Question

From an inclined plane two particles are projected with same speed at same angle $θ$, 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 the time of collision

Hint:

From an inclined plane two particles are projected with same speed at same angle $θ$, one up and other down the plane as shown in figure, which of the following statements is/are correct? \includegraphics[width=0.5\linewidth]6phy.png \labelfig:placeholder

Answer 1

The Correct Option is A

Step 1: Formula
Time of flight on an incline for 'up' is $T_{1}=\frac{2u \sin(\alpha-\beta)}{g \cos \beta}$ and for 'down' is $T_{2}=\frac{2u \sin \theta}{g \cos \theta}$.
Step 2: Analysis
Substituting $\alpha=2\theta$ and $\beta=\theta$, both yield $T = \frac{2u}{g} \tan \theta$.
Step 3: Collision
The relative acceleration is zero, and relative velocity is directed along the line joining them, leading to a mid-air collision.
Final Answer: (A)