Block 1 of mass m1 slides along a frictionless floor and into a one-dimensional elastic collision with stationary block 2 of mass m2 = 3.0m1. Prior to the collision, the center of mass of the two-block system had a speed of 5.0 m/s. Afterward, what are the speeds of (a) the center of mass and (b) block 2?

Answers: (a) 2.0 m/s (b) 4 m/s

Method:

(a) By conservation of momentum, the velocity of the center of mass is unchanged, i. e., 2.0 m/s.

(b) The velocity of the center of mass = (m1v1+m2v2) / (m1+m2)

Since the second mass is initially at rest, vcom = m1v1 / (m1+m2)

Therefore, the initial v1 = vcom (m1+m2) / m1 = 2.0 m/s x 6 = 12 m/s

Since the second mass is initially at rest, v2f = v1i (2m1 / m1+m2) = 12 m/s (2/6) = 4 m/s