A) 40
B) 50
C) 70
D) 80
Correct Answer: A
Solution :
From law of conservation of momentum, Final momentum = initial momentum \[{{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}}=mv\] \[7{{v}_{1}}+3{{v}_{2}}=0\] \[\therefore \] \[{{v}_{2}}=-\frac{7{{v}_{1}}}{3}\] So, the two pieces are moving in opposite directions. Increase in KE = 1680 ie, \[\frac{1}{2}{{m}_{1}}v_{1}^{2}+\frac{1}{2}{{m}_{2}}v_{2}^{2}-0=1680\] \[\frac{1}{2}\times 7{{({{v}_{1}})}^{2}}+\frac{1}{2}\times 3{{\left( -\frac{7{{v}_{1}}}{3} \right)}^{2}}=1680\] \[7v_{1}^{2}+3\times \frac{49v_{1}^{2}}{9}=1680\times 2\] \[7v_{1}^{2}+\frac{49v_{1}^{2}}{3}=3360\] \[v_{1}^{2}=\frac{3360\times 3}{70}=144\] \[\therefore \] \[{{v}_{1}}=\sqrt{144}=12\,m/s\] \[\therefore \] \[{{v}_{2}}=-7\times \frac{12}{3}=-28\,m/s\] Relative velocity \[={{v}_{1}}-{{v}_{2}}\] = 12 - (-28) = 40 m/s
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