A gun mass 3kg fires a bullet of mass 30g.the bullet takes 0.003sec to move through the barrel of the gun and acquires a velocity of 100m/sec.calculate
(A) velocity with which the gun recoils.
(B) the force exerted on gunman due to recoil of the gun.
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For gun m1=3kg v1=?
For bullet m2=30g=0.03kg v2=110m/s
By law of conservation of momentum
M1*v1=m2*v2
3*v1=0.03*110
V1=0.03*110/3
V1=1.1m/s
Aof bullet=v-u/t
110-0/0.03=3666.6m/s2
F=ma=0.03*3666.6=109.9N
For bullet m2=30g=0.03kg v2=110m/s
By law of conservation of momentum
M1*v1=m2*v2
3*v1=0.03*110
V1=0.03*110/3
V1=1.1m/s
Aof bullet=v-u/t
110-0/0.03=3666.6m/s2
F=ma=0.03*3666.6=109.9N
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