Question

A 15.0-g bullet is fired at a muzzle velocity of 3250 m/s from a high-powered rifle with a mass of 4.75 kg and barrel of length 75.0 cm. (a) How long is the bullet in the barrel? (b) What is the force on the bullet while it is in the barrel? (c) Find the impulse exerted on the bullet while it is in the barrel. (d) Find the bullet’s momentum as it leaves the barrel.

I've seen this question a number of times being answered and it's still not making sense????????????

Answer 1

Solution)

Acc to Conservation of Momentum

M1V1 = M2V2

So,

(15/1000)*(3250) = (4.75)*V2

velocity recoil =10.263 m/s

a) acc to kinematics equation

0.5*a*t^2 = S...........(1)

V=at

a = V/t = 3250/t........(2)

substituting (2) in (1)

We get

0.5*(3250/t)*t^2 = 0.75

0.5(3250t) = 0.75

t = 4.61*10^-4 s = 0.461 ms (Ans)

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b) Force on the bullet =ma = (15/1000)*(3250/4.61*10^-4)

= 10.58*10^4 N (Ans)

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c)impulse ,P = F*t = 10.58*10^4*4.61*10^-4 = 48.7738 N-s (Ans)

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d) momentum of bullet = mV = (15/1000)*(3250) = 48.75 Kg-m/s (Ans)

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Good luck!:)