Physics Friday 30: A Speeding Bullet

A wooden block of mass M rests on a horizontal surface, with the coefficient of kinetic friction for these two being μk. Now, suppose a bullet of mass m is fired horizontally into the block; the bullet embeds itself inside the block, and the block is driven back across the surface by a distance d. What is the velocity of the bullet just before striking the block?

We have all the information we need. First, while the block, with bullet embedded, is sliding backward after the initial velocity is imparted, it is being decellerated by the force of kinetic friction f=-μkN, where N is the normal force, and the minus sign is because the force will be in the direction of decreasing velocity. We further see that the normal force will be equal to the weight of the block and bullet, (M+m)g. Now, using Newton’s second law, we have that the decelleration due to the frictional force is .
Now, using the time independent kinematic equation , with our final velocity being zero and our displacement being d, we have: , and so the velocity of the block with the bullet imbedded immediately after the collision and embedding is .

Now, to find the bullet’s pre-collision velocity, we do not need to know any of the exact forces involved in the collision; instead we simply use conservation of momentum: this is a perfectly inelastic collision. The momentum following the collision is . Before the collision, the block is at rest, so all the momentum is that of the bullet . Conservation of momentum says these are equal, and so:


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