A) \[\frac{1}{2}mg(h+2d)\]
B) \[\frac{1}{2}mg(h+4d)\]
C) \[\frac{1}{2}mg(h-d)\]
D) \[\frac{1}{2}mg(h-2d)\]
Correct Answer: B
Solution :
Final velocity of first body of mass m in falling from a height h \[{{v}^{2}}=0+2gh\] \[\Rightarrow \] \[v=\sqrt{2gh}\] Using the law of conservation of momentum initial momentum = final momentum \[mv+0=(m+m)v'\] \[2\sqrt{2gh}=2mv'\] \[\Rightarrow \] \[v'=\frac{\sqrt{2gh}}{2}=\sqrt{\left( \frac{gh}{2} \right)}\] Thus, work done against resistive force \[=\frac{1}{2}\times (2m)v{{'}^{2}}+(2m)gd\] where, \[d=\]distance with which the combined mass moves on the ground \[v'=\]velocity of combined mass \[\therefore \]Work done \[=m\left( \frac{gh}{2} \right)+2mgd\] \[=mg\left( \frac{h}{2}+2d \right)\] \[=\frac{1}{2}mg(h+4d)\]
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