The potential energy of a system of two masses varies inversely as the distance(r) between 1
them, i.e., `V(r)alpha 1//r`. When the two billard touch each other, P.E. Becomes zero i.e.,at r=R+R=2R, V(r) =0. Out of the given graphs, curve (v) only satisfies these two condition. Therefore, all other curves cannot possibly describe the eleastic collision of two billiard balls.

Correct Answer – C
The potential energy of a system of two masses varies inversely as the distance (r) between them.
i.e. `V(r)prop 1/r`
When the two billiard balls touch each other potential energy becomes zero, i.e at r=R+R=2R,V(r)=0
Out of the given graphs, only curve (c ) satisfies these conditions. Therefore, all other curves cannot possibly describe the elastic collision of two billiard balls.

The potential energy of a system of two masses varies inversely as the distance (r) between them i.e.

V(r) < \(\frac{1}{r} \) When the two balls touch each other, PE becomes zero i.e. at r – R + R = 2R , v (r) = 0. Out of the given graphs only the curve (V) satisfies these two conditions.

The potential energy of a system of two masses varies inveresly as the distance ® between them, i.e., `V(r)prop(1)/(r)`. When the twor billiard balls touch eachother,P.E. becomes zero i.e. at `r=R+R=2R,V(r )=0`. Out of the given curve (v) only satisfies these two conditions. Therefore, all other curves cannot possibly describe the elastic collision of two biliard balls.