Physics help, dynamics?

Question

A bullet (mass m) hits a stationary block of wood (mass mb) and becomes embeddedin it. The co-efficient of kinetic friction between the block and the floor is uk. As a result of the impact the block slides a distance of D before stopping. What was the velocity v of the bullet?

Answer 1

You need two physics principles to solve your problem: the law of conservation of momentum and the work-kinetic energy theorem.

The law of conservation of momentum requires that the initial momentum equals the final momentum of the system. The initial momentum pi is solely carried by the bullet,

pi = m*vb

where m is the mass of the bullet and vb is the speed of the bullet before impact. On the other hand, the final momentum pf is carried by both the bullet and the wood,

pf = (m+mb)*vc

where mb is the mass of the block of wood and vc is the common speed of the bullet and the wood after the collision

Momentum conservation requires that pi = pf so that you get the equality

m*vb = (m+mb)*vc. (equation-1)

Of course vb and vc are still both unknown at the moment. We need another equation to solve them.

The other equation follows from the work kinetic energy theorem which states that the total work done on a given body equals the change in its kinetic energy. When a constant force F is acted on a given body over some distance d, the work W done by F is given by

W = Fd.

Now if vi is the speed of the body just before the force acted and vf is the speed of the same body after F acted over some distance d, then the work kinetic theorem yields the relationship

Fd = vf^2 * M/2 - vi^2 M/2

where M is the mass of the body.

For your problem vi = vc, the common speed of the block and the bullet; and vf = 0, the wood stopped sliding. Also M=m+mb. The force is F = (m+mb)*g*uk and the displacement d is d = D. Combining everything, the work kinetic energy theorem gives you

-(m+mb)*g*uk*D = -(m+mb)*vc^2/2.

The left hand of the equation is negative because friction is in the direction opposite to the displacement. This gives the common speed

vc = sqrt(2*g*uk*D).

Substituting this back in equation-(1) yields

m*vb = (m+mb)*sqrt(2*g*uk*D).

Finally solving for vb gives

vb = (m+mb)*sqrt(2*g*uk*D)/m.

Answer 2

This can be solved by conservation of energy. The initial energy of the system is that of the bullet alone, 0,5*m*vb^2. The energy required to stop the block and the bullet is ∫Fds, where F is the stopping force, in this case friction, which is constant. So the stopping energy is F*D, and F = µk*(mb+m). Therefore

µk*(mb+m)*D = 0.5*m*vb^2

vb^2 = 2*µk*D*(mb+m)/m

vb = √[2*µk*D*(mb+m)/m]