.... We are told that all the subsequent collisions involving the balls and
floor are **elastic**. We are asked to determine the maximum height to
which the small sphere will rise on the rebound.

The problem does not mention any radii, but if we did know the radius
of each of the spheres, would it be valid to **bypass conservation of
linear momentum** calculations altogether and use a single conservation
of energy

It is a contradiction because an elastic collision is defined as one in which both KE and momentum are conserved. Only one equation is not enough to define the outcome of the collision.

Knowing the radii doesn't help, by the way, to solve any such problem, since the balls are usually considered of uniform density and the center of mass coincides with the center of gravity and with the point of application of all the vectors.

If you know the formulae, put any value you like, and solve for a particular example, unless they are asking for a general formula.

Since this is a homework question, nobody is allowed to give you more details.

First is the normal force exerted on the floor by the ball (indirectly due to gravity).

Gravity is not the only effect here. The ball is moving downward when it reaches the floor. In order to stop moving downward, the ball has to undergo an acceleration upward. This acceleration requires a force: $a = F/m$. The strength of this force and acceleration depends on the speed of the ball (how much acceleration is required) and the materials (how far the ball can deform).

It may be helpful to think of the ball as a spring. The force the ball exerts depends on the deformation, just like the force exerted by a spring depends on the deformation.

When the ball first touches the floor, the normal force is small, so the ball is still accelerating downward. After a certain amount of deformation, the ball is exerting a force equal to its weight. But this doesn't stop the ball, it just stops the acceleration. The ball continues to move downward.

As it does it compresses further. The normal force exceeds the weight and the ball begins to slow down. This continues until the acceleration removes all of the ball's downward velocity. The spring forces are still greater than gravity though, so the upward acceleration continues. This is sufficient to give it an upward velocity that allows it to bounce off the floor.

So, the normal reaction should always be more than the weight of a falling body, to account for that extra velocity that remains, even after the net force acting on the body is 0.

Well, at the instant the net force is zero, the normal force would be equal to the weight. But the downward velocity that remains will cause the ball to move closer to the floor and for the forces to increase with time.

For a stationary object kept on a floor, the normal force is exactly equal to weight, as there is no 'extra downward velocity'. Are these assertions correct ?

Right. The system is in equilibrium. The weight is exactly opposed by the normal force, so the object does not accelerate.

## Best Answer

OK, I think I know. I didn't pay special attention, but it's possible that at the lower bounce it had a higher horizontal speed, so that potential + kinetic energy still decrease.