# Question #14068

Sep 17, 2017

Because large mass is harder to accelerate.

#### Explanation:

Think about kicking a football. You can give it a large change in velocity - therefore a large acceleration. Now think about kicking a heavy bowling ball with the same kicking motion and effort. The bowling ball will not move off with near the same velocity. Why?

Because a larger mass is harder to accelerate. From that fact, I might argue that a more massive object should fall more slowly. The more massive object has more inertia, therefore is more sluggish.

Actually the two affects of the greater mass cancel each other out so they fall with equal acceleration.

1. Because of the greater mass, the more massive one has a greater force of gravity.
2. But also, because of the greater mass, the more massive one is not accelerated as easily.

Effect number 1 and effect number 2 cancel each other out.

The result is that in vacuum, if you drop a feather and a hammer simultaneously, they fall at the same rate. This was demonstrated by Commander David Scott on the moon.
https://nssdc.gsfc.nasa.gov/planetary/lunar/apollo_15_feather_drop.html

I hope this helps,
Steve

Sep 17, 2017

Mathematically-speaking, the mass cancels out in the equation of motion if we assume that gravity is the only force acting (so no air resistance).

#### Explanation:

It can be experimentally observed (in a vacuum) that the acceleration due to gravity near the surface of the Earth is about $g = 9.8$ meters per second per second (in saying this I am implicitly assuming that the acceleration is independent of the object...so I'm kind of "cheating"...but I thought it was worth it so that you actually see the mass canceling out in the equation below).

If $m$ is the mass of the object, in kg, then the product $m \cdot g$ is the gravitational force on the object (in Newtons). This can also be thought of as the weight of the object.

If this is the only force acting (no air resistance), then Newton's 2nd Law says that this equals the product of the mass $m$ and the (downward) acceleration $a$. In other words, $m \cdot g = m \cdot a$. But now the value of $m$ cancels out, leaving a constant acceleration $a = g$, independent of the mass.

Oct 16, 2017

There's a great video demonstrating this effect spectacularly here: