Question

P Two sticky spheres are suspended from light ropes of length L that are attached to the ceiling at a common point. Sphere AA has mass 2m and is hanging at rest with its rope vertical. Sphere BB has mass mm and is held so that its rope makes an angle with the vertical that puts B a vertical height HH above A. Sphere B is released from rest and swings down, collides with sphere A, and sticks to it. In terms of H,H, what is the maximum height above the original position of A reached by the combined spheres after their collision?

Answer 1

The maximum height above the original position of A reached by the combined spheres after their collision can be determined using conservation of energy.

Step-by-step explanation:

The maximum height above the original position of A reached by the combined spheres after their collision can be determined using conservation of energy.

Before the collision, the potential energy of B was mgh, where m is the mass of B, g is the acceleration due to gravity, and h is the vertical height of B above A.

After the collision, when the spheres reach their maximum height, all of the potential energy is converted into kinetic energy. The kinetic energy of the combined spheres is given by (m + 2m)v^2/2, where v is the velocity of the combined spheres at that height.

Since energy is conserved, we can equate the initial potential energy to the final kinetic energy: mgh = (m + 2m)v^2/2. Solving for v and substituting h with H, we get v = sqrt(2gh/(1+2m/m)).

Therefore, the maximum height above the original position of A reached by the combined spheres after their collision is H + v^2/(2g), where v is the velocity of the combined spheres and g is the acceleration due to gravity.