question_answer

A bomb of mass 9kg explodes into 2 pieces of mass 3kg and 6kg. The velocity of mass 3kg is 1.6 m/s, the K.E. of mass 6kg is                                     [AIEEE 2002]

A)                 3.84 J    

B)                 9.6 J

C)                 1.92 J    

D)                 2.92 J

Correct Answer: C

Solution :

                    As the bomb initially was at rest therefore Initial momentum of bomb = 0 Final momentum of system = \[{{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}}\] As there is no external force \ \[{{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}}=0\] Þ \[3\times 1.6+6\times {{v}_{2}}=0\] velocity of 6 kg mass \[{{v}_{2}}=0.8\,m/s\] (numerically) Its kinetic energy\[=\frac{1}{2}{{m}_{2}}v_{2}^{2}\]\[=\frac{1}{2}\times 6\times {{(0.8)}^{2}}=1.92\,J\]