Question

A bullet of mass 20 g is horizontally fired with a velocity of 150m/s from a pistol of mass 2kg what is the recoil velocity

Answer 1

(m1) Mass of rifle 2kg

(m2) Mass of bullet 20g

(v2) Velocity of bullet 35m/s

(v1) recoilvelocity of bullet ?

According to law of conservation of momentum

Momentum of rifle=momentum of bullet

M1v1=m2v2

2*v2=20*35

=20*35/2 =350m/s

Answer 2

Answer:

$\huge\colorbox{yellow}{Given\:-}$

• m1 = 20g = 0.02kg

• m2 = 2kg

• v1 = 150 ms-¹

$\huge\colorbox{yellow}{To\:Find\:-}$

The recoil velocity of the pistol

$\huge\colorbox{yellow}{Solution\:-}$

Total momenta of the pistol and bullet before the fire when the gun is at rest

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

$= 0kg \: \: ms^{ - 1}$

Total momenta after it is fired

$0.02kg \times ( + 150 \: {ms}^{ - 1})$

$+ 2kg \times v \: {ms}^{ - 1}$

$= (3 + 2v)kg \: {ms}^{ - 1}$

$\large\underbrace\mathrm\red{According\:to \: law \: of \: conservation \: of \: momentum}$

$\large{\boxed{Total \: \: momenta \: \: after \: \: fire = Total \: \: momenta \: \: before \: \: fire}}$

$3 + 2v = 0$

$\boxed{\pink{ = > v = - 1.5 \: {ms}^{ - 1} }}$

$\huge\colorbox{yellow}{Thank\:You}$