Question

A bullet of mass 50 g is moving with a velocity of 500 ms1. It penetrates 10 cm into 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

Here, m = 50 g = 50 � 10-3 kg, u = 500 ms-1, s = 10 cm, v = 0, KE = ?, F = ?

(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

1) Mass of bullet (m) = 50g

= $\dfrac{50}{1000}$ = 0.05 kg

Initial velocity (u) = 500 m/s

s = 10 cm = 0.1 m

Final velocity (v) = 0 m/s

Kinetic Energy (K.E.) = ?

Now..

Kinetic Energy = $\dfrac{1}{2}$ mv²

= $\dfrac{1}{2}$ × 0.05 × 500 × 500

= $\textbf{6250 J}$

2) Regarding Force = ?

v² - u² = 2as

a = $\dfrac{ {v}^{2} \:-\:{u}^{2} }{2s}$

a = $\dfrac{ {0}^{2}\:-\: {500}^{2} }{2\:\times\:0.1}$

a = $\dfrac{-250000}{0.2}$

a = -1250000 m/s²

F (regarding force) = ma

= 0.05 × (-1250000)

= $\textbf{62500 N}$