A bomb of mass `3.0 kg` explodes in air into two pieces of masses `2.0 kg` and `1.0 kg`. The smaller mass goes at a speed of `80 m//s`. The total energy impared to the two fragments is :

Options

A. 1.07 kJ
B. 2.14 kJ
C. 2.4 kJ
D. 4.8 kJ

Answer

D

Explanation

From law of conservation of momentum, when no external force acts upon a system of two (or more) bodies then the total momentum of the system remains constant.
Momentum before explosion = momentum after explosion.
Since bomb `v` at rest, its velocity is zero, hence
`mv = m_1 v_1 + m_2 v_2`
`3 xx 0 = 2 v_1 + 1 xx 80`
`rArr v_1 = - (80)/(2) = - 40 m//s`
Total energy impared is
`KE = (1)/(2) m_1 v_1^2 + (1)/(2) m_2 v_2^2`
=`(1)/(2) xx 2 xx (-40)^2 + (1)/(2) xx 1 xx (80)^2`
=`1600 + 3200 = 4800 J`
=`4.8 kJ`.