A ball of density d is dropped on to a horizontal solid surface. It bounces elastically from the sur

Q: Why is the time of fall in air exactly $t_1/2$ for the solid surface case?

Because the collision with the solid surface is perfectly elastic and instantaneous, there is no energy loss. The upward motion is a perfect time-reversal of the downward motion, making the fall time exactly half of the total round-trip time.

Q: Does the buoyant force change as the ball sinks deeper into the liquid?

No, because the ball is fully submerged. The buoyant force depends only on the volume of the submerged object and the density of the liquid ($F_B = V d_L g$), both of which remain constant once the ball is completely inside the liquid.

Q: Why does the volume of the ball not matter in the final time expression?

Both the mass of the ball ($m = Vd$) and the net force acting on it ($F_{\text{net}} = V(d_L - d)g$) are directly proportional to the volume $V$. When calculating acceleration ($a = F/m$), the volume terms cancel out completely.

Q: Why is the motion not considered Simple Harmonic Motion (SHM)?

In SHM, the restoring force must be proportional to the displacement ($F \propto -x$). Here, the force is constant in each medium (gravity in air, net buoyancy in liquid) and changes step-wise at the boundary, which violates the definition of SHM.

Q: What happens to the ball in part (c) if the liquid has a bottom?

Since $d = d_L$, the net force inside the liquid is zero, so the ball moves at a constant speed. It will continue to sink at this constant speed until it hits the bottom of the container.

Question

A ball of density $d$ is dropped on to a horizontal solid surface. It bounces elastically from the surface and returns to its original position in a time $t_1$. Next, the ball is released and it falls through the same height before striking the surface of a liquid of density $d_L$.

(a) If $d < d_L$, obtain an expression (in terms of $d$, $t_1$ and $d_L$) for the time $t_2$ the ball takes to come back to the position from which it was released.

(b) Is the motion of the ball simple harmonic?

(c) If $d = d_L$, how does the speed of the ball depend on its depth inside the liquid? Neglect all frictional and other dissipative forces. Assume the depth of the liquid to be large.

Sigma Percentile · Accepted Answer

### The Setup: A Tale of Two Media

Imagine dropping a ball from a height $H$ above a surface. In the first scenario, the surface is a solid floor. The ball falls under gravity, hits the floor, and bounces back up elastically. Because the collision is perfectly elastic, no kinetic energy is lost. The trip down is a perfect mirror image of the trip up.

If the total round-trip time is $t_1$, then by symmetry, the time of fall through the air is exactly half of that:

$$t_{	ext{fall}} = \frac{t_1}{2}$$

Using the standard kinematic equation $v = u + gt$, and knowing the ball starts from rest ($u=0$), we can find the velocity $v$ of the ball just as it strikes the surface:

$$v = g \left(\frac{t_1}{2}ight)$$

Now, let's change the game. Instead of a solid floor, we place a deep pool of liquid of density $d_L$ at the same height. The ball is released from the same height $H$. It falls through the air, taking the same time $t_{	ext{fall}} = t_1/2$, and strikes the liquid surface with the exact same velocity $v$.

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### Part (a): Navigating the Liquid Boundary

Once the ball enters the liquid, it experiences two competing forces:

1. **Gravity** pulling it down: $W = Vdg$
2. **Buoyancy** pushing it up: $F_B = Vd_Lg$

Since the density of the ball $d$ is less than the density of the liquid $d_L$, the upward buoyant force is greater than the downward gravitational force. This creates a net upward force, which acts as a constant retardation (deceleration) $a$:

$$F_{	ext{net}} = F_B - W = V(d_L - d)g$$

Using Newton's second law ($F = ma$), we find the retardation $a$:

$$a = \frac{F_{	ext{net}}}{m} = \frac{V(d_L - d)g}{Vd} = \left(\frac{d_L - d}{d}ight)g$$

Notice how the volume $V$ cancels out beautifully! The retardation depends purely on the ratio of the densities and gravity.

Inside the liquid, the ball slows down from its initial striking velocity $v$ to a complete stop ($v_f = 0$). The time $t$ taken to come to rest is:

$$t = \frac{v}{a} = \frac{g(t_1/2)}{\left(\frac{d_L - d}{d}ight)g} = \frac{dt_1}{2(d_L - d)}$$

Once the ball stops, the net upward force accelerates it back up to the surface. By symmetry, the time to rise back to the surface is also $t$, and it will emerge from the liquid with the same speed $v$. It then flies up through the air, taking another $t_1/2$ to reach its original release height.

Thus, the total round-trip time $t_2$ is the sum of the times spent in air and liquid:

$$t_2 = t_{	ext{fall}} + t_{	ext{liquid, down}} + t_{	ext{liquid, up}} + t_{	ext{rise}}$$

$$t_2 = \frac{t_1}{2} + t + t + \frac{t_1}{2} = t_1 + 2t$$

Substituting our expression for $t$:

$$t_2 = t_1 + 2\left[ \frac{dt_1}{2(d_L - d)} ight] = t_1 \left[ 1 + \frac{d}{d_L - d} ight]$$

Simplifying the fraction inside the bracket:

$$t_2 = \frac{t_1 d_L}{d_L - d}$$

This is our final elegant expression for the total time $t_2$.

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### Part (b): The Nature of the Oscillation

Is this motion Simple Harmonic Motion (SHM)?

For a motion to be SHM, the acceleration must be directly proportional to the displacement from a fixed equilibrium position ($a \propto -x$) at every instant.

In this system, the acceleration is constant in magnitude within each medium:
- In air: $a = -g$ (downward)
- In liquid: $a = +\left(\frac{d_L - d}{d}ight)g$ (upward)

The acceleration changes abruptly at the boundary ($y = 0$) and does not vary linearly with displacement. Therefore, while the motion is periodic, it is **NOT simple harmonic**.

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### Part (c): The Perfect Balance

What happens if the density of the ball is exactly equal to the density of the liquid ($d = d_L$)?

Let's look at our retardation formula:

$$a = \left(\frac{d_L - d}{d}ight)g = 0$$

Because the densities are equal, the buoyant force exactly balances the weight of the ball. The net force on the ball inside the liquid is zero!

According to Newton's first law, an object in motion with no net force acting on it will continue to move at a constant velocity. Therefore, the ball will not slow down; it will continue to sink deeper into the liquid at a **constant speed** equal to its striking velocity:

$$v = \frac{gt_1}{2}$$

Visualized Solution

Analyzing the Elastic Bounce on Solid Surface

Calculating the Velocity at the Liquid Surface

Analyzing Forces Inside the Liquid

Deriving Net Retardation $a$

Calculating Penetration Time $t$

Total Round-Trip Time $t_{2}$

Simplifying the Expression for $t_{2}$

Analyzing Simple Harmonic Motion (Part b)

Analyzing the Case $d = d_{L}$ (Part c)

The Sigma Insight: Buoyancy and Archimedes' Principle

The Setup

Part (a)

Part (b)

Part (c)

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