Solution to Question 12

Physics Solution: Impulse, Deformation & Restitution

1. Initial System Analysis

We have two bodies moving in opposite directions:

Body A: $ m_A = 5.00 \text{ kg} $, $ u_A = +4.00 \text{ m/s} $
Body B: $ m_B = 10.0 \text{ kg} $, $ u_B = -0.50 \text{ m/s} $

2. Calculating Period of Deformation (Rigorous Method)

The period of deformation ends when the two bodies achieve a common velocity (maximum deformation). We can calculate this time interval by finding the impulse required to reach this state and equating it to the area under the Force-time graph.

Step A: Find Common Velocity ($ v_{common} $)

At maximum deformation, relative velocity is zero.

$$ v_{common} = \frac{m_A u_A + m_B u_B}{m_A + m_B} $$ $$ v_{common} = \frac{5(4) + 10(-0.5)}{5 + 10} = \frac{20 – 5}{15} = \frac{15}{15} = 1 \text{ m/s} $$

Step B: Calculate Impulse of Deformation ($ J_D $)

This is the change in momentum of one body (e.g., Body A) from the start until maximum deformation.

$$ J_D = | \Delta p_A | = | m_A (v_{common} – u_A) | $$ $$ J_D = | 5 (1 – 4) | = | 5(-3) | = 15 \text{ Ns} $$

Step C: Relate to Graph Area to find Time ($ t_{def} $)

The impulse of deformation corresponds to the area of the graph from $ t=0 $ to the peak force $ F_{max} = 150 \text{ N} $.

$$ J_D = \text{Area of first triangle section} $$ $$ 15 = \frac{1}{2} \times \text{base} \times \text{height} $$ $$ 15 = \frac{1}{2} \times t_{def} \times 150 $$ $$ 15 = 75 t_{def} $$ $$ t_{def} = \frac{15}{75} = 0.20 \text{ s} $$

Conclusion: The calculated period of deformation is 0.20 s. This matches the graph’s peak and option (a).

3. Calculating Coefficient of Restitution (e)

The coefficient of restitution is the ratio of the Impulse of Restitution ($ J_R $) to the Impulse of Deformation ($ J_D $).

From the graph, the restitution phase is from $ t=0.20 $ to $ t=0.30 \text{ s} $.

$$ J_R = \text{Area of second triangle section} = \frac{1}{2} \times (0.30 – 0.20) \times 150 $$ $$ J_R = \frac{1}{2} \times 0.10 \times 150 = 7.5 \text{ Ns} $$

$$ e = \frac{J_R}{J_D} = \frac{7.5}{15} = 0.5 $$

Conclusion: The coefficient of restitution is 0.5. This matches option (b).

4. Calculating Final Velocities

Total Impulse on Body B is $ J_{total} = J_D + J_R = 15 + 7.5 = 22.5 \text{ Ns} $. The force acts in the positive direction for B.

$$ m_B (v_{B,final} – u_B) = J_{total} $$ $$ 10 (v_{B,final} – (-0.5)) = 22.5 $$ $$ 10 v_{B,final} + 5 = 22.5 $$ $$ 10 v_{B,final} = 17.5 \implies v_{B,final} = 1.75 \text{ m/s} $$

Body B moves at 1.75 m/s in the reverse direction of its initial motion (which was negative). This matches option (d).

Checking Body A (Force opposes motion):

$$ 5(v_{A,final} – 4) = -22.5 \implies v_{A,final} = -0.5 \text{ m/s} $$

Body A reverses direction. Option (c) is incorrect as it claims “original direction”.

Final Answer

The correct conclusions are (a), (b), and (d).