# Gravitational Energy transfered to Kinetic question

1. Jul 15, 2009

### rayfish

1. The problem statement, all variables and given/known data
So this was a long question recently given on my exam, and I had only a slight idea of how to do it.
A block of wood is 1.05m above the ground, and theres a ledge supporting it, and its just sitting there. A pendulum that weighs 200g. When it is pulled back, it is 0.73m above the ledge, and when it is pulled back, and released, it hits the block. The pendulum continues after hitting the block to a point where it is 0.25m above the ledge. The block then travels a certain distance (data is displayed below)

This is given information
Height of Pendulum above ledge : 0.73m
Mass of Pendulum: 0.2kg
Mass of Block: 0.085kg
Height of Block above the ground: 1.05m
Height of Pendulum after hitting the block at max height: 0.25m

There is also given data results:
1.7m
1.65
1.57
1.51
1.48
1.47
1.62

And the question was to determine an equation to predict what will happen and compare to the observations that were given. The results should have slight variations, but stay roughly similar. We were told to ignore all friction, and that it is a closed system.

2. Relevant equations
Gravitational Energy: mgh
Kinetic Energy: 0.5mv^2
Conservation of mass

3. The attempt at a solution]
http://img107.imageshack.us/i/ccf1507200900001.jpg/

2. Jul 16, 2009

### LowlyPion

Welcome to PF.

Well what happens at impact?

You have a collision of some sort, and the pendulum regains a specific height below the original release point. As you show this gives you a change in energy, but I think you have calculated the wrong change. The pendulum starts at .73m above and ends .25m higher. That translates into a loss of potential energy of (.200)*(9.8)*(.73 - .25) = .94J. Now you know that energy had to go somewhere.

If it is an elastic collision then it all goes into the block as kinetic energy, which in this case would be some horizontal velocity of the mass of the block after impact. From that you predict through kinematics the horizontal distance it will travel from the ledge at impact 1.05 m below.

And if it is not elastic, then results may vary.

3. Jul 16, 2009

### cepheid

Staff Emeritus
That's what I thought too, but if so, what are the heights that the OP listed supposed to be, and why do they start higher than 1.05 m?

4. Jul 16, 2009

### rayfish

The 1.05 is the block that is originally sitting on a ledge 1.05m above the ground.
I've worked on it a bit more, and it seems like no pendulum knowledge is required.

The .73 Height represents what energy is there to begin with. The .25 is to represent what is left, and subtracting the two will give you Kinetic energy transferred to the block.

By knowing all this information, how do I derive a formula that will be able to represent the input + data given? Will I have to create a constant? This is the part where I'm most confused about. Thanks!

p.s. I forgot to add, the teacher told us to assume that it is a closed system, and that no energy is lost. So during the impact between the block and pendulum, we are supposed to assume complete energy transfer, and no energy loss.

Last edited: Jul 16, 2009
5. Jul 16, 2009

### cepheid

Staff Emeritus
Once the block is launched, isn't it just a projectile motion problem?

Can you explain more clearly what your data are? You have a dependent variable, but no independent variable.

6. Jul 16, 2009

### rayfish

Yep it becomes a projectile motion. My teacher was asking the class, to determine a formula that will be able to roughly calculate the same data as he gave us. The independent variable is the height the pendulum is dropped above the ledge. In this case, he performed a lab where he roughly dropped it at 0.73m above the ledge, and the data displayed is what he received. Our job was to determine a formula, where we can input height and have an output of distance traveled, theoretically.

The data that I gave is the distance in meters that the block of wood traveled after it was launched. This is the x distance.

7. Jul 16, 2009

### cepheid

Staff Emeritus
Okay. And each one corresponds to a different trial? You see that's what I meant by independent variable. It wasn't clear why there were multiple values -- what they were a *function* of. It could have been time, for all I knew. But now I know that what is being varied is just, "the trial being performed." Okay, good.

EDIT: What added further to my confusion was that the values seemed to decrease monotonically (excluding the last one) suggesting that we were looking at some kind of progression or trend. Now I know that that isn't the case.

So, if it is a projectile motion problem, and you have the initial kinetic energy of the block, how would you proceed? Does this experiment assume that the pendulum strikes the block at the instant that it is vertical? If so, then you know what the initial direction of the block is too.

Last edited: Jul 16, 2009
8. Jul 16, 2009

### rayfish

Yep, each one corresponds to a different trial.

The experiment assumes it strikes it perfectly flat, and the whole system is linear.

I calculated the various values, such as kinetic energy on the block, the velocities, and the times, but how can I unite them under one equation where it is like "d(h) = .... ?" D being the distance it travels, h being the height above the ledge it is dropped. My physics teacher hasn't really talked about this.

The only thing I can think of that is related to my course is derivation of functions.. where A certain value is proportional to something else, and then introduction of a constant to create a equation. Am I heading in the right direction?

9. Jul 16, 2009

### LowlyPion

Since it is to be a complete transfer of energy, then you know the energy directly from the height the pendulum is released and where it ends up.

Since this energy is 1/2 mv2 of the block after impact then you know velocity, and from that it's just like horizontally launching a block from a ledge as to where it strikes the floor 1.05 m below.

The time to fall is pretty straight forward since the height it drops is fixed.

So ... combine the time to fall with the horizontal velocity imparted and ...