A) 288 J
B) 144 J
C) 128 J
D) 64 J
Correct Answer: A
Solution :
Here: Total mass m = 12 kg Initial velocity of total mass v = 0 mass of the first piece \[{{m}_{1}}=4kg\] mass of the second piece \[{{m}_{2}}=8kg\] velocity of second piece \[{{\upsilon }_{2}}=6\,\,m/s\] Now, applying law of conservation of Momentum \[m\upsilon ={{m}_{1}}{{\upsilon }_{1}}+{{m}_{2}}{{\upsilon }_{2}}\] \[12\times 0=4\times {{\upsilon }_{1}}+8\times 6\] \[4{{\upsilon }_{1}}=-48\,\,\,\Rightarrow \,\,{{\upsilon }_{1}}=-12\] (Minus sign corresponds the motion in opposite direction) Hence, kinetic energy of the 4 kg mass \[=\frac{1}{2}{{m}_{1}}u_{1}^{2}=\frac{1}{2}\times 4\times {{(12)}^{2}}=288\,J\]
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