A) 0.16 J
B) 1.00 J
C) 0.67 J
D) 0.34 J
Correct Answer: C
Solution :
[c] Initial kinetic energy of the system \[K.{{E}_{i}} =\frac{1}{2}m{{u}^{2}}+\frac{1}{2}M{{(0)}^{2}}=\frac{1}{2}\times 0.5\times 2\times 2+0=1J\] For collision, applying conservation of linear momentum \[\operatorname{m}\times u=\left( m+M \right)\times v\] \[\therefore 0.5\times 2= (0.5+1) \times v\Rightarrow v=\frac{2}{3}m/s\] Final kinetic energy of the system is \[\operatorname{K}.{{E}_{\operatorname{f}}} =\frac{1}{2}\left( m+\operatorname{M} \right){{v}^{2}}=\frac{1}{2}\left( 0.5+1 \right)\times \frac{2}{3}\times \frac{2}{3}=\frac{1}{3}\operatorname{J}\] \[\therefore \]Energy loss during collision \[=\left( 1-\frac{1}{3} \right)\operatorname{J}=0.67J\]
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