If two balls have the same volume, but ball A has twice as much mass as ball B, which one will have the greater density?

Question

If ball C is 3 times the volume of ball D and ball D has 1/3 the mass of ball C, which has the greater density?

If two balls have the same mass, but ball P is twice as large as ball Q, which one will have the greater density?

If ball X is twice as big as ball Y and weighs only half as much as ball Y, then which one will have the greater density?

GPT-4o mini · Answer

To determine the density of each pair of balls, we can use the formula for density, which is given by:

\[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \]

Let the volume of both balls be \( V \).
Let the mass of Ball A be \( M_A = 2M_B \) (since Ball A has twice the mass of Ball B).

The densities are:

Since \( \text{Density}_A = \frac{2M_B}{V} \) and \( \text{Density}_B = \frac{M_B}{V} \), it follows that:

\[ \text{Density}_A = 2 \times \text{Density}_B \]

Conclusion: Ball A has the greater density.

Let the volume of Ball D be \( V_D \). Then, Ball C's volume is \( V_C = 3V_D \).
Let the mass of Ball C be \( M_C \) and the mass of Ball D be \( M_D = \frac{1}{3}M_C \).

The densities are:

For Ball C: \( \text{Density}_C = \frac{M_C}{3V_D} \)
For Ball D: \( \text{Density}_D = \frac{M_D}{V_D} = \frac{\frac{1}{3}M_C}{V_D} = \frac{M_C}{3V_D} \)

Since \( \text{Density}_C = \text{Density}_D \):

Conclusion: Balls C and D have the same density.

Let the mass of both balls be \( M \).
Let the volume of Ball Q be \( V_Q \). Then, volume of Ball P is \( V_P = 8V_Q \) (since Ball P is twice as large in each dimension, volume scales as \( 2^3 = 8 \)).

The densities are:

Since \( \text{Density}_P = \frac{M}{8V_Q} \) and \( \text{Density}_Q = \frac{M}{V_Q} \):

\[ \text{Density}_P = \frac{1}{8} \times \text{Density}_Q \]

Conclusion: Ball Q has the greater density.

Let the volume of Ball Y be \( V_Y \). Then, volume of Ball X is \( V_X = 8V_Y \) (as Ball X is twice as big in each dimension).
Let the mass of Ball Y be \( M_Y \). Then, the mass of Ball X is \( M_X = \frac{1}{2}M_Y \) (since Ball X weighs only half as much as Ball Y).

The densities are:

For Ball X: \( \text{Density}_X = \frac{M_X}{8V_Y} = \frac{\frac{1}{2}M_Y}{8V_Y} = \frac{M_Y}{16V_Y} \)
For Ball Y: \( \text{Density}_Y = \frac{M_Y}{V_Y} \)

Comparing the two:

\[ \text{Density}_X = \frac{1}{16} \times \text{Density}_Y \]

Conclusion: Ball Y has the greater density.

To summarize: