Elastic collision between particles of equal mass

[davidpotts]

Homework Statement

Prove that in an elastic collision between two identical bodies, one of which is initially at rest, the angle between their velocities after collision is pi/2, except for the case of central impact.

Homework Equations

Assume particle 1 travels in the x direction with speed v1(before) and strikes particle 2, which is at rest. After the collision, particle 1 makes an angle theta1 with the x axis, and particle 2 makes angle theta2. We need to show that theta1 + theta2 = pi/2.

Conservation of linear momentum
mv1(before) = SUM(mvi(after))
x component: mv1(before) = mv1(after) cos(theta1) + mv2(after) cos(theta2)
y component: 0 = mv2(after) sin(theta2) - mv1(after) sin(theta1)

Conservation of kinetic energy
.5mv1(before)^2 = .5mv1(after)^2 + .5mv2(after)^2

The Attempt at a Solution

I'm afraid I really don't know what to do with this. I notice that if theta1 + theta2 form a right angle, then by the Pythagorean theorem,

v1(after)^2 + v2(after)^2 = [v1(after) sin(theta1) + v2(after) sin(theta2)]^2
= v1(after)^2 sin^2(theta1) + (2)(v1(after))(v2(after)) sin(theta1) sin(theta2) + v2(after)^2 sin^2(theta2)

But this doesn't really look very useful. Hint please?

 

[PeterO]

Homework Statement

Prove that in an elastic collision between two identical bodies, one of which is initially at rest, the angle between their velocities after collision is pi/2, except for the case of central impact.

Homework Equations

Assume particle 1 travels in the x direction with speed v1(before) and strikes particle 2, which is at rest. After the collision, particle 1 makes an angle theta1 with the x axis, and particle 2 makes angle theta2. We need to show that theta1 + theta2 = pi/2.

Conservation of linear momentum
mv1(before) = SUM(mvi(after))
x component: mv1(before) = mv1(after) cos(theta1) + mv2(after) cos(theta2)
y component: 0 = mv2(after) sin(theta2) - mv1(after) sin(theta1)

Conservation of kinetic energy
.5mv1(before)^2 = .5mv1(after)^2 + .5mv2(after)^2

The Attempt at a Solution

I'm afraid I really don't know what to do with this. I notice that if theta1 + theta2 form a right angle, then by the Pythagorean theorem,

v1(after)^2 + v2(after)^2 = [v1(after) sin(theta1) + v2(after) sin(theta2)]^2
= v1(after)^2 sin^2(theta1) + (2)(v1(after))(v2(after)) sin(theta1) sin(theta2) + v2(after)^2 sin^2(theta2)

But this doesn't really look very useful. Hint please?

In might be better if you use V for the initial velocity, then v1 ad v2 for the two parts of the velocity afterwards.
Conservation of momentum
mV = mv1 + mv2

divide by mass and

V = v1 + v2

You could draw the triangle that shows that vector sum

Now:
Conservation of kinetic energy
.5mV^2 = .5mv1^2 + .5mv2^2

divide by .5m

V^2 = v1^2 + v2^2

so here is an interesting connection between the sides of your triangle. Note: the squaring removes the vector nature, so this expression relates the magnitudes only.

 

[davidpotts]

Thanks for this reply -- it certainly helped to reduce the clutter in what I wrote!

Obviously, from V^2 = v1^2 + v2^2, we can see that IF theta1 + theta2 = pi/2, then the magnitude V is the length of the hypotenuse of the right triangle formed from sides v1 and v2. And the converse is also true: if we could somehow know that the distance between the endpoints of the vectors v1 and v2 is V, that would show that theta1 + theta 2 = pi/2.

So now we want to show that the distance between the endpoints of the vectors v1 and v2 is V.

We can find this distance by adding the squares of the differences between the x and y components of v1 and v2:

distance^2 = [v2sin(theta2) - (-v1sin(theta1)]^2 + [v2cos(theta2) - v1cos(theta1)]^2
= v2^2sin^2(theta2) + 2v1sin(theta1)v2sin(theta2) + v1^2sin^2(theta1) + v2^2cos^2(theta2) - 2v1cos(theta1)v2cos(theta2) + v1^2cos^2(theta1)
= v1^2(sin^2(theta1) + cos^2(theta1)) + v2^2(sin^2(theta2) + cos^2(theta2)) + 2v1sin(theta1)v2cos(theta2) - 2v1cos(theta1)v2cos(theta2)
= v1^2 + v2^2 + 2v1sin(theta1)v2cos(theta2) - 2v1cos(theta1)v2cos(theta2)

(The last transition is because sin^2(theta) + cos^2(theta) = 1.)

Now, we want distance^2 = V^2, and we've seen that V^2 = v1^2 + v2^2. So we can get what we want if 2v1sin(theta1)v2cos(theta2) - 2v1cos(theta1)v2cos(theta2) = 0.

Is it true? Unfortunately, I don't know how to show this. Can somebody help? Am I even proceeding the right way?

 

[PeterO]

Thanks for this reply -- it certainly helped to reduce the clutter in what I wrote!

