From an inclined plane two particles $P, Q$ are projected with same speed at same angle $θ$ , one up and other down the plane as shown in
figure. Which of the following statement(s) is/are correct ?

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

The particles will collide with the inclined plane with same speed

The times of flight of each particle are same

If the particles collide in the mid air after time t then , t $< \frac{P Q}{2 u \sin θ}$

If the particles collide in mid air after time t then , t $> \frac{P Q}{2 u \sin θ}$

answer is B, D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Time to flight $P$ is $T_{1} = \frac{2 u \sin θ}{g \cos θ} = \frac{2 u \tan θ}{g}$

and time to flight $Q$ is $T_{2} = \frac{2 u \sin θ}{g \cos θ} = \frac{2 u \tan θ}{g} ∴ T_{1} = T_{2}$

Further acceleration of both the particles is g downwards. Therefore, relative acceleration between the two is zero or relative motion between
the two is uniform. Now relative velocity of $P$ with respect to $Q$ is towards $P Q$ . Therefore, collision will take place between the two in mid air.

( $u \cos θ . t - \frac{1}{2} g \sin θ t^{2}) + (u \cos θ t + \frac{1}{2} g \sin θ t^{2}) > P Q$