Physics Collisions
I am having a bit of trouble with question 1c in the OCR 2824 June 2008 paper - the question about collisions between three objects.
Could someone explain to me how we know straight away for part 2 that block 1 just bounces back and moves in the opposite direction. I can show that mathematically but I wouldn't know how to do it without setting up simultaneous equations etc... Is there a more intuitive way? Also for part i, I thought the answer would be along the lines of, 2 and 3 can't remain stuck as the collision is elastic, so 3 must move off and as it has the same mass as 1, its velocity must be the same to conserve momentum, but I feel like this isn't true since 2 and 3 are stuck together in part ii and this is still an elastic collision.
Thank you for any help anyone can give me. I know this is basic stuff but it is better to be sure than to regret asking.
If anyone has time, I am also having a bit of trouble with 2d since I thought the KE of the glider would be zero at t = 0 due to the velocity being zero, but the mark scheme says the graph should be a negative cosine graph... What am I thinking about wrong?
Could someone explain to me how we know straight away for part 2 that block 1 just bounces back and moves in the opposite direction. I can show that mathematically but I wouldn't know how to do it without setting up simultaneous equations etc... Is there a more intuitive way? Also for part i, I thought the answer would be along the lines of, 2 and 3 can't remain stuck as the collision is elastic, so 3 must move off and as it has the same mass as 1, its velocity must be the same to conserve momentum, but I feel like this isn't true since 2 and 3 are stuck together in part ii and this is still an elastic collision.
Thank you for any help anyone can give me. I know this is basic stuff but it is better to be sure than to regret asking.
If anyone has time, I am also having a bit of trouble with 2d since I thought the KE of the glider would be zero at t = 0 due to the velocity being zero, but the mark scheme says the graph should be a negative cosine graph... What am I thinking about wrong?
Original post by Bulbasaur10
I am having a bit of trouble with question 1c in the OCR 2824 June 2008 paper - the question about collisions between three objects.
Could someone explain to me how we know straight away for part 2 that block 1 just bounces back and moves in the opposite direction. I can show that mathematically but I wouldn't know how to do it without setting up simultaneous equations etc... Is there a more intuitive way? Also for part i, I thought the answer would be along the lines of, 2 and 3 can't remain stuck as the collision is elastic, so 3 must move off and as it has the same mass as 1, its velocity must be the same to conserve momentum, but I feel like this isn't true since 2 and 3 are stuck together in part ii and this is still an elastic collision.
Thank you for any help anyone can give me. I know this is basic stuff but it is better to be sure than to regret asking.
If anyone has time, I am also having a bit of trouble with 2d since I thought the KE of the glider would be zero at t = 0 due to the velocity being zero, but the mark scheme says the graph should be a negative cosine graph... What am I thinking about wrong?
Could someone explain to me how we know straight away for part 2 that block 1 just bounces back and moves in the opposite direction. I can show that mathematically but I wouldn't know how to do it without setting up simultaneous equations etc... Is there a more intuitive way? Also for part i, I thought the answer would be along the lines of, 2 and 3 can't remain stuck as the collision is elastic, so 3 must move off and as it has the same mass as 1, its velocity must be the same to conserve momentum, but I feel like this isn't true since 2 and 3 are stuck together in part ii and this is still an elastic collision.
Thank you for any help anyone can give me. I know this is basic stuff but it is better to be sure than to regret asking.
If anyone has time, I am also having a bit of trouble with 2d since I thought the KE of the glider would be zero at t = 0 due to the velocity being zero, but the mark scheme says the graph should be a negative cosine graph... What am I thinking about wrong?
It would be easier for people to help by you posting the exact question here rather than expecting people to go and search for the question in order to help you.
Where do you get that paper from? I can't find it...
However, from what I can understand:
* You could know if it moves in the opposite direction if it has smaller mass or velocity or both than the object that it collides with... Again, I don't know the question.
* If they stick together, the collision has to be inelastic so maybe in part ii they change the fact that it is elastic, they do that sometimes...
I don't know if that is useful, sorry and good luck
However, from what I can understand:
* You could know if it moves in the opposite direction if it has smaller mass or velocity or both than the object that it collides with... Again, I don't know the question.
* If they stick together, the collision has to be inelastic so maybe in part ii they change the fact that it is elastic, they do that sometimes...
I don't know if that is useful, sorry and good luck
Original post by Eimmanuel
It would be easier for people to help by you posting the exact question here rather than expecting people to go and search for the question in order to help you.
Sorry
I wasn't intending to be lazy... If you go onto the OCR physics a level page on the OCR website for the pre-2015 specification and click legacy papers you will find a folder with even older papers and in the June 2008 section you can find this paper.I haven't attached the folder with the paper in as something went wrong with the upload but I will try to get it up in a bit.I didn't expect it to be too much of a problem given that I named the paper and question... I guess since it is an older paper it is quite hard to find.
