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OverLord2011 [107]
3 years ago
9

A rifle with a weight of 25 N fires a 4.0 g bullet with a speed of 290 m/s.

Physics
1 answer:
Ugo [173]3 years ago
3 0

Answer:

Explanation:

Given

Weight of rifle W=25\ N

mass of rifle M=\frac{25}{10}=2.5\ kg

mass of bullet m=4\ gm

speed of bullet u=290\ m/s

As there is no net force on the bullet-rifle system therefore momentum is conserved

0=Mv+mu

v=-\frac{mu}{M}

v=-\frac{4\times 10^{-3}\times 290}{2.5}

v=-0.464\ m/s

i.e. opposite to the direction of bullet speed

(b)Weight of Man =750\ N

Combined weight of man and rifle=750+25=775\ N

mass of man-rifle system M'=\frac{775}{10}=77.5\ kg

Now man and rifle combinedly recoil

therefore

0=(M')v '+mu

v'=-\frac{mu}{M'}

v'=-\frac{4\times 10^{-3}\times 290}{77.5}

v'=0.0149\ m/s

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