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A bullet of mass 20 g is horizontally fi...

A bullet of mass 20 g is horizontally fired with a velocity 150 m/s from a pistol of mass 2 kg . What is the recoil velocity of the pistol ?

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Initial velocity, `u=150m//s`
Find velocity, `v=o` (since the bullet finally comes to rest)
Time takne to come to rest, `t=0.03s`. According to the first equation of motion
`V=u+at`
Acceleration of the bullet, a
`O=150+(axx0.03s)a=-150//0.03=-5000m//s^(2)`
(Negative sign indicates that the velocity of the bullet is decreasing )
According to the third equation of motion `v^(2)=u^(2)=2as`
`0=(150)^(2)+2(-5000)`
`=22500//10000`
`=2.25m`
Hence the distance of penetration of the bullet into the block is 2.25 m.
From Newton.s second law of motion.
Force = Mass `xx` Acceleration
Mass of the bullet, `m=10g=0.01 kg`
Acceleration of the bullet, `a=5000m//s^(2)`
`F=ma=0.01xx5000=50N`
Hence the magnitude of force exerted by the wooden block on the bullet is 50 N.
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