Question

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 1

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.