· The fore on bullet a gun is given by the formula Fa-a-be. a while it is in the

Question

· The fore on bullet a gun is given by the formula Fa-a-be. a while it is in the barrel of a is in Newtons b is in Newtons per seconds. t is time and is measured in seconds. Assume the force starts acting on the bullet at t = 0 and at t = T, the bullet exits the gun. a) Compute the total impulse imparted to the bullet by the gun. Express your answer in terms of a, b, T b) If the mass of the bullet is me, what is the speed with which the bullet comes out of the gun Call the impulse computed in part a) J and write your answer in terms of J regardless of what you computed J to be e) If the recoil speed of the gun is 100 times smaller. What is the mass of the gun in terms of the mass of the bullet? (Assume external forces acting on gun +bullet system can be neglected.)

Explanation / Answer

a)

FB = a – bt

At t=T ,

Force on bullet = FB = a – bt = a - bT

Impulse = J = FB/t = FB /T

J= (a-bT)/T

J = a/T – b

b)

J = mb(vbf – vbi)

Initially bullet is at rest, hence vi= 0 m/s

J = mbfvf

vbf = J/mb

c)

By law of conservation of linear momentum,

mg*vgf =mb*vbf ---------------(1)

given vbf =100vgf ------------------(2)

mg*vgf =mb*100*vgf

mg =mb*100

Thus mg is 100 times greater than mb

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