Question

1. A10 g bullet, moving with a velocity of 500 m/s.
enters a stationary piece of ice of mass 10 kg
and stops. If the piece of ice is lying on a
frictionless plane, then its velocity will be​

Answer 1

Answer:0m/s

As it will stop moving after some time

Answer 2

The velocity of the ice after collision is 0.5 m/s.

Explanation:

The mass of the bullet, m = 10g = 0.010 kg.

The initial velocity velocity of the bullet, v = 500 m/s.

The final velocity of the bullet, u = 0.

The mass of the ice, M = 10 kg.

The initial velocity of the ice, V = 0.

use the coservation of momentum as

m x v + M x V = m x u + M x U

Here u is the final velocity of the ice after collision.

substitute the values, we get

 $0.010kg\times500m/s+10\times 0 = 0.010 kg\times 0+10kg\times U$

$U=\frac{5}{10}$

U = 0.5 m/s.

Thus, the velocity of the ice  after collision is 0.5 m/s.

#Learn More: conservation of momentum.

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