A 50.0 g bullet strikes a 7.00 kg stationary wooden block. If the bullet becomes embedded in the block, and the block with the embedded bullet moves off at a velocity of 5.00 m/s, what was the initial velocity of the bullet?

Question

Steve · Accepted Answer

conserve momentum: .050v + 7.00(0) = 5.00(.050+7.00) v = 705 m/s

tutor · Answer

.050v + 7.00(0) = 5.00(.050+7.00) v = 705 m/s