question_answer

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-.imparted, to the two fragments is

A)  1.07 kJ                 

B)         2.14 kJ

C)  2.4 kJ                   

D)         4.8 kJ

Correct Answer: D

Solution :

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, \[mv={{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}}\] \[3\times 0=2{{v}_{1}}+1\times 80\]                 \[\Rightarrow \]               \[{{v}_{1}}=-\frac{80}{2}=-40\,m/s\]                 Total energy imparted is                                 \[KE=\frac{1}{2}{{m}_{1}}v_{1}^{2}+\frac{1}{2}{{m}_{2}}v_{2}^{2}\] \[=\frac{1}{2}\times 2\times {{(-40)}^{2}}+\frac{1}{2}\times 1\times {{(80)}^{2}}\] \[=1600+3200=4800\,J\] \[=4.8\,kJ\] Note Since velocity of second piece is negative, it indicateds that the piece is moving in opposite direction to the direction of the other piece.