Question

A large heavy sphere and a small light sphere are dropped together onto a flat surface from a height $h$. The radius of spheres is much smaller than the height $h$. The large sphere collides with the surface with velocity $V_0$ and immediately thereafter with the small sphere. The spheres are dropped so that all motion is vertical before the second collision, and the small sphere hits the larger sphere at an angle from its uppermost point as shown in the diagram. All collisions are perfectly elastic and there is no surface friction between the spheres. The angle is made by the velocity vector of the small sphere with the vertical just after the second collision in the frame of the large sphere.

• A. $\alpha$
• B. $2\alpha$
• C. Zero
• D. $\frac{3\alpha }{2}$

Answer 1

The correct option is A

$\alpha$

Step 1: Given Data

The height from which the balls are dropped is $h$.

The velocity at which the larger sphere collides with the ground is $V_0$.

Since the collisions are elastic, the velocity at which the larger sphere hits the smaller sphere is also $V_0$.

Let $V_2$ be the velocity after the collision.

From the figure, $\alpha$ is the angle between the line connecting the center of the two spheres and the vertical.

Step 2: Solution

When the two objects collide, the force acting on them is along the line connecting their center of mass.

The center of mass of a sphere is at the center of its symmetrical axis i.e., its center.

So the force on the smaller object is along the line joining centers of the spheres.

Therefore, the acceleration it gains and its subsequent velocity vector is also on the same line.

We know that the angle between the line joining the centers of the spheres, from the figure, is $\alpha$.

Therefore, the angle made by the velocity vector of the small sphere with the vertical just after the second collision in the frame of the large sphere is $\alpha$.

Hence, option A is correct.