Question

A particle of mass m comes down on a smooth inclined plane from point B at a height of h from rest. The magnitude of change in momentum of the particle between position A (just before arriving on horizontal surface) and C (assuming the angle of inclination of the plane as $\theta$ with respect to the horizontal) is :

• A. 0
• B. $2m\sqrt{\left(2gh\right)}sin\theta$
• C. $2m\sqrt{2gh}sin\left(\frac{\theta }{2}\right)$
• D. $2m\sqrt{\left(2gh\right)}$

Answer 1

The correct option is C $2m\sqrt{2gh}sin\left(\frac{\theta }{2}\right)$

Let velocity of the particle at the base of the inclined plane be $v$

Work-energy theorem : $W_g=\Delta K.E$

$mgh=\frac{1}{2}m{v}^2-0$

$⟹v=\sqrt{2gh}$

Momentum of the particle at point C, $\overrightarrow{P_c}=mv\hat{i}$

Momentum of the particle at point A, $\overrightarrow{P_A}=mvcos\theta \hat{i}-mvsin\theta \hat{j}$

Change in momentum between A and C, $\Delta \overrightarrow{P}=\overrightarrow{P_c}-\overrightarrow{P_A}$

$\Delta \overrightarrow{P}=mv(1-cos\theta )\hat{i}-mvsin\theta \hat{j}$

$|\Delta \overrightarrow{P}|=\sqrt{m^2{v}^2(1-cos\theta {)}^2+m^2{v}^{2}si{n}^2\theta }$

$|\Delta \overrightarrow{P}|=mv\sqrt{1+co{s}^2\theta -2cos\theta +si{n}^2\theta }$

$|\Delta \overrightarrow{P}|=mv\sqrt{2(1-cos\theta )}$ $(\because 1-cos\theta =2si{n}^2\frac{\theta }{2})$

$⟹$ $|\Delta \overrightarrow{P}|=2mvsin(\frac{\theta }{2})$ $=2m\sqrt{2gh}sin(\frac{\theta }{2})$