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s.h_math s.h_math
wrote...
Posts: 294
Rep: 3 0
2 years ago











The questions listed 1-4, are the questions I need help on. This is my lab I’m doing in class. I have finished the calculations but I am having a hard time with completing these questions. If it’s possible if I could get some help how to correctly answer these questions. It’s always greatly appreciated.

Thank you so much!

Post Merge: 2 years ago



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wrote...
Educator
2 years ago
It's extremely hard to read your questions. Could you repost the last 4 questions bigger?
s.h_math Author
wrote...
2 years ago Edited: 2 years ago, bio_man
For sure I can!
Post Merge: 2 years ago




Post Merge: 2 years 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?
wrote...
Educator
2 years ago
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?

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

vf=m1m1+m2v1

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

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

KEf=12(m1+m2)|vf|2
s.h_math Author
wrote...
2 years ago
Thanks a lot sir!

Have you also posted the answers to the other questions?
wrote...
Educator
2 years ago Edited: 2 years ago, bio_man
4. When the KE of one of the two objects is halved in magnitude, what happens to its momentum?



So, if the kinetic energy is halved, linear momentum is 12 of its original value.

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.



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.
wrote...
Educator
2 years ago
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.

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.
s.h_math Author
wrote...
2 years ago





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!
wrote...
Educator
2 years ago Edited: 2 years ago, bio_man
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?
s.h_math Author
wrote...
2 years ago Edited: 2 years ago, s.h_math
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: 2 years ago

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

Post Merge: 2 years 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.
wrote...
Educator
2 years 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!

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

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.

m1=m2
v1 = speed of puck moving
v2=0 = speed of puck at rest

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

v1after=m1m2m1+m2v1before

v1after=0m1+m2v1=0

Notice how the speed is 0.

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

V2after=2m1m1+m2v

V2after=2m12mv

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.
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