Question

A ball strikes a horizontal floor at 45°. If 25% of its kinetic energy is lost in the collision, then the coefficient of restitution is:

• A. \( \frac{1}{2} \)
• B. \( \frac{1}{\sqrt{2}} \)
• C. \( \frac{1}{2\sqrt{2}} \)
• D. \( \frac{1}{4} \)

Hint:

The coefficient of restitution describes how much of the kinetic energy is retained after a collision.

Answer 1

The Correct Option is B

We are tasked with finding the coefficient of restitution (\(e\)) when a ball strikes a horizontal floor at an angle of \(45^\circ\) and loses \(25%\) of its kinetic energy during the collision. 
Step 1: Understand the problem The ball strikes the floor at an angle of \(45^\circ\). \(25%\) of its kinetic energy is lost during the collision. The coefficient of restitution \(e\) is defined as the ratio of the relative velocity after the collision to the relative velocity before the collision, along the line of impact. 
Step 2: Analyze the collision When the ball strikes the floor at \(45^\circ\), its velocity can be resolved into two components:
1. Horizontal component: \(v_x = v \cos(45^\circ) = \frac{v}{\sqrt{2}}\),
2. Vertical component: \(v_y = v \sin(45^\circ) = \frac{v}{\sqrt{2}}\).
During the collision:
The horizontal component \(v_x\) remains unchanged because there is no force acting horizontally. The vertical component \(v_y\) changes due to the collision. If \(e\) is the coefficient of restitution, the vertical velocity after the collision is \(v_y' = -e v_y\). Step 3: Kinetic energy before and after the collision The initial kinetic energy (\(K_i\)) is: \[ K_i = \frac{1}{2} m v^2 \] The final kinetic energy (\(K_f\)) is: \[ K_f = \frac{1}{2} m (v_x^2 + v_y'^2) \] Substitute \(v_x = \frac{v}{\sqrt{2}}\) and \(v_y' = -e \frac{v}{\sqrt{2}}\): \[ K_f = \frac{1}{2} m \left( \left(\frac{v}{\sqrt{2}}\right)^2 + \left(-e \frac{v}{\sqrt{2}}\right)^2 \right) \] Simplify: \[ K_f = \frac{1}{2} m \left( \frac{v^2}{2} + \frac{e^2 v^2}{2} \right) = \frac{1}{2} m \cdot \frac{v^2 (1 + e^2)}{2} \] \[ K_f = \frac{1}{4} m v^2 (1 + e^2) \] Step 4: Use the energy loss condition The ball loses \(25%\) of its kinetic energy, so the final kinetic energy is \(75%\) of the initial kinetic energy: \[ K_f = 0.75 K_i \] Substitute \(K_i = \frac{1}{2} m v^2\) and \(K_f = \frac{1}{4} m v^2 (1 + e^2)\): \[ \frac{1}{4} m v^2 (1 + e^2) = 0.75 \cdot \frac{1}{2} m v^2 \] Simplify: \[ \frac{1}{4} (1 + e^2) = \frac{3}{8} \] Multiply through by 8: \[ 2 (1 + e^2) = 3 \] Solve for \(e^2\): \[ 2 + 2e^2 = 3 \implies 2e^2 = 1 \implies e^2 = \frac{1}{2} \] Take the square root: \[ e = \frac{1}{\sqrt{2}} \] Step 5: Match with the options The coefficient of restitution is \(e = \frac{1}{\sqrt{2}}\), which matches option (2). Final Answer: \(\boxed{2}\)