# Hard Energy Physics Problem

1. Dec 5, 2011

### fobbz

1. The problem statement, all variables and given/known data

http://img694.imageshack.us/img694/1115/pendulumcutquestion.jpg [Broken]

2. Relevant equations
Ek=1/2 mv2
Ep=mgh

3. The attempt at a solution

Basically I looked at the problem firstly as the pendulum has reached the lowest point, therefore having an Ep of 0 and an Ek . Then I took a final point, where the string is cut. At this point there is Ek and Ep is there not?

The height would therefore be from the bottom of the swing, to the place where it's cut, which equals

6.4 - 6.4cos(30)

So now I have this

Ek= Ek2 + Ep
0.5mv2 = 0.5m(vsin30)2 + mgh

Am I on the right track?

What do I do after this? I don't know what my next move should be...

Should I use the point where the string is cut , and set it equal to the ball at the top of the height's energy?

Last edited by a moderator: May 5, 2017
2. Dec 5, 2011

### Staff: Mentor

It's hard to say if you're on the right track or not, since you haven't provided a clear statement of what some of your formulas are meant to represent. Also you've used the variable 'v' on both sides of an energy equation, so what does the formula mean?

I think if you can put a number to the velocity of the bob at position Y then you'll be making excellent progress. What's the KE of the bob at point Y?

3. Dec 5, 2011

### gnulinger

You're almost there, but your energy equation is a little off. Total energy is conserved, so find the speed at the height 6.4 - 6.4cos(30). (You start off with all potential energy, and end up with potential and kinetic energy.) Once you know the speed, you can find the vertical velocity, because you know that the path of the ball with be orthogonal to a line drawn from the ball to the point where the string is fixed.

4. Dec 6, 2011

### fobbz

Okay so this is what I've done.

I've taken the Ep (energy potential) when the ball is at the top (position X) and set it equal to the Ek (energy kinetic) and the new Ep2 at position Y, Thus obtaining the following:

Ep = Ep2 + Ek
mg(6.4) = mg(6.4-6.4cos(30)) + 0.5mv2

Solving for v, I get 9.9m/s

Now, Im looking at the problem starting at position Y and resulting at the max point of h.

So

Ep2 + Ek = Ep3 + Ek2

Ek2 = .5(9.9sin30)2

And solving for h I get 4.6

Is this right? Or does Ek2 not exist since the ball is no longer traveling in the y direction?!

5. Dec 6, 2011

### JHamm

You can find the energy when the pendulum is in position X since the only energy is potential, the diagram also gives you the length of the pendulum and the angle at which the rope is cut therefore you can find the potential energy at this point; the difference in these potentials is the kinetic energy at this point, from this and the angle you have you can find the vertical velocity and a kinematics equation will give you your max height (or another energy conservation application if that's more your style)

6. Dec 6, 2011

### fobbz

Okay so I did this, and it makes perfect sense to me! I used the V'2 = Vi2 -2ad formula and I got 3.43m which I believe is correct.

How would I use another energy conservation application though to solve for the answer? I don't really know what to have as an equation.

7. Dec 6, 2011

### gnulinger

Just ignore any horizontal displacement. Imagine that you have a ball that is tossed up with some vertical velocity. How high does it go above where it is released? You might be making this problem more complicated than it is.

8. Dec 6, 2011

### fobbz

I know the concept, I mean I just don't really understand it written out in an equation.

When the ball is cut from the string, and it's at the highest point, it has potential energy and kinetic energy does it not? I mean it is still moving? In the X at least... How can we not consider this part of the question?

I apologize if I sound stupid, maybe I'm missing something...

9. Dec 6, 2011

### JHamm

You don't sound stupid at all, in fact what you said is entirely true :)
Try this, if I have a ball with a vertical velocity of $1ms^{-1}$ and it reaches a height $h$ then I take this same ball and fire it with a velocity of $2ms^{-1}$ at an angle of $30^o$ how high will it go?

10. Dec 6, 2011

### fobbz

It will go the same height as the last one, as 2sin(30) = 1 m/s !

So basically, I understand that. It's just, now, what do we take from this into the question?