A bullet of mass `10g` travelling horizontally with a velocity of `150 ms^(-1)` strikes a stationary wooden block and come to rest in `0.03 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,

Answer

2.25 m; 50 N

Explanation

Initial velocity , u=150m/s
Final velocity , v=0 ( since the bullet finally comes to rest )
Time taken to come to rest , t=0.03s
According to the first equation of motion, v=u+at
Acceleration of the bullet, a
`0=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
=2.25 m
Hence, the distance of penetration of the bullet into the block is 2.25 m
From Newtons's second law of motion:
Force, F= Mass `xx` Acceleration
Mass of the bullet, m=10g=0.01g
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 50N.