Question

A 0.15 kg arrow moving with an initial velocity of 30 m/s passes through a 3.0 kg block initially at rest on a frictionless surface. If the final speed of the arrow is 25 m/s, what is the speed of the block?

Answer 1

.25 m/s

Step-by-step explanation:

We shall use the law of conservation of momentum . We do not use law of conservation of mechanical energy because in such cases there is loss of energy.

Momentum = mass x velocity

Initial momentum of arrow and block

= .15 x 30 +0 = 4.5 J

If final velocity of block be V

Total final momentum of both arrow and block

= .15 x 25 + 3 V

= 3.75 +3V

So

3.75 + 3 V = 4.5

V = .25 m/s