question_answer

A wedge of mass M = 4m lies on a frictionless plane. A particle of mass m approaches the wedge with speed v. There is no friction between the particle and the plane or between the particle and the wedge. The maximum height climbed by the particle on the wedge is given by :- [JEE Main 9-4-2019 Afternoon]

A) \[\frac{2{{\text{v}}^{2}}}{7g}\]                            

B) \[\frac{{{\text{v}}^{2}}}{g}\]

C) \[\frac{\text{2}{{\text{v}}^{2}}}{5g}\]     

D) \[\frac{{{\text{v}}^{2}}}{2g}\]

Correct Answer: C

Solution :

Applying Linear momentum conservation \[mv=(m+M){{v}_{c}}\] \[{{v}_{c}}=\frac{\text{v}}{5}\] applying work energy theorem \[-mgh=\frac{1}{2}(m+M){{v}_{c}}^{2}-\frac{1}{2}m{{\text{v}}^{2}}\] Solve,\[h=\frac{2{{\text{v}}^{2}}}{5g}\]