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 :

Question

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 : (A) (2v2/7g) (B) (v2/g) (C) (2v2/5g) (D) (v2/2g)

Tardigrade Admin · Accepted Answer

Applying Linear momentum conservation mv = (m + M) vc vc = (v/5) applying work energy theorem - mgh = (1/2) (m + M) vc2 - (1/2) mv2 solve , h = (2v2/5g)

Q. 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 :

$\frac{2 v ^{2}}{7 g}$

$\frac{v ^{2}}{g}$

$\frac{2 v ^{2}}{5 g}$

$\frac{v ^{2}}{2 g}$

Applying Linear momentum conservation
$m v = (m + M) v_{c}$
$v_{c} = \frac{v}{5}$
applying work energy theorem
$- m g h = \frac{1}{2} (m + M) v_{c}^{2} - \frac{1}{2} m v^{2}$
solve , $h = \frac{2 v ^{2}}{5 g}$

Q. 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 :

7g2v2​

gv2​

5g2v2​

2gv2​