posted by Sarah on .
A 50g bullet strikes a stationary 1.50kg block of wood paced on a horizontal surface just in front of the gun. The bullet becomes embedded in the block and the impact drives them a distance of 10m before they come to rest. If the coefficient of kinetic friction between the block and the surface is .520, what was the speed of the bullet?
Momentum is conserved during collision
use that to find v of total mass after collision in terms of initial Vi of bullet
(.050+1.5) v = .050 Vi
work done by friction = (1/2) m v^2 = u m g*10
where m is .050+1.5