Question

a bullet of mass 30g is fired from a gun of mass 9kg. if the velocity of the bullet is 400m/s, then the recoil velocity of the gun will be?

a) 1.33 m/s in the direction of bullet

b) 0.75m/s opposite to the direction of the bullet

c) 0.75m/s in the direction of bullet

d) 1.33m/s in the opposite direction of the bullet
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Answer 1

Given : a bullet of mass 30g is fired from a gun of mass 9kg.  the velocity of the bullet is 400m/s,

To Find: the recoil velocity of the gun will be

Solution:

Recoil velocity is the backward velocity experienced by a shooter when  shoots a bullet.

Initial velocity of gun and bullet = 0

u₁  and v₁   are = 0

u₁ = Initial velocity of gun = 0

u₂  = Initial velocity of  bullet = 0

mass of gun = m₁ = 9 kg = 9000 g

mass of bullet = m₂ =  30 g

v₁ =  Recoil  velocity of gun  ( back ward velocity )

v₂  = velocity of  bullet = 400m/s

Conservation of momentum :

m₁u₁ + m₂u₂  = m₁(-v₁) + m₂v₂

( note v₁ is taken as -ve because v₁ is recoil velocity hence backward )

=> 9000 * 0 + 30*0  = 9000(-v₁) + 30(400)

=> v₁ = 1.33

Hence 1.33m/s in the opposite direction of the bullet

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Answer 2

In these kind of questions, it is best to apply CONSERVATION OF LINEAR MOMENTUM principle :

$P_{initial} = P_{final}$

$\implies m_{g}u_{g} + m_{b}u_{b} = m_{g}v_{g} + m_{b}v_{b}$

$\implies 0 + 0 = 9v_{g} + ( \dfrac{30}{1000} \times 400)$

$\implies 9v_{g} = - 12$

$\implies v_{g} = - 1.33 \: m {s}^{ - 1}$

So, velocity of gun is 1.33 m/s opposite to direction of bullet.