Momentum with Rings

By: Catherine Medeiros

For this blog, I refused to go outside. I don't go outside unless its above 65 degrees and sunny. These were not the conditions over the weekend, so I decided to stay inside. Using a candle to keep warm and looking at my championship ring in sadness, I came up with an idea. The candle is a lot heavier than the ring box, so it will show conservation of momentum. The candle weighs about 623 grams so .623 kilograms and the ring box weighs 76 grams or .076 kilograms.


Here is the video of the candle hitting the ring box.

I put the video into Logger Pro and marked the points. This is what I got.

Using Logger Pro, I graphed these points to get this graph. 

I was then able to find the initial and final velocity for both the ring box and the candle.

Velocity initial candle = -0.917

Velocity final candle = 0

Velocity initial ring box = 0

Velocity final ring box = -0.282

The data does not show that the final velocity of the candle and initial velocity of the ring box equals zero; however, the ring box starts from rest and the candle ends in rest. Therefore, both of these values must equal zero. 

One can then see that there is a conservation of momentum in this demonstration by using the formula

m1v1 + m2v2 = m1v1f + m2v2f

(plug into the equation the numbers)

.623(-0.917) + .076(0) = .623(0) + .076(-0.282)

-0.571291 = -0.021432

There is a large difference in the numbers. The reason for this large difference is the friction 

between the candle and the table as well as the ring box and the table. However, if both masses were

frictionless, there would be no difference in the numbers.

To find the kinetic energy lost, first you must find the kinetic energy before the collision and the kinetic 

energy after the collision. The equation for kinetic energy is KE = 1/2mv^2

Therefore: 

KE initial = 1/2mv^2 (candle) + 1/2mv^2 (ring box)

The ring box initial velocity is zero so it is just 1/2mv^2 of the candle.

This equals

1/2mv^2 = 1/2(.623)(-0.917)^2 = 0.2619

KE final = 1/2mv^2 (candle) + 1/2mv^2 (ring box)

The candle final velocity is zero so it is just 1/2mv^2 of the ring box.

This equals

1/2mv^2 = 1/2(.076)(-0.282)^2 =  0.00302

Then subtract KE final from the KE initial so

0.2619 - 0.00302 = 0.25888

The kinetic energy lost in the system is 0.25888 Joules.