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

Question

Arun · Answer

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 · Answer

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

Kushagra Madhukar · Answer

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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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

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

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