Question

A ball of mass \( m \) moving at a speed \( v \) makes a head-on collision with an identical ball at rest. The kinetic energy at the balls after the collision is \( \frac{3}{4} \) of the original. What is the coefficient of restitution?

• A. \( \frac{1}{\sqrt{3}} \)
• B. \( \frac{1}{\sqrt{2}} \)
• C. \( \sqrt{2} \)
• D. \( \sqrt{3} \)

Hint:

The coefficient of restitution is determined by the relative velocities of the objects involved in a collision. In inelastic collisions, some kinetic energy is lost, which can be used to find \( e \).

Answer 1

The Correct Option is B

In an elastic collision, both momentum and kinetic energy are conserved.
However, the coefficient of restitution \( e \) is defined by the relative velocities of the balls before and after the collision: \[ e = \frac{v_2' - v_1'}{v_1 - v_2} \] where \( v_1 \) and \( v_2 \) are the velocities of the balls before the collision, and \( v_1' \) and \( v_2' \) are the velocities of the balls after the collision. Given: - Before the collision, the first ball is moving at \( v \), and the second ball is at rest. - The kinetic energy after the collision is \( \frac{3}{4} \) of the original. From the conservation of kinetic energy: \[ \frac{1}{2} m v^2 = \frac{1}{2} m {v_1'}^2 + \frac{1}{2} m {v_2'}^2 \] Since the total kinetic energy is reduced to \( \frac{3}{4} \), we can write: \[ \frac{1}{2} m {v_1'}^2 + \frac{1}{2} m {v_2'}^2 = \frac{3}{4} \cdot \frac{1}{2} m v^2 \] Simplifying, we find: \[ {v_1'}^2 + {v_2'}^2 = \frac{3}{4} v^2 \] Now, applying the coefficient of restitution formula and solving the system, we find: \[ e = \frac{1}{\sqrt{2}} \]
Thus, the correct answer is (b).