Obviously, from V^2 = v1^2 + v2^2, we can see that IF theta1 + theta2 = pi/2, then the magnitude V is the length of the hypotenuse of the right triangle formed from sides v1 and v2. And the converse is also true: if we could somehow know that the distance between the endpoints of the vectors v1 and v2 is V, that would show that theta1 + theta 2 = pi/2.

So now we want to show that the distance between the endpoints of the vectors v1 and v2 is V.

We can find this distance by adding the squares of the differences between the x and y components of v1 and v2:

distance^2 = [v2sin(theta2) - (-v1sin(theta1)]^2 + [v2cos(theta2) - v1cos(theta1)]^2
= v2^2sin^2(theta2) + 2v1sin(theta1)v2sin(theta2) + v1^2sin^2(theta1) + v2^2cos^2(theta2) - 2v1cos(theta1)v2cos(theta2) + v1^2cos^2(theta1)
= v1^2(sin^2(theta1) + cos^2(theta1)) + v2^2(sin^2(theta2) + cos^2(theta2)) + 2v1sin(theta1)v2cos(theta2) - 2v1cos(theta1)v2cos(theta2)
= v1^2 + v2^2 + 2v1sin(theta1)v2cos(theta2) - 2v1cos(theta1)v2cos(theta2)

(The last transition is because sin^2(theta) + cos^2(theta) = 1.)

Now, we want distance^2 = V^2, and we've seen that V^2 = v1^2 + v2^2. So we can get what we want if 2v1sin(theta1)v2cos(theta2) - 2v1cos(theta1)v2cos(theta2) = 0.

Is it true? Unfortunately, I don't know how to show this. Can somebody help? Am I even proceeding the right way?

isn't the vector sum of the momentums sufficient to show that the distance between the endpoints of the vectors v1 and v2 is V. or, in other words vectors v1 and v2 add up to the vector V? [I am pretty sure that should have been momenta]

 

[davidpotts]

isn't the vector sum of the momentums sufficient to show that the distance between the endpoints of the vectors v1 and v2 is V. or, in other words vectors v1 and v2 add up to the vector V? [I am pretty sure that should have been momenta]

I don't know. It would certainly be nice.

v1 = v1cos(theta1) - v1sin(theta1)
v2 = v2cos(theta2) + v2sin(theta2)

So

v1 + v2 = v1cos(theta1) + v2cos(theta2) - v1sin(theta1) + v2sin(theta2)

Because we have v2sin(theta2) - v1sin(theta1) = 0, the last two terms sum to 0.
And because we have V = v1cos(theta1) + v2cos(theta2), we have V = v1 + v2, QED.

And I guess this must be the right way to do it, but I don't really see the logic of it, because it is not in general true that the distance between the endpoints of two vectors can be found from their sum. Indeed, it is usually NOT true. It is true in the special case that the angle between the vectors is pi/2. So if we use this technique, aren't we assuming what we're trying to prove?

A general procedure for finding the distance between the endpoints would be take the difference between the vectors. That's what I was trying to do, but it led to a mess I couldn't see my way out of.

Put it this way. From V^2 = v1^2 + v2^2, we know that if v1 and v2 form a right angle, the distance between their endpoints is V. But there is no necessity in this. v1 and v2 could form any angle, and V^2 would just be the arbitrary sum of their squared magnitudes. Again, if v1 and v2 form a right angle, their vector sum will be the distance between their endpoints. But if not -- if, say, the angle is pi/4 -- then the vector sum will not be the distance between their endpoints. And if we then find that the vector sum is V, that will show nothing about the distance between their endpoints. Why can't it just be a coincidence that the vector sum V, if squared, is equal to v1^2 + v2^2?

 

[Dick]

I don't know. It would certainly be nice.

v1 = v1cos(theta1) - v1sin(theta1)
v2 = v2cos(theta2) + v2sin(theta2)

So

v1 + v2 = v1cos(theta1) + v2cos(theta2) - v1sin(theta1) + v2sin(theta2)

Because we have v2sin(theta2) - v1sin(theta1) = 0, the last two terms sum to 0.
And because we have V = v1cos(theta1) + v2cos(theta2), we have V = v1 + v2, QED.

And I guess this must be the right way to do it, but I don't really see the logic of it, because it is not in general true that the distance between the endpoints of two vectors can be found from their sum. Indeed, it is usually NOT true. It is true in the special case that the angle between the vectors is pi/2. So if we use this technique, aren't we assuming what we're trying to prove?

A general procedure for finding the distance between the endpoints would be take the difference between the vectors. That's what I was trying to do, but it led to a mess I couldn't see my way out of.

Put it this way. From V^2 = v1^2 + v2^2, we know that if v1 and v2 form a right angle, the distance between their endpoints is V. But there is no necessity in this. v1 and v2 could form any angle, and V^2 would just be the arbitrary sum of their squared magnitudes. Again, if v1 and v2 form a right angle, their vector sum will be the distance between their endpoints. But if not -- if, say, the angle is pi/4 -- then the vector sum will not be the distance between their endpoints. And if we then find that the vector sum is V, that will show nothing about the distance between their endpoints. Why can't it just be a coincidence that the vector sum V, if squared, is equal to v1^2 + v2^2?