(edited 5 years ago)
Original post by salmanz01
Where do you get that paper from? I can't find it...
However, from what I can understand:
* You could know if it moves in the opposite direction if it has smaller mass or velocity or both than the object that it collides with... Again, I don't know the question.
* If they stick together, the collision has to be inelastic so maybe in part ii they change the fact that it is elastic, they do that sometimes...
I don't know if that is useful, sorry and good luck
However, from what I can understand:
* You could know if it moves in the opposite direction if it has smaller mass or velocity or both than the object that it collides with... Again, I don't know the question.
* If they stick together, the collision has to be inelastic so maybe in part ii they change the fact that it is elastic, they do that sometimes...
I don't know if that is useful, sorry and good luck
Thank you! I think that helps a lot. If you want to see the paper it is on the OCR a level physics page for the old spec,and if you click legacy papers on the past papers section it should come up with a folder containing older papers from June 2008 etc... I am sorry that I didn't attach it but something isn't working... I'm really sorry if you spent ages trying to find it
(edited 5 years ago)
Original post by Bulbasaur10
Thank you! I think that helps a lot. If you want to see the paper it is on the OCR a level physics page for the old spec, and if you click legacy papers on the past papers section it should come up with a folder containing older papers from June 2008 etc... I am sorry that I didn't attach it but something isn't working...
Ok so I saw the question.. in part i I think you just needed to use the fact that 1 and 2 stop and 3 starts moving.. also you can't use your answer as the question is "After the collision, cubes 1 and 2 remain at rest and cube 3 move with speed u. Explain these observations" and your answer doesn't explain why 1 and 2 stop
In part ii, you got confused by them saying 2 and 3 are glued together as in stick together after the collision but they mean literally glues together which means you consider them as 1 object with their masses added so their mass is 2m while 1's mass is m so it's smaller so it moves in the opposite direction and 2 and 3 move off together as they are "glued together" hence 1 object.
Hope that helps
PS: carefully read the questions cause they want to confuse you
Original post by Bulbasaur10
Sorry I wasn't intending to be lazy... If you go onto the OCR physics a level page on the OCR website for the pre-2015 specification and click legacy papers you will find a folder with even older papers and in the June 2008 section you can find this paper.
I haven't attached the folder with the paper in as something went wrong with the upload but I will try to get it up in a bit.I didn't expect it to be too much of a problem given that I named the paper and question... I guess since it is an older paper it is quite hard to find.
I wasn't intending to be lazy... If you go onto the OCR physics a level page on the OCR website for the pre-2015 specification and click legacy papers you will find a folder with even older papers and in the June 2008 section you can find this paper.I haven't attached the folder with the paper in as something went wrong with the upload but I will try to get it up in a bit.I didn't expect it to be too much of a problem given that I named the paper and question... I guess since it is an older paper it is quite hard to find.
You can always copy the link and paste the link if you cannot attach the file.
Original post by Eimmanuel
You can always copy the link and paste the link if you cannot attach the file.
It is an Adobe Reader file so I'm not sure if I can do that. Do you know how I could?
Original post by Bulbasaur10
It is an Adobe Reader file so I'm not sure if I can do that. Do you know how I could?
just screenshot it
Original post by salmanz01
just screenshot it
I guess so. Thanks for the help
. I'll do that next time I have a question and I can't get the exact file. I'm not sure why I didn't think of that...Original post by Bulbasaur10
It is an Adobe Reader file so I'm not sure if I can do that. Do you know how I could?
Place your cursor on top of the pdf file link and right-clicked on it. A small pop-up will appear. It would show "Copy link address". Click on it. The link is copied. Then press on "Ctrl" + "v" to paste the link in the post.
Original post by Eimmanuel
Place your cursor on top of the pdf file link and right-clicked on it. A small pop-up will appear. It would show "Copy link address". Click on it. The link is copied. Then press on "Ctrl" + "v" to paste the link in the post.
Thanks!
*Edit: Wait... I don't think that works in an Adobe Reader file like it does for most pdf files. I know you can usually do that but here it isn't working. Don't worry about it anyway.
(edited 5 years ago)
Original post by Bulbasaur10
Thanks!
*Edit: Wait... I don't think that works in an Adobe Reader file like it does for most pdf files. I know you can usually do that but here it isn't working. Don't worry about it anyway.
*Edit: Wait... I don't think that works in an Adobe Reader file like it does for most pdf files. I know you can usually do that but here it isn't working. Don't worry about it anyway.
I mean in a web browser...
Original post by Eimmanuel
I mean in a web browser...
That will just take you to the page with the files on, but there are quite a few folders to open to get to the file, and once you have opened it there is a folder which contains the adobe documents. The files, as far as I'm aware, do not open my web browser at least.
Original post by Bulbasaur10
That will just take you to the page with the files on, ....
The answer is NO.
Original post by Bulbasaur10
.....but there are quite a few folders to open to get to the file, and once you have opened it there is a folder which contains the adobe documents. ...
