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Center of Mass and Collision question

2013 · Q147
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Center of Mass and Collision question

2013 · Q147

NEETPhysicsCenter of Mass and CollisionMCQ+4 / −1
A person holding a rifle (mass of person and rifle together is 100 kg) stands on a smooth surface and fires 10 shots horizontally, in 5 s. Each bullet has a mass of 10 g with a muzzle velocity of 800 m s−-−1. The final velocity acquired by the person and the average force exerted on the person are :
  1. A
    −-−0.08 ms−-−1, 16 N
  2. B
    −-−0.8 ms−-−1, 8 N
  3. C
    −-−1.6 ms−-−1, 16 N
  4. D
    −-−1.6 ms−-−1, 8 N
View written solutionFree

Correct answer: B

According to law of conservation of momentum

MV + mnv = 0

⇒V−−mNvM−−0.01 kg×10×800 m/s100 \Rightarrow V - {{ - mNv} \over M} - {{ - 0.01\,kg \times 10 \times 800\,m/s} \over {100}}⇒V−M−mNv​−100−0.01kg×10×800m/s​

⇒−0.8 m/s \Rightarrow - 0.8\,m/s⇒−0.8m/s

According to work energy theorem,

Average work done = Change in average kinetic energy

i.e, Fav×Sav=12mVrms2{F_{av}} \times {S_{av}} = {1 \over 2}mV_{rms}^2Fav​×Sav​=21​mVrms2​

⇒FavVmax⁡t2=12mVrms22 \Rightarrow {{{F_{av}}{V_{\max }}t} \over 2} = {1 \over 2}m{{V_{rms}^2} \over 2}⇒2Fav​Vmax​t​=21​m2Vrms2​​

⇒Fav=8N \Rightarrow {F_{av}} = 8N⇒Fav​=8N

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