A) \[0.3\text{ }J\]
B) \[0.6\text{ }J\]
C) \[0.8\text{ }J\]
D) \[1.5\text{ }J\]
Correct Answer: B
Solution :
[b] Applying momentum conservation \[{{\operatorname{m}}_{1}}{{u}_{1}} + {{m}_{2}}{{u}_{2}} = {{\operatorname{m}}_{1}}{{v}_{1}} + {{m}_{2}}{{v}_{2}}\] \[0.1u+m\left( 0 \right)= 0.1\left( 0 \right) +m\left( 3 \right)\] \[0.1\text{ }u=3\text{ }m\] \[\frac{1}{2}0.1{{u}^{2}}=\frac{1}{2}m{{\left( 3 \right)}^{2}}\] Solving we get, \[u=3\] \[\frac{1}{2}k{{x}^{2}}=\frac{1}{2}K{{\left( \frac{x}{2} \right)}^{2}}+\frac{1}{2}\left( 0.1 \right){{3}^{2}}\] \[\Rightarrow \,\,\,\,\,\frac{3}{4}k{{x}^{2}}=0.9\,\,\Rightarrow \,\,\,\frac{3}{2}\times \frac{1}{2}k{{x}^{2}}=0.9\] \[\therefore \,\,\,\,\,\,\,\,\frac{1}{2}K{{x}^{2}}= 0.6 J\] (total initial energy of the spring)
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