In Figure (1), a 3.50 g bullet is fired horizontally at two blocks at rest on a
ID: 2146767 • Letter: I
Question
In Figure (1), a 3.50 g bullet is fired horizontally at two blocks at rest on a frictionless table. The bullet passes through block 1 (mass 1.45 kg) and embeds itself in block 2 (mass 2.02 kg). The blocks end up with speeds v1 = 0.560 m/s and v2 = 1.38 m/s (Figure (2)). Neglecting the material removed from block 1 by the bullet, find the speed of the bullet as it (a) enters and (b) leaves block 1.
In Figure (1), a 3.50 g bullet is fired horizontally at two blocks at rest on a frictionless table. The bullet passes through block 1 (mass 1.45 kg) and embeds itself in block 2 (mass 2.02 kg). The blocks end up with speeds v1 = 0.560 m/s and v2 = 1.38 m/s (Figure (2)). Neglecting the material removed from block 1 by the bullet, find the speed of the bullet as it (a) enters and (b) leaves block 1.Explanation / Answer
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 M2(and note that the block’s initial momentum is zero), then from the law of conservation of momentum: mv(i) + 0 = (m + M2)v(f) v(i) = (m + M2)v(f) / m = (0.00350kg + 2.02kg)(1.38m/s) / 0.00350kg = 797.83m/s Now, the collision with the first block (M1) was elastic, and the 797.83m/s from above is the bullets final velocity after emerging from M1: mv(i) + 0 = mv(f) + M1v(f) v(i) = [mv(f) + M1v(f)] / m = [(0.00350kg)(797.83m/s) + (1.45kg)(0.560m/s)] / 0.00350kg = 1029.83m/s