A) 1.07 kJ
B) 2.14 kJ
C) 2.4 kJ
D) 4.8 KJ
Correct Answer: D
Solution :
Both fragment will possess the equal linear momentum \[{{m}_{1}}{{v}_{1}}={{m}_{2}}{{v}_{2}}\] Þ \[1\times 80=2\times {{v}_{2}}\] Þ \[{{v}_{2}}=40\ m./s\] \ Total energy of system \[=\frac{1}{2}{{m}_{1}}v_{1}^{2}+\frac{1}{2}{{m}_{2}}v_{2}^{2}\] = \[\frac{1}{2}\times 1\times {{(80)}^{2}}+\frac{1}{2}\times 2\times {{(40)}^{2}}\] = 4800 J = 4.8 kJ
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