# a person holding a rifle mass of the person and rifle together is 100 kg stands on a smooth surface and fires 10 shots horizontally 5 secondeach bullet has a mass of 10 gram with a muzzle velocity of 800 metre per

Arun
25750 Points
6 years ago
By the law of momentum conservation:-
=>P(iniitial) = P(final)
=>mu1 + Mu2 = mv1 +Mv2
=>0 + 0 = nmu + (M-nm)v
=>10 x 10 x 10^-3 x 800 + (100 - 10 x 10 x 10^-3) x v
=>v = -0.80 m/s {-ve just indicating the direction is opposite to the direction of bullets}
By F = ∆P/∆t
=>F = {10 x 10 x 10^-3 x 800}/5 = 16N
Rubeen Sarvani
15 Points
5 years ago
Momentum of a system is constant
Mass of 10 bullets = 10 ×10=100g=0.1kg

MV=mv
MV-mv=0
100×V=(-0.1)×800
100×V=(-80)
V=(-0.8) m/s

F=∆P/∆T       (bullet)
=MV/T
=0.1kg × 800m/s     /5sec
=80/5
=16N
4 years ago
Dear student,

By the law of momentum conservation:-
=>P(iniitial) = P(final)
=>mu1 + Mu2 = mv1 +Mv2
=>0 + 0 = nmu + (M-nm)v
=>10 x 10 x 10^-3 x 800 + (100 - 10 x 10 x 10^-3) x v
=>v = -0.80 m/s {-ve just indicating the direction is opposite to the direction of bullets}
By F = ∆P/∆t
=>F = {10 x 10 x 10^-3 x 800}/5 = 16N

Thanks and regards,
Kushagra