Question

A 4.86 g bullet traveling horizontally with a speed of 306 m/s is fired into a wooden block with mass 1.37 kg, initially at rest on a level frictionless surface. The bullet takes 0.221 ms to pass through the block and emerges with its speed reduced to 147 m/s. a. How fast is the block moving just after the bullet emerges from it?

Answer 1

The block is moving 0.564 m/s after the bullet emerges it.

Step-by-step explanation:

Step 1: Data given

Mass of the bullet m = 4.86 grams = 0.00486 kg

Initial speed of the bullet u = 306 m/s

Mass of the wooden block m' = 1.37 kg

Final speed of the bullet v = 147 m/s

Initial speed of the block u'= 0 m/s

Final speed of the block = v'

Step 2: conservation law of momentum

mu + m'u' = mv + m'v'

0.00486*306 + 1.37 * 0 = 0.00486*147 + 1.37*v'

1.48716 = 0.71442‬ + 1.37 v'

0.77274 = 1.37 v'

v' = 0.564 m/s

The block is moving 0.564 m/s after the bullet emerges it.