M4 Elastic Collisions (Components of Velocity)

zetafunc. · 2012-05-24 07:34:00

Hi, I solved this problem simply by substituting the initial and final velocities and applying the principle of conservation of momentum, but there's something I don't understand about this problem.

"Two small smooth spheres A and B have equal radii. The mass of A is 2m kg and the mass of B is m kg. The spheres are moving on a smooth horizontal plane and they collide. Immediately before the collision the velocity of A is (2i 2j) ms[sup]-1[/sup] and the velocity of B is (3i j) ms[sup]-1[/sup]. Immediately after the collision the velocity of A is (i 3j) ms[sup]-1[/sup]. Find the speed of B immediately after the collision."

I thought that the components of velocity perpendicular to the line of centres of the spheres would be unchanged in the collision, at least, this is what it says in my M4 book. The vertical component of A's velocity before is -2j, but after colliding with B, it's vertical component is -3j. Why is this?

Thanks.

Bob · 2012-05-24 11:51:57

hi zetafunc,

I've been out all day so I've only just read this and it's too late for me to start solving it now.

I'll do it in the morning (BST).

But my initial thought is this: the line of centres is not the x axis so why shouldn't the j component change.

I'm thinking you need a coefficient of elasticity too. Which you could deduce from A's final velocity.

In the morning all of this may begin to look like complete rubbish but I will certainly try the question properly.

Bob

Bob · 2012-05-24 21:43:29

hi zetafunc

Ok, brain half working now.

Didn't need elacticity at all.

I got V of B after = -i + j

Then I thought I'd work out the line of centres, change the base vectors to // and perp to this line and re-work.

I'm not getting a consistent result, probably because of a silly arithmetical error.

Please say if you get that value for B so I can, at least, eliminate that as where the error is.

Thanks.

Bob

zetafunc. · 2012-05-24 23:17:04

Hi,

Thanks for the reply. Yes, I got that answer for v, leading to a speed of √2 m/s.

Bob · 2012-05-24 23:53:26

Thanks.

I've worked out why my change of base didn't work and the revised version has some nasty sums in it.

Couldn't find any online calculators for this so I'm making my own using Excel. It'll take a while.

But the line of centres is definitely not the x axis; that much I do know,

bob

Bob · 2012-05-25 04:47:59

hi zetafunc

Two pictures to explain this.

number 1 shows (but not accurately) the line of centres. Momentum is transferred along this line and not at right angles to it.

number 2 shows my excel sheet for calculating the change in momentum along the line of centres and perpendicular to it.

The first box shows the velocity components expressed in terms of i and j as usual.

Then I calculated the line of centre direction as the vector h and perpendicular to this vector k. (second box)

This was tricky. At first I thought I should just subtract one velocity vector from the other. But the magnitudes are different and this matters. If we model the objects as spheres then it makes a difference how big they are relative to each other. I decided to assume they are of equal size so I divided each velocity by its magnitude to get an equally weighted vector for each. Then I subtracted these to get h, and reversed the components and made one swap sign to get k.

Then I wrote the velocities in terms of the new vectors, h and k. (third box)

Then I calculated the change in momentum in the h and k directions. The k change in momentum is zero. (final box).

Bob

anonimnystefy · 2012-05-25 05:05:11

May I just ask what the problem is?

Bob · 2012-05-25 05:16:53

post #1

Bob

anonimnystefy · 2012-05-25 05:19:22

So the only problem is that the vertical speed changes? Isn't that normal?

Bob · 2012-05-25 05:44:09

zetafunc was expecting that the momentum wouldn't change at right angles to the line of centres.

I was trying to prove that it doesn't..

Bob

anonimnystefy · 2012-05-25 05:50:22

Well maybe it doesn't change to the right angles when they are going on the line connecting the centers,but first,that is not the case here,and,second,the vertical speed is not necessarily perpendicular to the line connecting the centers.

zetafunc. · 2012-06-26 07:41:26

Hi,

Sorry for the late reply. I did the M4 exam a few weeks ago and it was very difficult. I don't think I did too well unfortunately.

Really sorry I couldn't reply to this thread sooner. I did read it but I had no time to respond at all.

According to the examiner's reports for these questions, candidates make the assumption that the line of centres is the x-axis, and that's because of the way they draw the collisions in the book. They say that candidates should instead take the dot product of the velocity and the line of centres when necessary.

Anonimnystefy is correct -- I assumed, wrongly, that the vertical component would be perpendicular to the line of centres, but that's only the case if the line of centres is the x-axis!

