Question

Two particles $P_1$ and $P_2$ are projected simultaneously one with velocity $v$ (along the plane) upon a smooth inclined plane of inclination ${30}^0$ to the horizontal and the other with a velocity $\frac{2v}{\sqrt{3}}$ at an angle ${60}^0$. The two particles will have relative velocity zero, after a time $t$ seconds.
Match the following lists.

Answer 1

For partical $P_1$

Its initial velocity is v along the plane and acceleration is $-gSin\theta$ along the surface of plane only.

So its initial velocity is v and velocity after time t is $v-gSin30°t$ $=v-\frac{g}{2}t$

Particle 1 will always move along the plane of surafce only because its net acceleration and velocity are in opposite direction so direction on velocity will not change it will first decrease to zero and then will start incresing in downward direction along the plane. I.e. 30° with horizontal means velocity perpendicular to plane will be zero.

For particle $P_2$

Second particle have velocity at 60° with horizontal so when we want to find velocity along the plane and perpendicular to the plane breaks into two plane along the 30° with horizontal that will be $v_{2}Cos30°=\frac{2v}{\sqrt{3}}\times \frac{\sqrt{3}}{2}=v$ and similarly perpendicular to the plane will be $v_{2}Sin30°=\frac{v}{\sqrt{3}}$

So best option will be

1 goes to C

2 goes to D

3 goes to B

4 goes to A

5 goes to B