Question

A 10 gram bullet travelling at 200 m per second strike and remain amended in a 2 kg target which is originally at rest but free to move at what speed does the target move off

Answer 1

Answer:

0.995 m/s

Explanation:

Given

The mass of the bullet = 10 gm = 0.01 kg

speed of the bullet v = 200 m/s

Mass of the target M = 2kg

Let the velocity of the target be V

Since after hitting the target the bullet is embedded into the target, the combined mass of target and bullet will be M+m

Using the principle of conservation of linear momentum

Momentum of the bullet before impact = momentum of the bullet and target after impact

$\implies mv = (M+m)V$

$\implies V = \frac{mv}{M+m}$

$\implies V = \frac{0.01\times 200}{2+0.01}$

$\implies V = \frac{2}{2.01}$

$\implies V = 0.995 \text{ m/s}$

Hope this helps.

Answer 2

Explanation:

Hope it might help you

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