Question

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A bullet of mass 20g is horizontally fired with velocity
$150m \: {s}^{ - 1}$
from a pistol of mass 2kg. What is the recoil velocity of the pistol ?

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Answer 1

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Answer 2

$\huge{\mathfrak{Required Answer}}$

We have the mass of bullet,

$m_{1} = 20g( = 0.02kg)$

and the mass of pistol,

$m_{2} = 2kg$

,initial velocities of the bullet

$( u_{1}) \: and \: ( u_{2}) = 0$

respectively. The final velocity of bullet,

$v_{1} = + 150m \: {s}^{ - 1} .$

The direction of bullet is taken from left to right. Let v be the recoil velocity of pistol. Total momenta of the pistol and bullet before the fire, when the gun is rest

$= (2 + 0.02)kg \times 0m \: {s}^{ - 1} \\ = 0kg \: m \: {s}^{ - 1}$

Total momenta if the pistol and bullet after it is fired

$= 0.02kg \times ( + 150 m \: {s}^{ - 1} + 2kg \times v \: m \: {s}^{ - 1} \\ = (3 + 2v)kg \: m \: {s}^{ - 1}$

According to the law of conversation of momentum

Total momenta after the fire = Total momenta before the fire 3+2v = 0

$= v = - 1.5m \: {s}^{ - 1}$

Negative sign indicates that the direction in which the pistol would recoil is opposite to that of the bullet,that is, right to left.

$<marquee direction = "right" > Hope it helps you Regards Miss.PJ$