A gun of mass 3 kg fires a bullet of mass 30 g. The bullet takes 0.003 seconds to move through the barrel of the gun, and acquires a velocity of 100 m/s.
Calculate:
a) the velocity at which the gun recoils
b) the force exerted on the gunman due to recoil of the gun.
I want the solution as well as the answer.
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Mass of gun m1=30g=0.03kg
mass of bullet m2= 3kg
Velocity of bullet v1= 100m/s
velocity of gun or recoil v2 = ? (to be calculated)
by the law of conservation of momentum
m1v1=m2v2
0.03x100=3xv2
v2=1.1m/s
acceleration of bullet= v-u/t
110-0/0.03=3666.6m/s2
f=ma
0.03x3666.6=109.9N
this is the answer
mass of bullet m2= 3kg
Velocity of bullet v1= 100m/s
velocity of gun or recoil v2 = ? (to be calculated)
by the law of conservation of momentum
m1v1=m2v2
0.03x100=3xv2
v2=1.1m/s
acceleration of bullet= v-u/t
110-0/0.03=3666.6m/s2
f=ma
0.03x3666.6=109.9N
this is the answer
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