Question

A cricket ball hits the ground with a speed of \(5\,\text{m s}^{-1}\) making an angle \(30^\circ\) with the horizontal. If the coefficient of restitution is \(0.2\), then the speed of the reflected ball in \(\text{m s}^{-1}\) is

• A. \(\sqrt{22}\)
• B. \(\sqrt{15}\)
• C. \(\sqrt{19}\)
• D. \(\sqrt{11}\)

Hint:

For a collision with a smooth horizontal surface: \[ \text{Horizontal velocity remains unchanged} \] while \[ v_{\text{vertical after}}=e\,v_{\text{vertical before}}. \] Always resolve the velocity into components before applying the coefficient of restitution.

Answer 1

The Correct Option is C

Step 1: Resolve the velocity into horizontal and vertical components.
The ball strikes the horizontal ground with speed \[ u=5\,\text{m s}^{-1} \] at an angle \[ 30^\circ \] with the horizontal. Horizontal component: \[ u_x=u\cos30^\circ \] \[ u_x=5\left(\frac{\sqrt3}{2}\right) \] \[ u_x=\frac{5\sqrt3}{2} \] Vertical component (towards the ground): \[ u_y=u\sin30^\circ \] \[ u_y=5\left(\frac12\right) \] \[ u_y=\frac52 \]

Step 2: Apply coefficient of restitution.
For collision with a smooth horizontal ground, the horizontal component remains unchanged. Therefore, \[ v_x=\frac{5\sqrt3}{2} \] The vertical component after collision becomes \[ v_y=e\,u_y \] where \[ e=0.2=\frac15 \] Hence, \[ v_y=\frac15\cdot\frac52 \] \[ v_y=\frac12 \] and its direction is upward.

Step 3: Find the speed after reflection.
The speed of the reflected ball is \[ v=\sqrt{v_x^2+v_y^2} \] \[ v=\sqrt{\left(\frac{5\sqrt3}{2}\right)^2+\left(\frac12\right)^2} \] \[ v=\sqrt{\frac{75}{4}+\frac14} \] \[ v=\sqrt{\frac{76}{4}} \] \[ v=\sqrt{19} \]

Step 4: Final conclusion.
Therefore, the speed of the reflected ball is \[ \boxed{\sqrt{19}\,\text{m s}^{-1}} \]