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\[=\frac{1}{2}{{m}_{2}}v_{2}^{2}\]\[=\frac{1}{2}\times 6\times {{(0.8)}^{2}}=1.92\,J\]You need to login to perform this action.
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