(a) use conservation of momentum, note that after collision, there is a common velocity v1=v2=v
m1u1+m2u2=m1v1+m2v2=(m1+m2)v
m1,m2=masses of bullet and block
u1,u2=initial velocities of bullet & block
v1,v2=(common) velocity of bullet and block after collision.
substituting values,
8*220+250*0=(8+250)*v
solve for v to get: v=880/129 m/s
Kinetic energy right after impact:
KE=(1/2)(m1+m2)v^2
=6003.1 J (approx.)
After the block has moved up the incline over x metres, potential energy
PE = (m1+m2)gx(sin(30°)
=258*9.81*x*(0.5)
=1265.49x
Equate sum of KE and PE at bottom and top of incline:
1265.49x = 6003.1 J
x=4.74 m (approx.)
An 8.00g bullet is fired at 220m/s into a 250g wooden block that is initially at rest. The bullet remains in the block and after the collision the two slide up a 30 degree incline.
a) Solve for the velocity of the bullet block after the collision.
b) Determine the distance along the incline the bullet block travel if the incline is frictionless
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