Question

A bullet of mass m1 is fired horizontally into a wooden block of mass m2 resting on a horizontal surface. The coefficient of kinetic friction between block and surface is μk. The bullet remains embedded in the block, which is observed to slide a distance s along the surface before stopping.

What was the initial speed of the bullet?

Take the free-fall acceleration to be g.

Answer 1

Given:

• The mass of the bullet is, m1

• The mass of the wooden block is, m2

• The velocity of the wooden block just before the collision is u2=0

• The distance slides by the wooden block before coming to rest is ,s

• The coefficient of friction between block and surface is μk

Let

• u1 is the velocity of the bullet just before the collision.

• v be the velocity of the bullet plus wooden block system just after the collision.

Now from the principle of conservation of momentum;

Momentum of the system before collision = Momentum of the system after collisionMomentum of the system before collision = Momentum of the system after collision

⟹m1u1+m2u2=(m1+m2)v

v=m1u1/(m1+m2) Eqn.(1)

The frictional force acting on the wooden block just after collision is;

f=μk(m1+m2)g

Now from the energy conservation, the kinetic energy of the wooden block plus bullet system just after collision is equal to the magnitude of the work done by the friction.

1/2(m1+m2)v²=μ(m1+m2)g×S

⟹v²=2μkgS

⟹{m1u1/(m1+m2)}²= 2μkgS (from, equation 1, v=m1u1/(m1+m2))

Therefore, the initial speed of the bullet is ,

u1=√(2μkgS)×(m1+m2)/m1

u1=√(2μkgS)(1+m2/m1)