A) 8 cm
B) 40 cm
C) 80 cm
D) 4 cm
E) None of these
Correct Answer: E
Solution :
The moment 1 kg hits the platform, \[1(v)+0=(1+3){{v}_{1}}\Rightarrow |(\sqrt{2gh})|=4{{v}_{1}}\] \[\Rightarrow \]\[{{v}_{1}}=\frac{\sqrt{2gh}}{4}=\frac{\sqrt{2\times 10\times 100}}{4}=\sqrt{\frac{{{10}^{3}}}{8}}m/s\] (Compression due to masses is negligible.) Using energy conservation principle, \[\frac{1}{2}Mv_{1}^{2}=\frac{1}{2}k{{x}^{2}};\] \[x=\sqrt{\frac{M'}{k}{{v}_{1}}}=\sqrt{\frac{4}{\frac{5}{4}\times {{10}^{6}}}\times \frac{{{10}^{3}}\times {{10}^{4}}}{4}}cm\]\[=2cm\] *None of the given options is correct.
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