bullet momentum = .017 v
total mass = .017+1.2 = 1.217 kg
final momentum = initial momentum
.017 v = 1.217 V'
V' = .014 v
normal force on table = m g
= 1.217 (9.81)
friction force = .23*1.217*9.81
= 2.75 Newtons
work done = 2.75 *8.6
= 23.6 Joules
so
23.6 =(1/2)mV'^2
47.2 = 1.217 (.014 v)^2
v^2 = 198001
v = 445 m/s
a bit above the speed of sound :)
A 17 g bullet strikes and becomes embedded in a 1.20 kg block of wood placed on a horizontal surface just in front of the gun. If the coefficient of kinetic friction between the block and the surface is 0.23, and the impact drives the block a distance of 8.6 m before it comes to rest, what was the muzzle speed of the bullet?
1 answer