# Ballistic Pendulum - Kinetic Energy vs Gravitational Potential Energy

1. Feb 21, 2010

### Kamilan

Hello,

I have been trying to get my head around the concepts of energy, work and momentum. I am trying to build some testing equipment to measure the "transfer of energy" of a moving object striking a pendulum type mass, but am unsure if my physics is correct.

The image attached is my idea for the testing equipment, and below is my working. Now, I am unsure is if I should now take into account rotational physics or the horizontal component.

Length h (maximum height of the mass)

h = R - x

x = R$$\times$$ cos a therefore,

h = R - ( R $$\times$$ cos a)

Kinetic Energy of the Ball

1/2 $$\times$$ mvmv

Potential Energy of Mass

mgh.

There will be energy losses due to heat, sound, vibration and possibility the rebound of the ball, but does;

mgh roughly = 1/2 $$\times$$ mvmv ?

Am I right in my logic, or do I need to take into account the vertical movement as well?

I wish to place foam pads in front of the mass and look at the height at which the mass is elevated too compared to an impact with no foam pads.

My understanding of the principle of work is that it takes energy to move something in the direction of the force. Even though the force applied is in the vertical, we are still getting horizontal movement.

Can someone share some light on the matter please?

Cheers

K

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2. Feb 21, 2010

### Kamilan

that is supposed to be mv sqaured

3. Feb 22, 2010

Looks good.but U will need to calculate the curved trajectory circumference between the two objects to get a distance. Use the distance to calculate acc vel of the object. so you can calculate the terminal velocity on impact , so the moment of impact you will know how fast it is going

To find the curved trajectory R X pi\a= trajectory circumference

4. Feb 22, 2010

### Bob S

It works on the principle of momentum transfer. If the bullet has a mass m and velocity v, and the pendulum mass is M and recoil velocity V, then

m·v = (M+m)·V,

if the bullet lodges in the pendulum, and does not bounce off.

Bob S

5. Feb 22, 2010

### Kamilan

Hello,

I probably should have stated the know variables which are;

Mass of Ball
Mass of the Mass (LOL)
Average Velocity of Ball
And Height.

Below is an image of a more refined idea. We know that the some of the kinetic energy will be transfered into the mass in the form of kinetic energy rotating the mass where that energy will turn into potential energy at the peak height of rotation, then back into kinetic energy on the way down.

By measuring the maximum height the mass attains, we can then work out the amount of work done by the ball to move the ball to that height.

Now... by using the coefficient of restitution we can calculate the velocity of the ball when rebounding off the mass (as the mass is solid and heavy)

Thus, we can now calculate the initial momentum of the system, and the momentum of the ball after rebound.

The problem however now lies in calculating the equivalent force required to move the mass to that height.

Anyone shed some light on the matter?

Cheers

K

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• ###### 23-02-2010 12-25-10 p.m..jpg
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6. Feb 22, 2010

### Kamilan

this may come as a stupid question, but if we know the average velocity, do we need to know the speed upon impact?

Cheers

K

P.S.

For the test equipment, I thought that the time of impact may be crucial. I was thing of using a sensor (don't know which type yet) to just measure the amount of time the ball is in contact with the Mass.

If we know this time, can force be calculated by some awesome algebraic playing with the formula of Impulse?

7. Feb 22, 2010

### Bob S

The momentum transfer is just the impulse, or the time integral of the force (mass times deceleration). If the ball falls to the ground after hitting the mass (coefficient of restitution = 0) then

MV=mv

If the ball bounces back with a 100% coefficient of restitution, then MV =~2mv for a large M/m ratio. The kinetic energy of the recoiling mass is ½MV2.

So Mgh = ½MV2 = (MV)2/2M

Bob S

Last edited: Feb 22, 2010
8. Feb 23, 2010

Yes because, the velocity of the object will change with weight and the distance of the curved circumference. To calculate this you will need determine the coefficient between the ball and the mass on the pendulum. Because the theoretical experiment could be on mars. So the gravity would affect the accelerated velocity of the ball. Then distance travelled by weight by the accelerated velocity will give you the terminal velocity on impact. This will be the main factors to calculate the kinetic exchange on the pendulum.

9. Feb 23, 2010

to better explain myself. Using Newtons second law of physics M+M=M now take out gravity so a parabola does not affect the equation.

If I have an object with a weight of 10kg=A.
if I throw it at another object=B.
weighing the same amount, the kinetic exchange is half that of object A.

If object A is travelling at 5m\sec as it hit object B the velocity of B would be 2.5m\sec.
If object A has a velocity 5m\sec moving in a direction towards y
And object B has a velocity of 2m\sec towards y
B velocity would =3.5m\sec
But the problem is can you know the velocity of the object by only throwing it.
You know the distance between the two objects, but can you find the terminal velocity of the object when it hits. And remember mass is not like light, its velocity is not a constant.
An average velocity is not an accurate formula. talking about engineering a sensor you need to be exact.

10. Feb 23, 2010

### Bob S

I don't understand your math. If object B has half the kinetic energy of object A, doesn't object B have 1/√2 the velocity of object A? Seriously, kinetic energy is not a conserved quantity; linear momentum is. Furthermore, the amount of momentum transferred to object B from object A depends on the coefficient of restitution.

Bob S

11. Feb 23, 2010

Why in the environment of zero gravity would coefficient of restitution affect the transfer of kinetic energy? The object bouncing wouldn’t affect the kinetic exchange. The first impact is the factor of the exchange, the object connecting several times would be a different factor. But what I’m explaining is that velocity exchange in relation to mass is half in this example. And in the example of object B reaching a velocity of 3.5m\sec because the current velocity of B =2m\sec when object A impacts witch would be 5-2\2=1.5+current velocity of B=3.5m\s because the object A doesn’t impact B at 5M\sec it impacts at 3m\sec because of the current velocity of object B. And in regards to liner momentum, because the vector of both objects is y. The angular momentum is irrelevant because the object hits the centre, dispersing the exchange equally. I can understand you’re confusion, because it wasn’t supposed to be formidable explanation. In fact the objects would stay connected in a real world.
I wasn’t explaining weights of the objects and the affect on the velocity, Only the velocity exchanged prompting this thread to divert to calculating the acc vel.

12. Feb 23, 2010

### Bob S

Please write down the equations for the transfer of kinetic energy of a bullet of mass m and velocity v to a much larger stationary mass M >>m, with a coefficient of restitution =0 and =1.

Thanks
Bob S

13. Feb 25, 2010