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
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?
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