A tennis ball is dropped on a horizontal smooth surface. It bounces back to its original position af

Q: Why isn't the first part of the graph a straight line? Doesn't velocity increase linearly?

You are absolutely right that velocity increases linearly ($v = gt$). However, the graph is plotting **Kinetic Energy**, not velocity. Since $K = \frac{1}{2}mv^2$, we have to square the velocity. Squaring a linear function ($t$) gives a quadratic function ($t^2$), which graphs as a parabola, not a straight line.

Q: How do we know if the parabola opens upwards or downwards?

Look at the equation $K = \frac{1}{2}mg^2t^2$. The coefficient in front of $t^2$ is $\frac{1}{2}mg^2$, which is a positive number. In mathematics, a quadratic equation $y = ax^2$ opens upwards if 'a' is positive. Physically, it means the rate at which kinetic energy is increasing is itself increasing.

Q: Why can't the collision part just be a sharp 'V' shape dropping to zero?

A sharp 'V' shape would imply that the acceleration (and thus the force) is constant during the collision. But the problem explicitly states the force is proportional to compression ($F = -kx$). This varying force causes Simple Harmonic Motion, where velocity changes sinusoidally. The square of a sine wave is a smooth curve, not a sharp point.

Q: Wait, if the ball stops at maximum compression, shouldn't the graph stay at zero for a moment?

The ball only stops for an infinitesimally small instant in time—exactly at the point of maximum compression. It doesn't rest there. Because the spring force is at its absolute maximum when the ball is stopped, it immediately accelerates the ball back upwards. Therefore, the kinetic energy touches zero and immediately starts rising again.

Q: Does the graph look the same after the bounce?

Yes, the shape is the same (a parabola), but reversed in time. After the bounce, $v = v_0 - gt$. The kinetic energy is $K = \frac{1}{2}m(v_0 - gt)^2$. This is still a quadratic equation with a positive coefficient for the $t^2$ term, so it's still an upward-opening parabola, but this time the value of K is decreasing as the ball goes higher.

Question

A tennis ball is dropped on a horizontal smooth surface. It bounces back to its original position after hitting the surface. The force on the ball during the collision is proportional to the length of compression of the ball. Which one of the following sketches describes the variation of its kinetic energy $K$ with time $t$ most appropriately? The figures are only illustrative and not to the scale.

Sigma Percentile · Accepted Answer

### The Physics of a Bouncing Ball

Have you ever watched a tennis ball bounce in slow motion? To the naked eye, it seems like an instantaneous event—the ball hits the ground and immediately shoots back up. But physics tells us a much richer, more intricate story. In this problem, we are tasked with plotting the kinetic energy of a bouncing ball as a function of time. To do this accurately, we must break the ball's journey into three distinct phases: the free fall, the collision, and the rebound.

### Phase 1: The Free Fall

Imagine the ball the moment it is released from your hand. It begins its descent under the sole influence of gravity. This is a classic case of free fall. According to the first equation of kinematics, the velocity $v$ of the ball at any time $t$ is given by:

$$v = gt$$

Now, we want to find the kinetic energy $K$. The formula for kinetic energy is $K = \frac{1}{2}mv^2$. Substituting our expression for velocity into this equation, we get:

$$K = \frac{1}{2}m(gt)^2 = \frac{1}{2}mg^2t^2$$

Look closely at this relationship. The mass $m$ and the acceleration due to gravity $g$ are constants. Therefore, the kinetic energy is directly proportional to the square of time ($K \propto t^2$). In mathematics, a relationship where the dependent variable is proportional to the square of the independent variable represents a **parabola**. Because the coefficient is positive, this parabola opens upwards. As the ball falls, its kinetic energy doesn't just increase; it increases at an accelerating rate. This immediately tells us that the first segment of our graph must be a curve that is concave upwards.

### Phase 2: The Collision (The Hidden Spring)

Here is where the problem gets truly fascinating. The ball hits the ground. But it doesn't just magically reverse direction. The problem gives us a crucial piece of information: **"The force on the ball during the collision is proportional to the length of compression of the ball."**

Does that sound familiar? It should! A force proportional to displacement is the exact definition of **Hooke's Law** ($F = -kx$). During the brief moment of impact, the tennis ball acts exactly like a spring. It compresses, storing its kinetic energy as elastic potential energy, comes to a momentary halt, and then expands, converting that potential energy back into kinetic energy.

Because the ball is obeying Hooke's Law, its motion during the collision is **Simple Harmonic Motion (SHM)**. In SHM, the velocity varies sinusoidally with time, typically modeled as $v(t) = v_0 \cos(\omega t)$.

If we plug this sinusoidal velocity into our kinetic energy formula, we get:

$$K = \frac{1}{2}mv_0^2 \cos^2(\omega t)$$

This $\cos^2$ function creates a very specific shape on our graph. It is a smooth, bell-like curve. The kinetic energy smoothly drops to zero at the point of maximum compression and then smoothly rises back up as the ball expands. It is **not** a sharp, jagged V-shape or a triangle. The transition is mathematically smooth.

### Phase 3: The Rebound

Finally, the ball loses contact with the ground and begins its upward journey. Once again, it is in free fall, but this time it has an initial upward velocity $v_0$ and gravity is acting against its motion. Its velocity at any time $t$ after the bounce is:

$$v = v_0 - gt$$

Squaring this to find the kinetic energy, we get:

$$K = \frac{1}{2}m(v_0 - gt)^2$$

Once again, we have a quadratic relationship with time. The graph will be another parabola opening upwards. As the ball climbs higher, its velocity decreases, and its kinetic energy follows this parabolic curve back down to zero at the apex of its flight.

### Conclusion

By carefully analyzing the physics of each phase, we've deduced the exact shape of the kinetic energy graph. It must start as an upward-opening parabola, transition into a smooth, bell-shaped dip to zero during the collision, and finish as another upward-opening parabola. Matching this physical reality to the given options, we find that only one sketch perfectly captures this beautiful sequence of mathematical curves.

Select Answer:

Visualized Solution

Visualizing the Setup

Phase 1: Free Fall Kinematics

Kinetic Energy during Free Fall

Shape of the Free Fall Graph

Phase 2: The Collision

Force during Collision

Kinetic Energy during Collision

Phase 3: The Bounce

Shape of the Bounce Graph

The Sigma Insight: Kinetic Energy, Potential Energy and Power

The Physics of a Bouncing Ball

Phase 1

Phase 2

Phase 3

Conclusion

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