1. Motion before collision

The ball is dropped from rest, so during free fall its speed increases under gravity.

If it is dropped from height $h$, then before collision: $K = \frac{1}{2}mv^2 = mg\times (\text{distance fallen})$ So as the ball falls, its kinetic energy increases.

Also, since for free fall: $v = gt \quad \Rightarrow \quad K = \frac{1}{2}m g^2 t^2$ Hence, before touching the ground, $K$ increases parabolically upward with time.

2. During collision with the smooth horizontal surface

We are told that the contact force is proportional to compression $x$ of the ball: $F \propto x$ So the ball behaves like a spring during compression and restitution: $F = kx$ This means the interaction is elastic-like.

At the instant the ball first touches the ground, compression starts. The ball's kinetic energy is converted into elastic potential energy of deformation.

• As compression increases, speed decreases.
• At maximum compression, the ball momentarily comes to rest.
• Therefore, $K=0$ at maximum compression.

Because spring force varies linearly with compression, the compression-decompression motion is part of simple harmonic motion, so the velocity varies sinusoidally and kinetic energy varies smoothly.

Thus during collision, $K$ decreases smoothly to zero and then increases smoothly again.

Important: the slope of the $K$ vs $t$ graph changes continuously; there is no sharp corner if $F \propto x$.

3. After collision

The ball rises back to its original height, so just after losing contact it has the same speed as just before impact.

Then while moving upward against gravity, its speed decreases, so kinetic energy decreases.

Using upward motion under gravity: $v = u-gt$ Hence, $K = \frac{1}{2}m(u-gt)^2$ So after collision also, $K$ decreases parabolically with time until it becomes zero at the top.

4. Overall shape of $K$ vs $t$

So the graph must have these features:

1. Starts from $K=0$.
2. Increases parabolically during fall.
3. During collision, decreases smoothly to $0$ at maximum compression.
4. Then increases smoothly again during restitution.
5. After leaving the surface, decreases parabolically back to $0$ at the original height.
6. Since the ball returns to the original position, the graph is symmetric in energy scale before and after collision.

The correct sketch is the one showing:

• two outer parabolic parts,
• a smooth dip to zero during contact,
• no discontinuity.

This corresponds to Option B.

5. Comparison with stored answer

Stored correct answer: B

Derived answer: B

So the derived answer agrees with the stored answer.