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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
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
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
Dear student,Please find the solution to your problem below. 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
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