question_answer

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 correct?

A)  The time of flight of each particle is the same

B)  The particels will collide the plane with same speed.

C)  Both the particles strike the plane perpendicular.

D)  the particles will not collide in mid air even if projected simultaneously and time of flight of each particle is greater then the time of collision.

Correct Answer: A

Solution :

Since their relative velocity is along the line joining them and relative acceleration is zero hence they collide when their time of flight is greater then time of collision. Here \[\beta =2\theta ,\,\,\beta =\theta \] 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 light 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 is zero or relative motion between the two is uniform. The relative velocity of A w.r.t. B is toward AB, therefore collision will take place between the two in mid air.