Question

A gun of mass 3 kg fires a bullet of mass 30g.the bullet takes 0.003 s to move through the barrel of the gun and aquries a velocity of 100 m/s calculate

Answer 1

$\huge\underline\green{\sf Answer :-}$

$\large{\boxed{\sf Force(F)= -1000N}}$

$\huge\underline\green{\sf Solution :-}$

Given :-

FOR BULLET :-

$\sf{m_{1}=0.03kg}$

$\sf{u_{1}=0m/s}$

$\sf{v_{1}=100m/s}$

FOR RIFFLE :-

$\sf{m_{2}=3kg}$

$\sf{u_{2}=0m/s}$

$\sf{v_{2}=?}$

ACCORDING TO LAW OF CONSERVATION :-

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

$\large{\sf 0 + 0 = 0.03 \times 100 +3v_{2}}$

$\large{\sf 0 = 3 \times 3v_{2}}$

$\large{\sf v_{2}={\frac{-3}{3}}}$

$\large{\boxed{\sf v_{2}=-1m/{s}^{2}}}$

We have to calculate force :-

$\large{\boxed{\sf F=m{\frac{v_{2}-u_{2}}{t}}}}$

$\large\implies{\sf {\frac{3(-1-0)}{0.003}}}$

$\large\implies{\sf {\frac{-3}{3}}\times 1000}$

$\huge\red{\boxed{\sf F=-1000N}}$

Answer 2

❤HEY❤

Refer to the attachment!!