# Why is acceleration inversely proportional to mass?

Apr 28, 2018

acceleration equals to the force applied divided by mass

#### Explanation:

an object moving at a velocity of x carries the force of its mass times its speed.

when you apply a force onto an object, the increase in speed of it would be affected by its mass. Think of it this way: you apply some force onto an iron ball, and apply the same force on a plastic ball (they are of equal volume). Which one moves faster, and which one moves slower? The answer is obvious: the iron ball will accelerate slower and travel slower, while the plastic ball is faster.

The iron ball has a greater mass, so the force which makes it accelerate is deduced more. The plastic ball has a smaller mass, so the force applied is divided by a smaller number.

I hope this helps you a bit.

Apr 28, 2018

Assuming we're using $F = m a$, then it's because, when one goes up, the other must go down in order to keep the equation balanced.

#### Explanation:

Say we wish to keep a force $F$ exerted by an object constant. If the mass $m$ of the object doubles, what must happen to the object's acceleration $a$ to keep $F$ unchanged?

The answer is: the object's acceleration must be halved.

$F = m \cdot a$

and if we double the mass to $2 m$, the RHS as a whole has doubled. Thus, the LHS also doubles, meaning we get double the force:

$2 F = 2 m \cdot a$

This is an example of direct proportionality between $F$ and $m$. If $m$ doubles, $F$ responds by doubling as well.

But we want to keep the force the same; we don't want $2 F$, we want $F$. So we need to divide the LHS by 2. And to do that, we must divide the RHS by 2 as well. So either the mass $2 m$ goes back down to $m$, or the acceleration $a$ gets cut to $\frac{1}{2} a$.

$F = 2 m \cdot \frac{1}{2} a$

This is an example of inverse proportionality. When the force is taken as a constant, if mass doubles, acceleration must be halved.

### Note:

You can also see the inverse relation between $m$ and $a$ by solving $F = m a$ for one or the other.

$F = m a \text{ "=>" "a=F/m" " <=>" } a = F \left({m}^{-} 1\right)$

$\textcolor{w h i t e}{F = m a} \text{ "=>" "m=F/a" "<=>" } m = F \left({a}^{-} 1\right)$

It's now easy to see mathematically that $a$ and $m$ are inversely proportional, because each is a multiple of the other's inverse (that multiple being $F$ itself).