Conservation of momentum lab

It's extremely hard to read your questions. Could you repost the last 4 questions bigger?

For sure I can!

Post Merge: 7 months ago

Post Merge: 7 months ago

I also have to explain why the percentage of difference for x-component is higher. After putting some thought, I’m not sure about what to really use to explain why it’s so  high?

Thanks

Questions

1. Compare the total momentum before the collision with the total momentum after the collusion on both the x-axis and y-axis. Calculate the percentage difference. Discuss the possible errors that may have been the causes of the small discrepancies between the total momentum before and the total momentum after the collision.

2. If the two pucks are of identical mass with one puck being at rest at the center of the air table and the second puck is to head-on elastically collide with the one at rest, what will be the speed of the colliding puck and the puck at rest after the collision? Explain your answer.

3. Give an example of a collision in which all the KE of the participating objects is lost. What happens to their total momentum after the collision?

4. When the KE of one of the two objects is halved in magnitude, what happens to its momentum?

That is an example of inelastic collision. For example, a car moving at a certain speed smashes into a rock (or wall) that doesn't move. The wall remains intact, while the car is crushed by the impact. In this case, all the kinetic energy of the car is lost to the environment, transferred into other forms such as heat, sound, etc.

Before they collide, they have a combined energy:

E = 1/2 m1 v1^2 + 1/2 m2 v2^2

And they have a combined momentum:

p = m1v1 + m2v2

After the collision, there is a single combined mass (the rock, car combined) (m1 + m2). Since momentum is conserved in an inelastic collision (because the total momentum of both objects before and after the collision is the same), this combined car and wall have a momentum equal to the total initial momentum (p = (m1 + m2)*vf). Therefore:

m1v1 + m2v2 =  (m1 + m2)*vf

\(v_f=\frac{m_1}{m_1+m_2}v_1\)

Since the wall is stationary, don't worry about m2v2.

The final energy of the system could be found using the formula:

\(KE_f=\frac{1}{2}\left(m_1+m_2\right)\left|v_f\right|^2\)

Thanks a lot sir!

Have you also posted the answers to the other questions?

So, if the kinetic energy is halved, linear momentum is \(\frac{1}{\sqrt{2}}\) of its original value.




Please watch this video, it explains it perfectly. The only difference is that the masses are different in the video. If you still don't understand after watching, please write back.

What were you expecting here when you looked at the total momentum before the collision and the momentum after the collision? Were you expecting them to be the same? Was this lab done in real-life or it was a computer generated simulation? If it was done in real-life, did you consider forces due to friction? Tell me more about the lab because I couldn't read the original pictures you had posted.

Hi, yes this lab was done in real life, here is the lab clear pictures, I’m so sorry, I didn’t realize they were so unclear!

Ok, thanks for providing that. I was hoping you could give me a run down in your own words what you did, but it's okay, I'll just try answering this on my own.

First, there are two types of collisions:

Inelastic collisions: momentum is conserved,
Elastic collisions: momentum is conserved and kinetic energy is conserved

I'm assuming what happened here was an elastic collision, so you were expecting the momentums to be relatively the same before and after the collision. But, after calculating the percentage difference, you found that they were more different than you anticipated. Generally  (and without reading the experiment), non-constant velocities or frictional losses (coefficient of friction not accounted for) or measurement errors make the differences between them. These probably weren't accounted for as control variables when the experiment was conducted. Do you think this is right?

I believed reactively the same but he said he has nothing to do with the puck or the air table or the person them self and force

Post Merge: 7 months ago

Sir, also for number two, I’m not sure how to explain my answer without calculations.

Post Merge: 7 months ago

For number one, I have got the answer it was to switch the x and y axis. It is just number two I’m having trouble with.

Oh, that's good news!


So according to the video. If an inelastic collision were to occur where both masses are the same, and one of pucks is at rest.

\(m_1=m_2\)
\(v_1\) = speed of puck moving
\(v_2=0\) = speed of puck at rest

The puck originally moving will loss 100% of its speed.

\(v_{1_{after}}=\frac{m_1-m_2}{m_1+m_2}v_{1_{before}}\)

\(v_{1_{after}}=\frac{0}{m_1+m_2}v_1=0\)

Notice how the speed is 0.

For the puck that was stationary, its speed can be calculated like this:

\(V_{2_{after}}=\frac{2m_1}{m_1+m_2}v\)

\(V_{2_{after}}=\frac{2\cdot m_1}{_{2\cdot m}}v\)

Cancel out, you get the speed of the original puck. Therefore, the speed of the puck at rest will be the same as the speed of the puck that was originally moving after the collision.