Question

A 10 g bullet is fired with 450 m/s into a 10 kg block that sits at rest on a wooden table 20 cm from the edge of the table. The bullet gets embedded in the block (perfectly inelastic collision). The block, with the embedded bullet, then slides to the edge of the table and drops down with some initial velocity while leaving the edge of the table. The coefficient of kinetic friction between the block and the surface of the table is k = 0.2. If the table is 1 m high, how far from the point directly below the table’s edge does the block land?

Answer 1

Solution:

mass of bullet

speed of bullet

The kinetic energy of the bullet when fired is given by,

.

On hitting the block with mass the bullet remakns embedded in the block. This combination moves through a distance of on the table before falling.

The Work done against friction is

The Kinetic energy with which the body leaves the edge of the table is given by,

.

Let the velocity on leavig the table be v. Thus,

  

  

The combination leaves with a horizontal velocity of 14.2m/s. Let the range be x. Height h=1m.

The time of falling is

  

So the range at which the combination will fall is,