A wedge of mass M = 4m lies on a frictionless | vedclass

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(C) Let the particle attain a maximum height $h$ on the wedge. At this point,the particle and the wedge move together with a common horizontal velocit

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let the particle attain a maximum height $h$ on the wedge. At this point,the particle and the wedge move together with a common horizontal velocity $V_f$.By the principle of conservation of linear momentum in the horizontal direction:$mv = (m + M)V_f$Since $M = 4m$,we have:$mv = (m + 4m)V_f = 5mV_f$$V_f = \frac{v}{5}$By the principle of conservation of mechanical energy:$\frac{1}{2}mv^2 = \frac{1}{2}(m + M)V_f^2 + mgh$$\frac{1}{2}mv^2 = \frac{1}{2}(5m)\left(\frac{v}{5}\right)^2 + mgh$$\frac{1}{2}mv^2 = \frac{1}{2}(5m)\left(\frac{v^2}{25}\right) + mgh$$\frac{1}{2}mv^2 = \frac{1}{10}mv^2 + mgh$$mgh = \frac{1}{2}mv^2 - \frac{1}{10}mv^2 = \frac{5-1}{10}mv^2 = \frac{4}{10}mv^2 = \frac{2}{5}mv^2$$h = \frac{2v^2}{5g}$

$A$ 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

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