A) 1.07 kJ
B) 2.14 kJ
C) 2.4 kJ
D) 4.8 kJ
Correct Answer: D
Solution :
By conservation of linear momentum momentum before collision = momentum after collision \[mv={{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}}\] Or \[3\times 0=2{{v}_{1}}+1\times 80\] \[{{v}_{1}}=-\frac{80}{2}=-40\,m/s\] Total energy of both parts \[KE=\frac{1}{2}{{m}_{1}}v_{1}^{2}+\frac{1}{2}{{m}_{2}}v_{2}^{2}\] \[=\frac{1}{2}\times 2\times {{(-40)}^{2}}+\frac{1}{2}\times 1\times {{(80)}^{2}}\] \[=1600+3200=4800J\] \[=4.8\text{ }kJ\]You need to login to perform this action.
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