Question

A bullet of mass 20 gram is horizontally fired with velocity 150 metre per second a pistol of mass 2kg what is the cal velocity of the pistol​

Answer 1

Solution:

Given:

$\bullet$ Mass (m1) = 0.02 kg

$\bullet$ Initial velocity (u1) = 0 m/s

$\bullet$ Final velocity (v1) = 150 m/s

And:

$\bullet$ Mass (m2) = 2 kg

$\bullet$ Initial velocity (u2) = 0 m/s

$\bullet$ Final velocity (v2) = v

Now:

By the law of conservation of momentum,

$\boxed{\sf{m_{1}u_{1} + m_{2}u_{2} = m_{1}v_{1} + m_{2}v_{2}}}$

Now:

By substituting the above values in this formula, we get:

$\sf{m_{1}u_{1} + m_{2}u_{2} = m_{1}v_{1} + m_{2}v_{2}}$

$\implies$ 0.02*0 + 2*0 = 0.02*150 + 2*v

$\implies$ 0 + 0 = 3 + 2v

$\implies$ 3 = 2v

$\implies$ 3/2 = v

$\implies$ 1.5 = v

Therefore:

Recoil velocity = 1.5 m/s

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Answered by: Niki Swar, Goa❤️

Answer 2

Answer:1.5m/s

Explanation:

As no external force is acting so conservation of momentam

M1V1= - M2V2

(20/1000)*150=2*V

V=1.5 which is opposite to the direction of bullet moving {action reaction pair} meaning of -ve sign