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,

Explanation

We will first calculate the acceleration (or rather retardation) of the bullet.
Now, Initial velocity,u = `150 ms^(-1)`
Final velocity,v = 0
And, Time,t=0.03 s
Now, v=u+at
So, `0=150+axx0.03`
`0.03a=-150`
`a=-150/0.03`
`a=-5000ms^(-2)`
Let us calculate the distance of penetration of bullet now.
We know that : `v^(2)=u^(2)+2as`
So, `(0)^(2)=(150)^(2)+2xx(-5000)xxs`
`0=22500-10000xxs`
`10000s=22500`
So, `s=22500/10000`
s=2.25m
Thus, the distance of penetration of the bullet into the block of wood will be 2.25 metres
We will now calculate the magnitude of force. Now:
Force,F=m`xx`a
`F=10/1000kgxx(-5000)ms^(-2)`
F=50 N
Thus, the magnitude of force exerted by the wooden block on the bullet is 50 newtons.