Question

A bullet if mass 50g is moving with a velocity of 500 m/s. It penetrates 10cm into a still target and comes to rest. Calculate:
i. The kinetic energy possessed by the bullet.
ii. The average retarding force offered by the target.

Answer 1

Retarding force

m = 50 g = 50 × 10-3 kg, u = 500 ms-1, s = 10 cm, v = 0

(i) KE = 1/2mv2 = 1/2×50 × 10-3 × (500)2 = 6250 J. 

(ii) Using v2  u2 = 2aS, we get 

a = {v2  u2}/{2aS} = {02 (500)2}/{2 × 10 ×10-2} 

= 1250000ms-2.

Therefore, retarding force, F = ma = 50 × 10-3 × 1250000 = 62500 N. 

Answer 2

By expression :K=1/2mv2

K=1/2*50/1000*500*500

then we get :-

K=6250J

Second part :-

V=0(as it comes to rest)

By expression :v2-u2=2aS

We get:a

a= -1250000(negative sign due to retardation)

So, neglect the negative sign,

hence,

F=ma

F=50/1000*1250000

By solving :-

F=62500N(as it is ask to find)