To solve this problem, we need to apply the principles of elastic collision. Here’s a step-by-step breakdown:
- Initially, ball A is moving with velocity \(v\), and balls B, C, and D are stationary.
- When two objects collide elastically, both momentum and kinetic energy are conserved.
- Since the balls are identical and the surface is smooth, we assume no external forces are acting on the system. Each ball has mass \(m\).
- Initial momentum: Since ball A is moving with speed \(v\) and balls B, C, D are at rest, the total initial momentum is: \(mv\).
- Initial kinetic energy: The kinetic energy for ball A is \(\frac{1}{2} mv^2\).
- When ball A hits ball B, it transfers all its momentum and energy to ball D (as ball A follows the linear arrangement of balls in contact, and such a transfer of linear momentum leads to only the last ball in the line moving when identical types collide).
- After the collision, balls A, B, C become stationary, thus conserving total initial momentum and energy: \(m \times v\).
- Momentum transferred to ball D is \(mv\), and ball D moves with speed \(v\).
- Therefore, for the energy and momentum conservation to be satisfied, the only possibility is that after the collision, ball D moves with velocity \(v\), while balls A, B, and C remain at rest.
This results in all balls A, B, and C coming to rest, while ball D moves with speed \(v\). Thus, the correct answer is:
Balls A, B, C will be at rest and D will move with speed \(v\).