Too bad a question like this didn't come up in the exam...

anonimnystefy · 2012-06-26 07:53:59

What is the scoring like?

zetafunc. · 2012-06-26 08:00:46

In the A-level Edexcel Maths exams over here, you have A, B, C, D, E and U grades. The papers are 90 minutes, each out of 75 marks, but the grade boundaries change depending on how the rest of the country do. In M4, because candidates regularly do poorly, boundaries are usually low. It's usually about 58/75 for an A grade in M4. The paper I sat though, I found very difficult and I don't think I did very well. It will probably be 48/75 for an A.

Unfortunately, there is the A* these days which requires that you get 90% in some of your modules. Definitely got nowhere near that in that M4 paper. Sigh... I could have performed so much better if I didn't panic. I'll have to retake it next year along with a couple other modules. I will never do 18 exams in one session ever again. I'm completely burnt out mentally.

I could post the paper up on here if you'd like to look at it?

anonimnystefy · 2012-06-26 08:05:24

Hi zetafunc.

That is a really low boundary.

Are there copies of the tests on the net?

zetafunc. · 2012-06-26 08:09:31

I agree. But I think for that particular paper, not many people will do well.

There are copies from teachers who have digital copies on their computer and post them online, but Edexcel don't officially publish them until around results day I think (in mid-August).

The link to the paper is here: sendspace . com / file /mjzp5j
(remove spaces)

anonimnystefy · 2012-06-26 08:19:58

Ok. Thanks for the link.

What is coefficient of restitution?

zetafunc. · 2012-06-26 08:25:37

The co-efficient of restitution for two particles 1 and 2 is the speed of separation (v[sub]2[/sub] - v[sub]1[/sub]) divided by the speed of approach (u[sub]1[/sub] - u[sub]2[/sub]).

anonimnystefy · 2012-06-26 08:33:01

And what are speeds of separation and approach?

zetafunc. · 2012-06-26 08:42:43

The speed of separation is the velocity of the second particle minus the velocity of the first, both after the collision has occurred.
The speed of approach is the velocity of the first particle minus the velocity of the second, both before the collision has occurred.

A better way to imagine it is if you have to particles A and B, with A moving toward B in the direction AB. The speed of approach is the velocity of A relative to B ([sub]A[/sub]v[sub]B[/sub]) after collision, and the speed of separation would be the velocity of B relative to A ([sub]B[/sub]v[sub]A[/sub]). Strictly speaking v is a vector but in one dimension it doesn't matter. In the case of Q1 you can just use this law by resolving parallel to the line of centres, with e = 3/4.

anonimnystefy · 2012-06-26 08:59:15

Ok. I will try the first question now.

Math Is Fun Forum

#1 2012-05-24 07:34:00

M4 Elastic Collisions (Components of Velocity)

#2 2012-05-24 11:51:57

Re: M4 Elastic Collisions (Components of Velocity)

#3 2012-05-24 21:43:29

Re: M4 Elastic Collisions (Components of Velocity)

#4 2012-05-24 23:17:04

Re: M4 Elastic Collisions (Components of Velocity)

#5 2012-05-24 23:53:26

Re: M4 Elastic Collisions (Components of Velocity)

#6 2012-05-25 04:47:59

Re: M4 Elastic Collisions (Components of Velocity)

#7 2012-05-25 05:05:11

Re: M4 Elastic Collisions (Components of Velocity)

#8 2012-05-25 05:16:53

Re: M4 Elastic Collisions (Components of Velocity)

#9 2012-05-25 05:19:22

Re: M4 Elastic Collisions (Components of Velocity)

#10 2012-05-25 05:44:09

Re: M4 Elastic Collisions (Components of Velocity)

#11 2012-05-25 05:50:22

Re: M4 Elastic Collisions (Components of Velocity)

#12 2012-06-26 07:41:26

Re: M4 Elastic Collisions (Components of Velocity)

#13 2012-06-26 07:53:59

Re: M4 Elastic Collisions (Components of Velocity)

#14 2012-06-26 08:00:46

Re: M4 Elastic Collisions (Components of Velocity)

#15 2012-06-26 08:05:24

Re: M4 Elastic Collisions (Components of Velocity)

#16 2012-06-26 08:09:31

Re: M4 Elastic Collisions (Components of Velocity)

#17 2012-06-26 08:19:58

Re: M4 Elastic Collisions (Components of Velocity)

#18 2012-06-26 08:25:37

Re: M4 Elastic Collisions (Components of Velocity)

#19 2012-06-26 08:33:01

Re: M4 Elastic Collisions (Components of Velocity)

#20 2012-06-26 08:42:43

Re: M4 Elastic Collisions (Components of Velocity)

#21 2012-06-26 08:59:15

Re: M4 Elastic Collisions (Components of Velocity)

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