Question

A body starts to slide from $P$, down an inclined frictionless plane $PQ$ having inclination $\alpha$ with horizontal and then ascends another smooth inclined plane $QR$ with an angle of inclination $2\alpha$. Neglecting impact at $Q$, we have:

• A. $t_{PQ}=t_{QR}$
• B. $t_{PQ}<t_{QR}$
• C. $h^{\prime }=2h$
• D. $h^{\prime }=h$

Answer 1

The correct option is D $h^{\prime }=h$

As the planes PQ and QR are frictionless and impact is neglected, so mechanical energy will conserve.
Thus, $mg{h}^{\prime }=mgh\Rightarrow h^{\prime }=h$