Finding Density and Volume of a Submerged Ball Using Kinetic Energy and Buoyancy

In this lesson, you’ll learn how to determine a submerged ball’s density (ρ_ball) and volume (V) by analyzing how its kinetic energy (K) changes with the liquid density (ρ_liq). This combines Archimedes’ principle with the work–energy theorem using clear physical reasoning

What You’ll Learn

• Identify the forces acting on a submerged object: weight and buoyant force
• Relate net force and work to kinetic energy change
• Interpret a K vs. ρ_liq graph to find where the ball and fluid densities match
• Calculate the ball’s volume from the intercept on the K-axis
• Convert between units: g/cm³ ↔ kg/m³, cm ↔ m
• Apply buoyancy and energy concepts in experimental-style problems

Key Concepts Covered

• Archimedes’ principle (buoyant force F_b = ρ_liq V g)
• Work–energy theorem (W = ΔK)
• Net force in fluids: (ρ_ball − ρ_liq) V g
• Linear graph interpretation: slope, intercept, and zero crossing
• Neutral buoyancy (ρ_ball = ρ_liq)
• Unit conversions and physical reasonableness
• Data interpretation from graphs

Why This Lesson Matters
In AP, IB, and JEE physics, graph-based data problems are common. Understanding how to extract physical quantities like density and volume from simple linear relationships is an essential lab and exam skill — helping you interpret real experimental data with confidence.

Prerequisite or Follow-Up Lessons

• Before: Archimedes’ Principle & Buoyancy
• After: Work–Energy Theorem in Fluids

Full Lesson: Determining Density and Volume from K vs. ρ_liq Graph

1) Forces Acting on the Ball

When the small solid ball is fully submerged:

• Weight (downward): W = ρ_ball × V × g
• Buoyant force (upward): F_b = ρ_liq × V × g

Thus, the net force on the ball is:
F_net = (ρ_ball − ρ_liq) × V × g

The direction of this force is downward when the ball is denser than the liquid (ρ_ball > ρ_liq).

2) Work and Energy Relationship

The ball starts from rest (K_initial = 0) and moves a distance x = 4.0 cm = 0.040 m while submerged.

From the work–energy theorem:
Work done by the net force = Change in kinetic energy

So,
K = F_net × x
K = (ρ_ball − ρ_liq) × V × g × x

This equation shows a linear relationship between K and ρ_liq:

• Slope: −V × g × x
• Vertical intercept: K when ρ_liq = 0 → K(0) = ρ_ball × V × g × x
• Zero crossing (K = 0): occurs when ρ_liq = ρ_ball (neutral buoyancy)

(a) Finding the Density of the Ball

From the provided K vs. ρ_liq graph, the kinetic energy becomes zero when
ρ_liq = 1.5 g/cm³

This means at that point, buoyant force equals the ball’s weight (no net force, no acceleration).

Therefore,
Density of the ball:
ρ_ball = 1.5 g/cm³ = 1500 kg/m³

(b) Finding the Volume of the Ball

At ρ_liq = 0, the intercept on the graph corresponds to
K = K(0) = 1.60 J

From the equation:
K(0) = ρ_ball × V × g × x

Rearranging for V:
V = K(0) / (ρ_ball × g × x)

Now substitute the known values:
K(0) = 1.60 J
ρ_ball = 1500 kg/m³
g = 9.8 m/s²
x = 0.040 m

V = 1.60 / (1500 × 9.8 × 0.040)
V ≈ 2.72 × 10⁻³ m³

Volume of the ball: V ≈ 2.72 × 10⁻³ m³

3) Quick Sanity Check

If the ball is roughly spherical,
V = (4/3)πr³ → r ≈ 0.087 m or about 8.7 cm radius (17 cm diameter).
That’s a reasonable size for a small solid sphere.

4) Key Insights

• When K = 0, net force = 0 → ρ_ball = ρ_liq (neutral buoyancy point).
• When ρ_liq = 0, all buoyant effects vanish → K(0) depends only on the ball’s own weight.
• The straight-line relationship between K and ρ_liq makes it easy to read both the density and the volume from one plot.

Final Answers

• (a) Density of the ball: ρ_ball = 1.5 g/cm³ = 1500 kg/m³
• (b) Volume of the ball: V = 2.72 × 10⁻³ m³