In a one-dimensional perfectly elastic collision between a 1 kg block moving at 3 m/s and a 2 kg block at rest, what are the final velocities?

• A. v1f = -1 m/s, v2f = 2 m/s
• B. v1f = 3 m/s, v2f = 0 m/s
• C. v1f = 1 m/s, v2f = -2 m/s
• D. v1f = 0 m/s, v2f = 3 m/s

Answer: (A)

In a one-dimensional perfectly elastic collision, both momentum and kinetic energy are conserved, and the relative speed of the two bodies is preserved in magnitude but reverses direction. That means the difference between the final velocities satisfies v1 − v2 = −(u1 − u2).

Here, m1 = 1 kg with u1 = 3 m/s, m2 = 2 kg with u2 = 0. The initial relative speed is u1 − u2 = 3, so the final relative speed must be v1 − v2 = −3. Momentum conservation gives 1·3 + 2·0 = 1·v1 + 2·v2, so v1 + 2v2 = 3.

Solve the system:

• v1 − v2 = −3

• v1 + 2v2 = 3

From the first, v1 = v2 − 3. Sub into the second:

(v2 − 3) + 2v2 = 3 → 3v2 − 3 = 3 → v2 = 2, so v1 = 2 − 3 = −1.

Interpretation: the lighter block rebounds backward at 1 m/s, and the heavier block moves forward at 2 m/s.