Question

A sphere A of mass \(m\) is thrown at 50 m/s along \(\tan^{-1}(3/4)\). At the topmost point of its trajectory, it collides centrally with sphere B (mass \(3m\)) at rest. Gravity \(g = 10\,\text{m/s}^2\). Coefficient of restitution e = 0.3. Find the speed (m/s) of sphere A immediately after collision (round to 1 decimal).

Hint:

At the top of the flight, only horizontal motion matters. Use restitution + momentum equations for 1D central collisions.

Answer 1

Correct Answer: 0.9

At the topmost point, vertical velocity is zero. Horizontal component:
\[ u_{Ax} = 50 \cos\left(\tan^{-1}\frac{3}{4}\right) = 50 \cdot \frac{4}{5} = 40\ \text{m/s} \] Sphere B is at rest. Let \(u_1 = 40\), \(u_2 = 0\). Masses: \(m_1=m\), \(m_2=3m\). Using 1D restitution equation:
\[ v_2 - v_1 = e(u_1 - u_2) \] \[ v_2 - v_1 = 0.3(40) = 12 \] Momentum conservation:
\[ m u_1 = m v_1 + 3m v_2 \] \[ 40 = v_1 + 3v_2 \] Solve simultaneously. From restitution: \(v_2 = v_1 + 12\). Substitute:
\[ 40 = v_1 + 3(v_1 + 12) \] \[ 40 = 4v_1 + 36 \] \[ 4v_1 = 4 \] \[ v_1 = 1\ \text{m/s} \] Thus speed of sphere A after collision:
\[ v_A = 1.0\ \text{m/s} \]