Question

A bullet of mass 50g is fired from a gun with velocity 50m/s w. r. t ground. If mass of the gun is 20kg then recoil velocity of the gun is

Answer 1

Answer:

recoil velocity = 0.125 $ms^{-1}$ direction opposite to bullet

Explanation:

for an isolated system the net change of momentum is zero.

so lets make the net momentum zero

small letter represent bullet and capital letter represent gun

1 is initial condition and 2 is final condition

$m_{1} v_{1} + M_{1}V_{1}$ = $m_{2} v_{2} + M_{2} V_{2}$

now initial gun is stationary wrt ground cause we are running and also bullet is stationary.

this implies that

$v_{1}$ and $V_{1}$ are zero

so the equation we have is

$m_{2} v_{2} + M_{2} V_{2}$ = 0

now put the values

$m_{2}$ = 50g = .05Kg

$v_{2}$ = 50

$M_{2}$ = 20

$V_{2}$ = ?

Put he values now

(20 * $V_{2}$ ) + (.05 * 50) = 0

$V_{2}$ = -$\frac{0.05 * 50}{20}$

$V_{2}$ = -0.125 $ms^{-1}$

recoil velocity = 0.125 $ms^{-1}$ direction opposite to bullet