question_answer

A bomb of mass 3 kg explodes in air and splits in two parts of 2 kg and 1 kg. Small peace moves with a speed of 80 m/s, then the total energy of both the pieces will be

A)  1.07 kJ          

B)  2.14 kJ

C)  2.4 kJ            

D)  4.8 kJ

Correct Answer: D

Solution :

 By conservation of linear momentum momentum before collision = momentum after collision \[mv={{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}}\] Or \[3\times 0=2{{v}_{1}}+1\times 80\] \[{{v}_{1}}=-\frac{80}{2}=-40\,m/s\] Total energy of both parts \[KE=\frac{1}{2}{{m}_{1}}v_{1}^{2}+\frac{1}{2}{{m}_{2}}v_{2}^{2}\] \[=\frac{1}{2}\times 2\times {{(-40)}^{2}}+\frac{1}{2}\times 1\times {{(80)}^{2}}\] \[=1600+3200=4800J\] \[=4.8\text{ }kJ\]