Question

A gun of mass 5 kg fires a bullet of mass 25g. The bullet takes 0.005 seconds to move through the barrel of the gun and acquires a velocity of 500 m/s. calculate; i) The velocity with which the gun recoils. ii) The force exerted on gun man due to recoil of the gun.​

Answer 1

Answer:

Recoil velocity of the gun => -2.5

Force exerted on gun man due the recoil of the gun = 2500N

Explanation:

Given :

$\sf{}Mass\ of\ the\ gun,m_1 = 5kg$

$\sf{}Mass\ of\ the\ bullet,m_2 = 25g = 0.025kg$

$\sf{}Initial\ velocity\ of\ gun,u_1 = 0m/s$

$\sf{}Initial \ velocity\ of\ bullet,u_2 = 0m/s$

$\sf{}Final\ velocity\ of\ the\ bullet,v_2 = 500m/s$

$\sf{}\sf{}Time\ taken\ by\ bullet\ to\ move\ through\ the\ barrel,t=0.005s$

To Find :

$\sf{}The\ velocity\ with\ which\ the\ gun\ recoils,v_1=\ ?$

$\sf{}Force\ exerted\ on\ gunman\ due\ to\ recoil\ of\ the\ gun,f=\ ?$

Solution :

i)According to the conservation law of momentum.

$\sf{}m_1u_1+m_2u_2=m_1v_1+m_2v_2$

$\sf{}5\times0+0.025\times0=5\times v_1+0.025\times 500$

$\implies \sf{}5\times0+0.025\times0=5\times v_1+0.025\times 500$

$\implies\sf{}0=5v_1+12.5$

$\implies \sf{}-12.5=5v_1$

$\implies \sf{}-\dfrac{12.5}{5}=v_1$

$\sf{}\therefore \sf{}v_1=-2.5$

2)We know Force,

$\sf{}f=m\dfrac{v-u }{t}$

$\sf{}\implies 5\times\dfrac{2.5-0}{0.005}$

$\sf{}\implies 5\times \dfrac{2.5}{0.005}$

$\sf{}\implies 5\times 500}$

$\sf{}\therefore 2500N}$

Answer 2

Answer:

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Explanation:

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