Question

A bullet of mass 20g moving with a speed of 120 m/s hits a thick muddy wall and penetrates into it. It takes 0.03 s to stop in the wall. Find  (a) The acceleration 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 (d) The distance covered by the bullet in the wall

Answer 1

given that- m= 20g=0.020kg, v=120m/s,t=0.03s Now, we know that,
v=u+ at
so, 0= 120+a(0.03)
a =-120/0.03
a= -4000m/s^2
and,
F= ma
so, F= 0.020×4000
=-8N
and, F'= 8N

Answer 2

Mass of the bullet =20g or 0.02kg
initial velocity of the bullet(u)=120m/s
final velocity(v)=0m/s (because the bullet eventually stops)
a) Acceleration = (v-u)÷t
    = (0-120)÷0.03
    = -120÷0.03
    = -4000m/s²→Acceleration
Or it can be said that the bullet is retarding at 4000m/s²


b) Force=mass × acceleration
    Force exerted by wall on bullet= 0.02kg×4000m/s²
    =80 Newton


c)Answer will be the same...80 Newton. Every action has an equal and opposite reaction. Just the direction will be opposite.


d) Distance covered = speed×time taken
    =120m/s×0.03 seconds
    =3.6 m
Hence, the dist. covered by the bullet =3.6m


Hope it helps :)