Question

A person holding a rifle, mass of person and rifle together is 100kg. He stands on smooth surface and fires 10 shots horizontally in 5 sec each bullet has a mass of 10 g with a muzzle velocity of 800 m per sec. The average force exerted on the person is

Answer 1

We have to use the Law of momentum conservation:

P ( initial ) = P ( final )

0 = n · m · u + ( M - n · m ) · v

where: n = 10, m = 10 g = 0.01 kg, u = 800 m/s, M = 100 kg.

0 = 10 · 0.01 kg · 800 m/s + ( 100 kg - 10 · 0.01 kg ) · v

v = - 80 kgm/s / 99.9 kgm/s

v = 0.8 m/s

Then : F = Δ P / Δ t = ( 10 · 0.01 kg · 800 m/s ) : 5 s = 16 N

Answer: The average force exerted on the person is 16 N.

Answer 2

Step-by-step explanation:

v= -0.8m/s

F=16N. .take the solution from pic