Perfectly plastic (completely inelastic) impact of identical balls on a smooth table: Ball A (mass 250 g) moves at 10 m/s and strikes an identical stationary ball B on a smooth horizontal table. If the impact is perfectly plastic (the balls stick together), what is the velocity of ball B just after impact?

Introduction / Context:
In a perfectly plastic (completely inelastic) collision, colliding bodies stick together and move with a common velocity after impact. On a smooth horizontal table, external impulses are negligible during the short impact, so horizontal momentum is conserved.

Given Data / Assumptions:

• Masses: m_A = m_B = 0.250 kg (identical).
• Initial velocities: u_A = 10 m/s, u_B = 0 m/s.
• Perfectly plastic impact ⇒ common velocity v after impact.
• No external horizontal impulse; momentum conserved along the line of motion.

Concept / Approach:
Use conservation of linear momentum for the system of the two balls. For perfectly plastic impact, kinetic energy is not conserved, but that does not affect the determination of the common velocity immediately after collision.

Step-by-Step Solution:

Verification / Alternative check:
Relative speed of separation is zero in a perfectly plastic collision, consistent with sticking together and moving at 5 m/s.

Why Other Options Are Wrong:

• 0 m/s: would violate momentum conservation.
• 10 m/s: would imply no momentum transfer to B.
• 2.5 m/s: corresponds to unequal masses or different conditions.
• “None of these”: incorrect since 5 m/s is valid.

Common Pitfalls:
Trying to conserve kinetic energy in inelastic collisions; forgetting identical masses make the arithmetic especially simple.

Final Answer:
5 m/s