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Conservation of Momentum

A person hold...

Question

A person holding a rifle (mass of person and rifle together is 100 kg) stands on a smooth surface and fires 10 shots horizontal, in 2 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

- 0.8 {m s}^{- 1}, 16 N

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- 0.8 {m s}^{- 1}, 8 N

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- 1.6 {m s}^{- 1}, 16 N

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- 1.6 {m s}^{- 1}, 8 N

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Solution

The correct option is B $- 0.8 {m s}^{- 1}, 16 N$
$P (i n i t i a l) = P (f i n a l)$

$0 = n \times m \times u + (M - n \times m) \times v$

where: $n = 10, m = 10 g = 0.01 k g, u = 800 m / s, M = 100 k g$

$0 = 10 \times 0.01 k g \times 800 m / s + (100 k g - 10 \times 0.01 k g) \times v$

$v = - \frac{80 k g m / s}{99.9 k g m / s}$

$v = - 0.8 m / s$

Then : $F = \frac{Δ P}{Δ t} = \frac{10 \times 0.01 k g \times 800 m / s}{5 s} = 16 N$

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A person holding a rifle (mass of person and rifle together is 100 kg) stands on a smooth surface and fires 10 shots horizontal, in 2 s. Each bullet has a mass of 10 g with a muzzle velocity of 800ms−1. The final velocity acquired by the person and the average force exerted on the person are

The correct option is B −0.8ms−1,16NP(initial)=P(final)0=n×m×u+(M−n×m)×vwhere: n=10,m=10g=0.01kg,u=800m/s,M=100kg0=10×0.01kg×800m/s+(100kg−10×0.01kg)×vv=−80kgm/s99.9kgm/sv=−0.8m/sThen : F=ΔPΔt=10×0.01kg×800m/s5s=16N