A 9 g bullet leaves the muzzle of a rifle with a speed of 432 m/s

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#1

A 9 g bullet leaves the muzzle of a rifle with a speed of 432 m/s.
What constant force is exerted on the bullet while it is traveling down the 0.5 m length of the barrel of the rifle

Answer:

Mass of bullet = 9 g
Final velocity of bullet = 432 m/s
Length of barrel = .5 m
By NEWTON’S second equation of motion, s = (u + v).t/2
where s is distance (= .5 m) , u( = 0) and v( = 432 m/s) are initial and final velocities respectively. and t is the time taken.

Therefore, t = 2.s/(u + v) = 2.(.5)/(0 + 432) = .0023 seconds.

Now, Acceleration = Change in Velocity/ time taken = 432/.0023 = 187826.087 m/s

And, Force = mass X acc. = .009 X 187826.087= 1690.4 Newtons