Head-on elastic collision on a smooth table: Ball A (mass = 250 g) moves at 10 m/s and strikes an identical stationary ball B. If the impact is perfectly elastic, what is the speed of ball B immediately after impact?

Introduction / Context:Elastic impacts between identical masses are a standard application of conservation of linear momentum combined with the coefficient of restitution e. For perfectly elastic collisions (e = 1), kinetic energy along the line of impact is conserved, and the bodies exchange velocities in a head-on collision.

Given Data / Assumptions:

• m_A = m_B = 0.25 kg.
• u_A = 10 m/s; u_B = 0 m/s.
• Smooth horizontal table; central (head-on) impact; e = 1.

Concept / Approach:For central elastic impact of equal masses, post-impact velocities swap: the moving mass stops and the stationary mass departs with the initial speed of the moving mass. This can be shown from momentum conservation and restitution.

Step-by-Step Solution:

Verification / Alternative check:Kinetic energy check: Initial KE = (1/2) m (10)^2; final KE = (1/2) m (10)^2 (all carried by B). Energy is preserved along the line of impact as expected for e = 1.

Why Other Options Are Wrong:0 m/s or 5 m/s contradict the conservation-plus-restitution solution; higher speeds like 15 or 20 m/s would require energy input, impossible here.

Common Pitfalls:Neglecting the restitution equation; confusing equal-mass elastic collisions with inelastic cases where speeds do not exchange perfectly.

Final Answer:10 m/s