I believe there is only one file, so there is a single link to the file. The "folder" does not matter. Open the "folder" to find the file link in the web browser under the OCR website and place the cursor on top of the link and right-clicked to copy the link.
Original post by Bulbasaur10
...The files, as far as I'm aware, do not open my web browser at least.
You can always enable the web browser to read pdf. Under your case, you have disable it and choose to save pdf file to your computer. This is unimportant in this case.
Original post by Eimmanuel
The answer is NO.
I believe there is only one file, so there is a single link to the file. The "folder" does not matter. Open the "folder" to find the file link in the web browser under the OCR website and place the cursor on top of the link and right-clicked to copy the link.
You can always enable the web browser to read pdf. Under your case, you have disable it and choose to save pdf file to your computer. This is unimportant in this case.
I believe there is only one file, so there is a single link to the file. The "folder" does not matter. Open the "folder" to find the file link in the web browser under the OCR website and place the cursor on top of the link and right-clicked to copy the link.
You can always enable the web browser to read pdf. Under your case, you have disable it and choose to save pdf file to your computer. This is unimportant in this case.
http://www.ocr.org.uk/Images/58044-past-paper-materials-2008-june-legacy.zip
I think this works? :?
Original post by Bulbasaur10
I am having a bit of trouble with question 1c in the OCR 2824 June 2008 paper - the question about collisions between three objects.
Could someone explain to me how we know straight away for part 2 that block 1 just bounces back and moves in the opposite direction. I can show that mathematically but I wouldn't know how to do it without setting up simultaneous equations etc... Is there a more intuitive way? Also for part i, I thought the answer would be along the lines of, 2 and 3 can't remain stuck as the collision is elastic, so 3 must move off and as it has the same mass as 1, its velocity must be the same to conserve momentum, but I feel like this isn't true since 2 and 3 are stuck together in part ii and this is still an elastic collision.
....
Could someone explain to me how we know straight away for part 2 that block 1 just bounces back and moves in the opposite direction. I can show that mathematically but I wouldn't know how to do it without setting up simultaneous equations etc... Is there a more intuitive way? Also for part i, I thought the answer would be along the lines of, 2 and 3 can't remain stuck as the collision is elastic, so 3 must move off and as it has the same mass as 1, its velocity must be the same to conserve momentum, but I feel like this isn't true since 2 and 3 are stuck together in part ii and this is still an elastic collision.
....
For (c)(i), you can make use of the result of (b) to explain what is going. (c) is like a Newton’s Cradle problem.
https://en.wikipedia.org/wiki/Newton%27s_cradle
From (b), you know that if an object 1 of mass m moves at a speed of u, collides elastically with another stationary object 2 of mass m, object 1 will transfer all its momentum and KE to object 2 and object 2 will move with speed of u while object 1 will remain at rest.
From (c), we are told that “Cube 3 is replaced in its original position close to cube 2.”, so there is a small between cube 3 and 2. When cube 1 collides with cube 2, both momentum and KE is passed to cube 2 and cube 2 will move with speed u. Then cube 2 collides with cube 3 and transfer both momentum and KE to cube 3 to cause cube 3 to move with speed u.
There is a similar question in which someone had asked and you visit the post with the link shown as follow.
https://www.thestudentroom.co.uk/showthread.php?t=5387636
For (c)(ii), although the question asked the candidate NOT to do any calculation, you can still do some calculations in your own practice to get an idea of what is going on. Under this situation where the collision is elastic, the transfer of momentum and KE is NOT complete because the mass is different. So cube 1 will “retain” some momentum and KE after the collision and get rebounced while the glued cube 2 and 3 will move to the right.
Original post by Bulbasaur10
Yes. It works.
Original post by Bulbasaur10
...
If anyone has time, I am also having a bit of trouble with 2d since I thought the KE of the glider would be zero at t = 0 due to the velocity being zero, but the mark scheme says the graph should be a negative cosine graph... What am I thinking about wrong?
If anyone has time, I am also having a bit of trouble with 2d since I thought the KE of the glider would be zero at t = 0 due to the velocity being zero, but the mark scheme says the graph should be a negative cosine graph... What am I thinking about wrong?
If you look at the diagram, the negative cosine graph is the pink curve which is very similar to the black curve. It is the black curve that the MS is referring to - negative cosine wave touching x-axis.
Original post by Eimmanuel
If you look at the diagram, the negative cosine graph is the pink curve which is very similar to the black curve. It is the black curve that the MS is referring to - negative cosine wave touching x-axis.
If you look at the diagram, the negative cosine graph is the pink curve which is very similar to the black curve. It is the black curve that the MS is referring to - negative cosine wave touching x-axis.
Right. I understand. That's what I had in mind I just didn't understand the MS...So it's just that it's been translated across somewhat. Thank you again - you've been a great help! Sorry about all my faffing around with the pdf...
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