Question

A diagram of two billiard balls colliding broken into before and after situations. The Before Collision situation has a billiard ball on the left labeled 3 meters per second with a velocity vector toward the right. The ball on the right is labeled 0 meters per second and does not have a velocity vector. The After Collision situation has a billiard ball on the left labeled 1 meters per second with a velocity vector toward the right. The ball on the right is labeled 2 meters per second and also has a velocity vector toward the right.
Two balls, each with a mass of 0.5 kg, collide on a pool table. Is the law of conservation of momentum satisfied in this collision? Explain why or why not.

Answer 1

Yes — momentum is conserved.

Compute total momentum before:
p_before = m v1 + m v2 = 0.5(3) + 0.5(0) = 1.5 kg·m/s (to the right).

Compute total momentum after:
p_after = 0.5(1) + 0.5(2) = 0.5 + 1.0 = 1.5 kg·m/s (to the right).

Since p_before = p_after, the law of conservation of momentum is satisfied. (Note: kinetic energy is not conserved here — initial KE = 2.25 J, final KE = 1.25 J — so the collision is inelastic even though momentum is conserved.)