Question

A particle of mass $m$ when projected with velocity $u$ at an angle $\theta$ to the horizontal has a range $R$. There are two walls at O and A separated by a distance $\alpha$. Now the particle is project from O with velocity $u$ and angle of projection $\theta$. Where will the particle land? Choose the correct option. (All the collisions are perfectly elastic.)

• A. If $\alpha =\frac{R}{2}$, it will retrace its path and land at O.
• B. If $\alpha =\frac{R}{2}$, it will land at A.
• C. If $\alpha =\frac{R}{3}$, it will land at O.
• D. If $\alpha =\frac{R}{3}$, it will land at $\frac{R}{6}$ distance from O.

Answer 1

Answer: (A)

If $\alpha =\frac{R}{2}$, it will retrace its path and land at O.

As all the collisions are perfectly elastic, the total horizontal distance covered by the projectile will be equal to its range, R.

If $\alpha =\frac{R}{2}$:

Then, after the ball hits the wall A at A${}_1$, it is at its maximum height,

where the vertical velocity $=v_y=0$

and the horizontal velocity gets reversed in direction because the collision is perfectly elastic, so it will retrace its path and land at O.

If $\alpha =\frac{R}{3}$:

When the ball collides with wall A at A${}_2$, the horizontal component of velocity will be reversed due to perfectly elastic collision but the vertical component will remain same, so it will follow path A${}_2$PO${}_2$ and hit the wall O at $O_2$. Now, at O${}_2$, again it will undergo perfectly elastic collision with the wall O and the horizontal component of its velocity will be reversed while the vertical component will remain unchanged, and it will finally land at A.