Question

A bullet hits the wall with a velocity of 20m/s and penetrates it up to a distance of 5cm. Find the deceleration of the bullet in the wall

Answer 1

We have,

Initial velocity, u = 20 m/s

Distance travelled, s = 5cm=0.05 m

Let the retardation produced be ‘a’. 

The final velocity is v = 0.

So,

v2 = u2+2as

=> 0 = 202  +2a(0.05)

=> a = -4000 m/s2

Thus the retardation/deceleration produced is 4000 m/s2

Answer 2

S= 0.05 m
Vi = 0
Vf = 20 m/s
A?
2as= vf2- vi2
A = vf2 - vi2 ÷ 2s
Vi= 0
So,
A = vf2 ÷ 2s
A = (20)2 ÷ 2 × 0.05
A = 400 ÷ 0.1
A = 4000m/s2 And deacceleration = -(a) = - ( 4000 ) Deacceleration = - 4000 m/s2