Impact of a ball on a fixed plane (smooth line of impact): A ball moving at 5 m/s strikes a fixed smooth plane at an angle of 45° to the line of impact (normal). After impact, its direction is again equally inclined to the line of impact. If the coefficient of restitution e = 0.5, what is the ball’s speed immediately after impact?

Introduction / Context:
This problem tests rigid-body impact decomposition into normal and tangential components, and the definition of the coefficient of restitution. The phrase “equally inclined to the line of impact” indicates that the direction of motion after impact makes the same angle with the line of impact (the common normal) as the incident direction, a frequent textbook condition used to relate components after impact when friction impulses are present.

Given Data / Assumptions:

• Initial speed v = 5 m/s.
• Angle to the line of impact (normal) before impact = 45°.
• Coefficient of restitution e = 0.5 (defined along the normal).
• “Equally inclined after impact” implies the post-impact angle to the normal also equals 45° (symmetric inclination), achievable if the tangential component scales consistently with the normal component due to frictional impulse during impact.

Concept / Approach:
Let u_n and u_t be the normal and tangential components before impact; v_n and v_t be those after impact. By restitution, the normal component magnitude scales as v_n = e * u_n (reversed direction). If the outgoing direction is equally inclined to the normal (45°), then the ratio v_t / |v_n| must equal tan 45° = 1, hence v_t = e * u_n as well. Since initially u_t = u_n at 45°, we infer v_t = e * u_t. Therefore every component is reduced by the same factor e, and the speed scales by e overall: v_after = e * v_before.

Step-by-Step Solution:

Verification / Alternative check:
If e = 1 (perfectly elastic), the same reasoning would yield v′ = v with equal angles, matching the familiar mirror reflection. For e = 0.5, the direction constraint forces proportional component reduction, hence v′ = 0.5 v.

Why Other Options Are Wrong:

• 0.5 and 1.5 m/s: too small; they do not satisfy the equal-angle constraint with e = 0.5.
• 3.5 and 4.5 m/s: these ignore the required proportional scaling of both components; they are inconsistent with the given condition.

Common Pitfalls:
Assuming the plane is perfectly smooth (no tangential impulse), which would keep tangential velocity unchanged and alter the outgoing angle; here the phrase “equally inclined” implies a tangential impulse acting to match angles.

Final Answer:
2.5 m/s