Parts \( \mathrm{A} \), \( \mathrm{B} \))
▶️ Answer/Explanation

(A)(i)

Block 1 has initial momentum \(p_{1i} = m(2v_{0}) = +2mv_{0}\), which points in the \(+x\)-direction with a length of 2 units.
Block 2 has initial momentum \(p_{2i} = 6m(-v_{0}) = -6mv_{0}\).
Therefore, the arrow for Block 2 before the collision must point leftward with a length of 6 units.
The total momentum of the system before the collision is \(p_{\text{sys}} = +2mv_{0} – 6mv_{0} = -4mv_{0}\).
Since the total momentum of an isolated system is conserved, the system momentum remains the same after the collision.
Thus, the arrows for the Two-Block System both before and after the collision point leftward and are 4 units long.

(A)(ii)

\(\int F \, dt = \Delta p\)
\(m_{1}v_{1i} + m_{2}v_{2i} = (m_{1} + m_{2})v_{f}\)
\(m(2v_{0}) + 6m(-v_{0}) = (m + 6m)v_{f}\)
\(-4mv_{0} = 7mv_{f}\)
\(v_{f} = -\dfrac{4}{7}v_{0}\)
\(\Delta p_{2} = m_{2}v_{f} – m_{2}v_{2i}\)
\(\Delta p_{2} = 6m\left(-\dfrac{4}{7}v_{0}\right) – 6m(-v_{0}) = \dfrac{18}{7}mv_{0}\)
\(\int_{0}^{t_{c}} F_{\text{max}} \sin(At) \, dt = \dfrac{18}{7}mv_{0}\)
\(F_{\text{max}} \left[ -\dfrac{\cos(At)}{A} \right]_{0}^{t_{c}} = \dfrac{18}{7}mv_{0}\)
\(-\dfrac{F_{\text{max}}}{A}(\cos(At_{c}) – \cos(0)) = \dfrac{18}{7}mv_{0}\)
\(\dfrac{F_{\text{max}}}{A}(1 – \cos(At_{c})) = \dfrac{18}{7}mv_{0}\)
\(\boxed{F_{\text{max}} = \dfrac{18mv_{0}A}{7(1 – \cos(At_{c}))}}\)

(B)

\(p_{i} = p_{f}\)
\(m_{1}v_{1i} + m_{2}v_{2i} = (m_{1} + m_{2})v_{f}\)
Case 1: The two-block system moves in the \(-x\)-direction after the collision (\(v_{f} = -v_{0}\))
\(mv_{1} + 6m(-v_{0}) = (m + 6m)(-v_{0})\)
\(mv_{1} – 6mv_{0} = -7mv_{0}\)
\(v_{1} = -v_{0}\)

Case 2: The two-block system moves in the \(+x\)-direction after the collision (\(v_{f} = +v_{0}\))
\(mv_{1} + 6m(-v_{0}) = (m + 6m)(v_{0})\)
\(mv_{1} – 6mv_{0} = 7mv_{0}\)
\(mv_{1} = 13mv_{0}\)
\(v_{1} = 13v_{0}\)
\(\boxed{v_{1} = 13v_{0}}\)

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