Question

A 4.25 g bullet traveling horizontally with a velocity of magnitude 375 m/s is fired into a wooden block with mass 1.20 kg , initially at rest on a level frictionless surface. The bullet passes through the block and emerges with its speed reduced to 128 m/s.How fast is the block moving just after the bullet emerges from it?

Answer 1

Answer:

The block moves with a speed of 0.875 m/s just after the emergence of the bullet from the block.

Solution:

Mass of the bullet, m = 4.25 g = 0.00425 kg

Initial velocity of the bullet, u = 375 m/s

Mass of the wooden block, M = 1.20 kg

Initial velocity of the block, v = 0

Final velocity of the bullet, u' = 128 m/s

Now,

Let the final velocity of the block be v'.

Using the principle of conservation of the momentum:

mu + Mv = mu' + Mv'

$0.00425\times 375 + 1.20\times 0 = 0.00425\times 128 + 1.20v'$

$0.00425(375 - 128) = 1.20v'$

v' = 0.875 m/s