A) \[h\]/4
B) \[h\]/2
C) \[h\]
D) independent of \[h\]
Correct Answer: B
Solution :
[b] \[{{v}_{0}}=\sqrt{2gh}\] before collision. Apply conservation principle of momentum, \[m{{v}_{0}}=(m+m)u\] \[\therefore u=\frac{{{v}_{0}}}{2}\] Applying conservation principle of momentum, \[\frac{1}{2}(2m){{u}^{2}}=2mgh'\] \[\therefore h'=\frac{{{u}^{2}}}{2g}\Rightarrow h'\Rightarrow =\frac{{{v}^{2}}_{0}}{4g}\] \[h'=\frac{2gh}{4g}\Rightarrow h'=\frac{h}{2}\]
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