Question

A gun of mass 3kg fires a bullet as mass 30g the bullet takes 0.003 to move through the barrel of the gun and acquires a velocity of 100m/s calculate
(1.) The velocity with which the gun recoils.
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Answer 1

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Answer 2

Solution:-

Given:-

$Mass \: of \: the \: gun \: (m _{1} )= 3 \: kg$

$Mass \: of \: the \: bullet \: ( m_{2} )= 30g = 0.03 \: kg$

$Velocity \: of \: the \: bullet \: ( v_{2} ) = 100 \:{ m/s}$

To Find:-

$Recoil \: velocity( v_{1})$

Formula:-

Formula:-$= > Law \: of \: conservation \: of \: momentum \: :- \\ m_{1}u_{1} + m_{2}u_{2} = m_{1}v_{1} + m_{2}v_{2} \\ = > acceleration \:(a) = \frac{(v - u)}{t} \\ = > Force = mass \times acceleration$

$But  \: ,  \: Initial  \: velocity \:  of \:  the \:  Gun  \: (u_{1})=  \: 0 \:  m/ s \\ And \: , \:  Initial  \: velocity \:  of \:  the  \: Bullet  \:(u_{2}) = \:  0 \:  m/s$

Because at First , the Gun and the Bullet is at rest.

So, Actual Formula :-

$m_{1}u_{1} + m_{2}u_{2} = m_{1}v_{1} + m_{2}v_{2} \\ = > m_{1} + 0 + m_{2} + 0 = m_{1}v_{1} + m_{2}v_{2} \\ = > 0 = m_{1}v_{1} + m_{2}v_{2} \\ = > m_{1}v_{1} = - m_{2}v_{2}$

By the Problem:-

Plot the value:-

$i) \: m_{1}v_{1} = - m_{2}v_{2} \\ = > 3 \times v_{1} = - 0.03 \times 100 \\ = > v_{1} = \frac{100 \times - 0.03}{ 3 } \\ = > - 1m/s$

ii)  The Force  exerted on Gun due to recoil of the Gun.

$ii) Initial  \: velocity  \: of  \: the \:  Gun  \: ( u_{1}) \: = \:  0 \: m/s \\Final  \: velocity  \: of \:  the \:  Gun \:  ( v_{1}) =  - 1 \: m/s \\time(t) = 0.003 \: s \\ acceleration \:(a)  =  \frac{(v - u)}{t}  =  \frac{ - 1 - 0}{0.003}  \\  =  >  \frac{ - 1000}{3} m/ {s}^{2}$

$Force = mass \times acceleration \\ = > 3 \times \frac{ - 1000}{3} \\ = > - 1000N$