A red sphere of mass m and a blue sphere of mass 5m are attached to the ceiling by massless strings of identical length forming twin pendulums of length L. The red sphere is drawn to the left so that its center of mass has been raised a distance h and is then released. As the red sphere moves back through equilibrium, it collides with the blue sphere head-on.

a. If the two spheres adhere together during the collision, the combined system swings to the right and reach a maximum height, hmax, inelastic. Find this height in terms of h.
b. If the two spheres collide elastically, the two spheres will rebound in opposite directions. What is the maximum height of the blue sphere, hmax,elastic, in this case? How does it compare to your answer in part (a)?

Explanation:

a ) masses in inelastic collision are m and 5m . The potential energy of red sphere will be converted into kinetic energy before collision with blue sphere.

mgh = 1/2 mv² , v is velocity at the lowest point of red sphere.

v = √ 2gh

Let V be common velocity after collision . We shall apply law of conservation of momentum to calculate V .

mv = 6m x V

V = v / 6

If hmax be the height upto which both of them go after collision , the kinetic energy of both the  ball will be spent to generate potential of both the masses

1/2 x 6m x V² = 6m x g x hmax

hmax = V² / 2g

= v² / (36 x 2g)

= 2gh / 72 g

= h / 36

b ) Let v₁ and v₂ be the velocity of m and 5m after collision . m will go in opposite direction .

Applying law of conservation of momentum

m v = 5m v₂ - m v₁

v = 5 v₂ - v₁

Since collision is elastic ,

relative velocity of approach = relative velocity of separation

v = v₁ + v₂

from the two equation above

5 v₂ - v₁ = v₁ + v₂

4v₂ = 2 v₁

v₁ = 2v₂

putting this value in equation above

2v₂ + v₂ = v

v₂ = v / 3

if hmax is the height upto which blue ball goes

1/2 5m x v₂² = 5m x g x hmax

hmax = v₂² / 2g

= v² / 18g

= 2gh / 18g

= h / 9 Ans , so in this case blue ball will go to greater height.