A bullet of mass 10 g travelling horizontally with a velocity of `150 ms^(-1)` strikes a stationary wooden block and comes to rest in 09.03 s. Calculate the distance of penetration of the bullet into the block. Also calculate the magnitude of the force exerted by the wooden block on the bullet.

Explanation

Initial velocity u=150 m/s
Find velocity v=o (since the bullet finally comes to rest)
Time taken to come to rest t=0.03 s . According to the first equation of motion
V=u+at
Acceleration of the bullet ,a
`O=150+(axx0.03s)a=-150//0.03 =-5000 m//s^2`
(Negative sign indicates that the velocity of the bullet is decreasing)
According to the third equatino of motion `v^2=u^2+2as`
`0=(150)^2+2(-5000)`
=22500/10000
=2.25 m
Hence the distance of penetration of the bullet into the block is 2.25 m.
From Newton.s second law of motion.
Force=Mass x Acceleration
Mass of the bullet m=10 g =0.01 kg
Acceleration of the bullet , `a=5000 m//s^2`
F =ma =0.1 x 5000 =50N
Hence the magnitude of force exerted by the wooden blcok on the bullet is 50N .