JEE Main · 2019 · 09 Apr 2019 · Shift 2 · Conservation of Momentum

A wedge of mass

M = 4 m

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:

(A) $\frac{2 v ^{2}}{7 g}$
(B) $\frac{v ^{2}}{g}$
(C) $\frac{2 v ^{2}}{5 g}$ ✓ Correct
(D) $\frac{v ^{2}}{2 g}$

Answer: (C)

Solution

At the instant the particle reaches its highest point on the wedge it is momentarily not sliding relative to the wedge, so the two move together with one common horizontal velocity. The floor is frictionless, so the horizontal momentum of the particle-plus-wedge system is conserved:

m v = (m + 4 m) v_{f}

gives

v_{f} = \frac{v}{5}

. Nothing dissipates energy either, so mechanical energy is conserved between the start and the top:

\frac{1}{2} m v^{2} = \frac{1}{2} (5 m) v_{f}^{2} + m g h

. Dividing by

m

and putting

v_{f} = \frac{v}{5}

\frac{1}{2} v^{2} = \frac{v ^{2}}{10} + g h

, so

g h = \frac{v ^{2}}{2} - \frac{v ^{2}}{10} = \frac{2 v ^{2}}{5}

and the greatest height climbed is

h = \frac{2 v ^{2}}{5 g}

JEE Main · 2019 · 09 Apr 2019 · Shift 2 · Single correct option

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JEE Main · 2019 · 09 Apr 2019 · Shift 2 · Conservation of Momentum

Solution

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