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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

Grade:12th pass

3 Answers

Arun
25758 Points
5 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
3 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
Kushagra Madhukar
askIITians Faculty 629 Points
2 years ago
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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