Question

Answer the following question.

From the terrace of a building of height H, you dropped a ball of mass m. It reached the ground with speed v. Is the relation mgh = `1/2"mv"^2` applicable exactly? If not, how can you account for the difference? Will the ball bounce to the same height from where it was dropped?

Answer 1

1. Let the ball dropped from the terrace of a building of height h have mass m. During free fall, the ball is acted upon by gravity (accelerating conservative force).
2. While coming down, the work that is done is equal to the decrease in potential energy.
3. This work done however is not entirely converted into kinetic energy but some part of it is used in overcoming the air resistance (retarding non-conservative force). This part of energy appears in some other forms such as heat, sound, etc.
4. Thus, in this case of an accelerating conservative force along with a retarding non-conservative force, the work-energy theorem is given as,
   Decrease in the gravitational
   P.E. = Increase in the kinetic energy + work done against non-conservative forces.
5. Thus, the relation mgh = `1/2"mv"^2` is not applicable when non-conservative forces are considered. The part of the energy converted to heat, sound, etc also needs to be added to the equation.
6. The ball will not bounce to the same height from where it was dropped due to the loss in kinetic energy during the collision making it an inelastic collision.