Q.
From an inclined plane two particles are projected with same speed at same angle $\theta$, one up and other down the plane as shown in figure, which of the following statements is/are correct?
A) The time of flight of each particle is the same
B) The particles will collide the plane with same speed
C) Both the particles strike the plane perpendicularly
D) The particles will collide in mid air if projected simultaneously and time of flight of each particle is less than of collision
Explanation
(a) $Here,\ \alpha = 2\theta,\ \beta = \theta$
[image:4_image105.emf]
$Time\ of\ flight\ of\ A\ is,$
$T_{1} = \frac{2u\sin(\alpha - \beta)}{g\cos\beta}$
$= \frac{2u\sin(2\theta - \theta)}{g\cos\theta}$
$= \frac{2u}{g}\tan\theta$
$Time\ of\ flight\ of\ B\ is,\ T_{2} = \frac{2u\sin\theta}{g\cos\theta}$
$= \frac{2u}{g}\tan\theta$
So, $T_{1} = T_{2}$. The acceleration of both the particles is $g$ downwards. Therefore, relative acceleration between the two I zero or relative motion between the two is uniform. The relative velocity of $A$ w.r.t. $B$ is towards$AB$, therefore collision will take place between the two in mid air.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.