# Homework Help: Falling Pennies and Force

1. Jul 29, 2007

### jumbogala

1. The problem statement, all variables and given/known data
Okay, this isn't actually a problem that my teacher assigned, but just thinking about it is confusing me enough to make me confused about the whole course, haha.

Here it is: If you drop a penny one inch above your hand, it isn't going to hurt much. However, drop it 3000 feet above your hand and you'll be in a lot of pain. Why?

2. Relevant equations
Okay, I would assume it's because the penny hits your hand with a lot more force the second time, right? But... that can't be right if F=ma.

The penny's mass doesn't change depending on the height you drop it from, and neither does its acceleration. So the force must be equal both times, right?

3. The attempt at a solution
So it's not force that causes the penny to hit your hand harder the second time...? *confused*

2. Jul 29, 2007

### EugP

Think of it this way. What is acceleration? It's change of velocity over time. Well if you have a short period of time, like if you drop a penny one inch above your hand, you wont have a big change in velocity. But if you drop it 3000 feet above your hand, it will take it longer, meaning there will be more change in velocity. In other words, if it doesn't have to travel a large distance, it wont have time to gain a lot of speed.

I hope this helped.

3. Jul 29, 2007

### mgb_phys

I in f=ma for the penny hitting your hand 'a' is the de-acceleration between it touching your hand and coming to rest after pressing into your hand.
Now the time taken for the penny to be stopped by your hand is roughly the same, it depends on the property of your skin, the end velocity is zero (the penny is stopped. What is different about the velocity of the two pennies when they are just about to touch your hand ?

4. Jul 30, 2007

### jumbogala

But for an object falling, doesn't its acceleration remain constant throughout the entire fall?

For the penny falling a short distance, if inital velocity = 0, and final velocity = 10, and its acceleration is 9.81 m/s/s, then the time it takes to fall will be 0.98 s.

For the penny falling a long distance, inital velocity = 0 and final velocity = 50 (hypothetically only, just as long as it's bigger than 10, because it has more time to get up to a velocity that high). Its acceleration, however, remains the same at 9.81 m/s/s - it just takes longer to fall: 5.1 s this time.

Right? There is a bigger change in velocity, but because the time interval is also bigger, the acceleration remains the same.

I guess my question now is, why does speed affect the fact that the penny hits harder when dropped from a higher distance, since the formula f=ma doesn't account for speed?

The long distance penny has a higher velocity, so the deceleration of this same penny is higher... using f=ma also means that that force is bigger. I think I kind of get it now!

So why isn't it the accceleration of the falling penny that is used to calculate this, but its deceleration after it hits my hand?

5. Jul 30, 2007

### nicktacik

"So why isn't it the accceleration of the falling penny that is used to calculate this, but its deceleration after it hits my hand?"

Think of this. Let's say you're driving a car at a constant 30m/s and you hit a wall. If it was the "acceleration of the falling penny that is used to calculate the force", then in this analogous situation there would be no force when you hit the wall, since your acceleration is zero. However, once you hit the wall you will decelerate, and there is the force.

6. Jul 30, 2007

### turdferguson

Because what hurts is when you decide to stop the penny with your hand. Thats the force and acceleration were concerned with

7. Jul 30, 2007

### jumbogala

Ohhh okay, I get it now! Thanks to everyone that replied!