Question

Particle A is released from a point P on a smooth inclined plane inclined at an angle $\alpha$ with the horizontal. At the same instant another particle B is projected with initial velocity $u$ making an angle $\beta$ with the horizontal. Both the particles meet again on the inclined plane. Find the relation between $\alpha$ and $\beta$

• A. $\alpha +\beta ={45}^{\circ }$
• B. $\alpha +\beta ={30}^{\circ }$
• C. $\alpha +\beta ={90}^{\circ }$
• D. $\alpha -\beta ={30}^{\circ }$

Answer 1

The correct option is C $\alpha +\beta ={90}^{\circ }$


Consider motion of B along the plane, initial velocity $=u\cos (\alpha +\beta )$ and acceleration $=g\sin \alpha$
Using second equation of motion along the inclined plane,
$\begin{array}{}\therefore OP=u\cos (\alpha +\beta t)+\frac{1}{2}g\sin \alpha t^2& \text{(1)}\end{array}$
For motion of particle A along the plane,
Initial velocity = 0
Acceleration $=g\sin \alpha$
Using second equation of motion along the inclined plane,
$\begin{array}{}\therefore OP=\frac{1}{2}g\sin \alpha t^2& \text{(2)}\end{array}$
From Equation (1) and (2), we get
$u\cos (\alpha +\beta )t=0$
so, either $t=0$ or $\alpha +\beta =\frac{\pi }{2}$
Thus, the condition for the particles to collide is $\alpha +\beta =\frac{\pi }{2}$