A 15.0 g bullet is fired at a muzzle velocity of 3250 m/s from a high powered ri

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 lenght 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 bullets momentum as it leaves the barrel.

Explanation / Answer

M1V1 = M2V2 (15/1000)*(3250) = (4.75)*V2 => velocity recoil of the gun: 10.263 m/s. a) time length bullet is in there: 0.5*a*t^2 = S...........(1) V=at => a = V/t = 3250/t........(2) substitute (2) in (1) 0.5*(3250/t)*t^2 = 0.75 => 0.5(3250t) = 0.75 => t = 4.61*10^-4 s = 0.461 ms b) force on the bullet while it is in the barrel: ma = (15/1000)*(3250/4.61*10^-4) = 10.58*10^4 N c)impulse on the bullet: P = F*t = 10.58*10^4*4.61*10^-4 = 48.7738 N-s d) bullet momentum as it leaves = mV = (15/1000)*(3250) = 48.75 Kg-m/s pl rate lifesaver :)

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