Ball dropped from height h bounces back to h/2 — find fraction of KE lost

Question

A ball is dropped from height $h$ and bounces back to a height $h/2$ . Find the fraction of kinetic energy lost during the collision with the ground.

Solution — Step by Step

The ball is dropped from rest at height $h$ .

Using energy conservation (from top to just before hitting ground):

KE_{before} = mgh

(All potential energy converts to kinetic energy; take the ground as zero PE level.)

Alternatively, using kinematics: $v_{before}^2 = 2gh$ , so $KE_{before} = \frac{1}{2}mv_{before}^2 = mgh$ .

The ball bounces back to height $h/2$ . At this maximum height after the bounce, all kinetic energy has converted back to potential energy.

Using energy conservation (from just after bounce to maximum height $h/2$ ):

KE_{after} = mg \cdot \frac{h}{2} = \frac{mgh}{2}

(The kinetic energy just after the bounce equals the potential energy at the maximum bounce height.)

The kinetic energy lost during the collision:

\Delta KE = KE_{before} - KE_{after} = mgh - \frac{mgh}{2} = \frac{mgh}{2}

The fraction of KE lost:

\text{Fraction lost} = \frac{\Delta KE}{KE_{before}} = \frac{mgh/2}{mgh} = \frac{1}{2}

\boxed{\text{Fraction of KE lost} = \frac{1}{2} = 50\%}

Why This Works

During the collision with the ground, kinetic energy is lost as sound, heat, and deformation of the ball. The ground collision is inelastic (not perfectly elastic — a perfectly elastic collision would return the ball to the original height $h$ ).

The initial KE just before impact = $mgh$ (all potential energy converted). The KE just after impact = $mg(h/2)$ (only enough to reach $h/2$ ).

The ratio of KE after to KE before = 1/2. So half the kinetic energy was lost in the bounce.

This is directly related to the coefficient of restitution $e$ :

e = \frac{v_{after}}{v_{before}} = \sqrt{\frac{h_{bounce}}{h_{drop}}} = \sqrt{\frac{h/2}{h}} = \frac{1}{\sqrt{2}}

Energy lost fraction = $1 - e^2 = 1 - \frac{1}{2} = \frac{1}{2}$ .

Alternative Method

Generalise: if a ball drops from $h$ and bounces to $h'$ :

e = \sqrt{h'/h}, \quad \text{Fraction of KE lost} = 1 - e^2 = 1 - \frac{h'}{h}

For $h' = h/2$ : fraction lost = $1 - 1/2 = 1/2$ .

This formula is worth memorising for JEE: fraction of KE lost = $1 - h'/h$ where $h'$ is the height after bounce.

Common Mistake

A common error is calculating the KE just BEFORE and AFTER the bounce as if they are equal (claiming no energy is lost). The energy is lost DURING the collision, not during free flight. The ball’s KE at the moment it leaves the ground is LESS than its KE at the moment it hits — this difference is the lost energy. Between drops (from $h$ to ground) and rises (from ground to $h/2$ ), energy conservation holds; the loss happens only during the ground contact.

For JEE, the coefficient of restitution $e = \sqrt{h'/h}$ allows you to predict successive bounce heights: $h_1 = e^2 h$ , $h_2 = e^4 h$ , etc. The total distance travelled is a geometric series: $d = h + 2h(e^2 + e^4 + ...) = h \cdot \frac{1+e^2}{1-e^2}$ . This is a standard JEE series-sum problem combined with physics.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

Try These Next