A bullet of mass 20g moving with a speed of `120 ms^-1` hits a thick muddy will and penetrates into it. It takes 0.03 to stop in the wall. Find (a) the accleration of the bullet in the wall. (b)the force exerted by the wall on the bullet , ( c) the force exerted by the bullet on the wall and (d) the distance covered by the bullet in the wall

Explanation

(a) The velocity of the bullet as it hits the wall is `u=120 ms^-1`
the velocity after 0.03s is v=0 So , using `v=u+at`
`0=(120 ms^-1)+a(0.03s)`
or `a=- 120/0.03 ms^-2=-4000 ms^-2`
(b) The force exerted by the wall on the bullet is
`F=ma=(20g) (-4000 ms^-2)`
`=(20/1000 kg) (-4000 ms^-2)=-80N`
The negative sign shows that the force by the wall on the bullet is in a direction opposite to that of velocity.
( c) From Newton third law, the force exerted by the bullet on the wall is also 80N, in the direction of the velocity.
(d) The distance covered by the bullet in the wall is
`s=ut+1/2 at^2=(120 ms^-1) (0.03s)+1/2(-4000 ms^-2)(0.0009 s^2)`
`=3.6m-1.8m=1.8m`