Impact time ratio and coefficient of restitution: A glass ball is projected along a smooth floor to strike a vertical wall. It returns to the original point of projection in twice the time it took to reach the wall. Assuming no floor friction, determine the coefficient of restitution between the ball and the wall.

Introduction / Context:
When a smooth ball strikes a rigid wall, only the normal (perpendicular) component of velocity changes according to the coefficient of restitution e. On a smooth horizontal floor, motion is effectively one-dimensional. Time and distance relationships before and after impact provide a clean way to compute e without needing masses or forces.

Given Data / Assumptions:

• Smooth (frictionless) floor, so speed along the floor is uniform between wall and point of projection.
• Time to reach wall = t₁.
• Time to return to the start after impact = 2 t₁ (stated).
• Distance from start to wall = L (same both ways).
• Pre-impact speed toward wall = u; post-impact speed away from wall = e*u.

Concept / Approach:
The coefficient of restitution e is defined along the line of impact as the ratio of the speed of separation to the speed of approach. Here that is simply e = (speed after rebound) / (speed before impact). Using kinematics on a frictionless floor gives times as distance over speed.

Step-by-Step Solution:

Verification / Alternative check:
If e halves the speed, the return trip should take double the time of the outgoing trip. This matches the statement, confirming e = 0.5.

Why Other Options Are Wrong:

• 0.25, 0.33, 0.40, 0.55: These would imply return times of 4, ~3, 2.5, or ~1.82 times t₁, respectively, none of which match the given “twice”.

Common Pitfalls:
Mistaking distance ratios for velocity ratios, or assuming energy methods are needed. Here, constant horizontal speed segments make the time ratio method simplest.

Final Answer:
0.50