Question

A gun of mass 3 kg fires a bullet of mass 30g .The bullet takes 0.003 second to move through the barrel of the gun and acquires a velocity pf 100m/s. Calculate the force exerted on gunman due to recoil of the gun

Answer 1

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Answer 2

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For gun m1=3kg

For bullet m2=30g=0.03kg,v2= 100m/s

By law of conservation of momentum m1*v1= m2 *v2 3*v1 =0.03*100 v1 =0.03*100/3

(i) The velocity with which the gun recoils=v1=1 m/s Acceleration of bullet=v-u/t 100-0/0.03=3333.33m/s2

(ii)The force exerted on gunman due to recoil of the gun=F=ma=0.03*3333.33=100 N