Frictionless SolutionFirst find the speed the bullet had whe

Solution

First, find the speed the bullet had when it makes a perfectly inelastic collision with the second block. Call the bullet’s mass m, and the second block M(and note that the block’s initial momentum is zero), then from the law of conservation of momentum:

mv(i) + 0 = (m + M)v(f)
v(i) = (m + M)v(f) / m
= (0.00350kg + 1.80kg)(1.40m/s) / 0.00350kg
= 721m/s

Now, the collision with the first block (M) was elastic, and the 721m/s from above is the bullets final velocity after emerging from M:

mv(i) + 0 = mv(f) + Mv(f)
v(i) = [mv(f) + Mv(f)] / m
= [(0.00350kg)(721m/s) + (1.20kg)(0.630m/s)] / 0.00350kg
= 937m/s