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Suppose a thermocol ball and a metal ball of same volume and surface area (but different masses, obviously) are dropped from the same height from rest. The acceleration due to gravity is 'g' and the air resistance is also same in both the cases, then why is it that the metal ball reaches the ground first?

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    $\begingroup$ What is a " thermocol ball"? $\endgroup$
    – mike stone
    Nov 21, 2022 at 13:49
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    $\begingroup$ @mikestone I believe is a brand of polystyrene $\endgroup$
    – Mauricio
    Nov 21, 2022 at 14:07
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    $\begingroup$ @VaiMan you are confusing force with acceleration. The forces are equal, the acceleration is not. $\endgroup$
    – Boson
    Nov 21, 2022 at 17:02

2 Answers 2

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Let $R$ be the resistive force.

Applying Newton's second law $mg-R = ma$ where $a$ is the acceleration of the ball.

Thus $a= g-\frac Rm$.

Assuming that the density of the metal ball is greater than that of the thermocol (made of expanded polystyrene beads) ball then the mass of the metal ball is greater than that of the thermocol ball, which in turn means that the second term on the right-hand side of the equation, $\frac Rm$, is smaller for a metal ball, and so the acceleration of the metal ball is greater this its time of travel is smaller than the thermocol ball.

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I offer a nearly-zero-mathmatics answer:

The acceleration due to gravity is $g$

Here is your misconception. $g$ is the acceleration due to gravity in free fall. When there is air resistance or other forces involved, you don't have perfect free fall, in fact you may have nothing like free fall, for example if the object is buoyant.

However, $g$ is still a really handy number to have lying around. Because we can use it in formulas like

$$F = mg$$

Which gives us the force due to gravity, AKA the weight. And that's pretty convenient when you consider that g has approximately the same value at all practical altitudes anywhere on earth - depending how precise you need to be.

The reason that $g$ is the same for all objects is that, if you double the mass of an object, you double the amount of force needed to accelerate it, but you also double the force due to gravity, so these two doublings cancel out and the resultant acceleration is the same. $F=mg$ is simply derived from $F=ma$ where $g$ represents a constant value of $a$ if there are no other forces present.

The ball with the higher mass will have the higher weight, weight is the force which is overcoming the air resistance, so it will accelerate more rapidly. It will also reach a higher terminal velocity.

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  • $\begingroup$ Thanks for the edit. It's about time I learnt to use Mathjax ;) $\endgroup$
    – Rodney
    Nov 23, 2022 at 19:24

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