Your logic there is wrong. Like in v1 = v1cos(theta1) - v1sin(theta1). You aren't dealing with the quantities as vectors. V^2=v1^2+v2^2 comes from conservation of kinetic energy. Do you know what a dot product is?

 

[ehild]

Homework Equations

Assume particle 1 travels in the x direction with speed v1(before) and strikes particle 2, which is at rest. After the collision, particle 1 makes an angle theta1 with the x axis, and particle 2 makes angle theta2. We need to show that theta1 + theta2 = pi/2.

Conservation of linear momentum
mv1(before) = SUM(mvi(after))
x component: mv1(before) = mv1(after) cos(theta1) + mv2(after) cos(theta2)
y component: 0 = mv2(after) sin(theta2) - mv1(after) sin(theta1)

Conservation of kinetic energy
.5mv1(before)^2 = .5mv1(after)^2 + .5mv2(after)^2

It is a bit shorter if you denote the original speed of the first ball by U, and the speeds after the collisions by v1 and v2. All your equations can be divided by m, and the energy equation multiplied by 2.Your equations become then

U=v1cosθ1+v2cosθ2
0=v1sinθ1-v2sinθ2
U²=v1²+v2²
Take the square of the first two equations and add them. Compare the result with the third one. Remember, you have to prove that θ1+θ2=pi/2, that is, cos(θ1+θ2)=0

ehild

 

[davidpotts]

It is a bit shorter if you denote the original speed of the first ball by U, and the speeds after the collisions by v1 and v2. All your equations can be divided by m, and the energy equation multiplied by 2.Your equations become then

U=v1cosθ1+v2cosθ2
0=v1sinθ1-v2sinθ2
U²=v1²+v2²
Take the square of the first two equations and add them. Compare the result with the third one. Remember, you have to prove that θ1+θ2=pi/2, that is, cos(θ1+θ2)=0

ehild

Excellent. Thanks for this. (And I like the coding of your notation too!)

So keeping your symbols, it goes like this:

U² = v₁²cos²θ₁ + 2v₁v₂cosθ₁cosθ₂ + v₂²cos²θ₂
0 = v₁²sin²θ₁ - 2v₁v₂sinθ₁sinθ₂ + v₂²sin²θ₂

So

U² = v₁²cos²θ₁ + 2v₁v₂cosθ₁cosθ₂ + v₂²cos²θ₂ + v₁²sin²θ₁ - 2v₁v₂sinθ₁sinθ₂ + v₂²sin²θ₂

Since sin²θ + cos²θ = 1 and U² = v₁² + v₂², we have

U² = U² + 2v₁v₂cosθ₁cosθ₂ - 2v₁v₂sinθ₁sinθ₂

Simplifying,

sinθ₁sinθ₂ = cosθ₁cosθ₂

Now in general, given f(Φ)f(Ψ) = g(Φ)g(Ψ), where Φ and Ψ range over multiple values and f(Φ) <> g(Φ), we can infer that f(Φ) = g(Ψ) and f(Ψ) = g(Φ). So sinθ₁ = cosθ₂ and sinθ₂ = cosθ₁. And this will be true just where θ₂ = pi/2 - θ₁, as is easy to see geometrically. Therefore, θ₁ + θ₂ = pi/2, QED.

You remark that this could be done by showing that cos(θ1+θ2)=0. I haven't done that, so there must be some other way to do it as well. But this gets the job done, which is good enough for me.

Concerning the idea that we could show that the distance between the endpoints of the vectors v₁ and v₂ = V (or in our new terminology, U) by showing that that is their vector sum, which was suggested by PeterO and supported by Dick, I think the procedure we just used can show that that won't work, as follows. Taking D to be the distance between the endpoints of v₁ and v₂, we have

D² = (v₂sinθ₂ - (-v₁sinθ₁))² + (v₂cosθ₂ - v₁cosθ₁)²
D² = v₁²sin²θ₁ + 2v₁v₂sinθ₁sinθ₂ + v₂²sin²θ₂ + v₁²cos²θ₁ - 2v₁v₂cosθ₁cosθ₂ + v₂²cos²θ₂
D² = U² + 2v₁v₂(sinθ₁sinθ₂ - cosθ₁cosθ₂)

If D² = U², we could proceed as above to show that θ₁ + θ₂ = pi/2. But of course, whether D = U is just what we are trying to discover. Again, if we knew that θ₁ + θ₂ = pi/2, we could show that the second term of the last equation = 0 and therefore that D = U, but of course that θ₁ + θ₂ = pi/2 is what we are trying to prove to begin with. So this procedure begs the question, in line with my argument in an earlier post.

 

[ehild]

Simplifying,

sinθ₁sinθ₂ = cosθ₁cosθ₂

It is the same as cosθ₁cosθ₂-sinθ₁sinθ₂=0

You know the formula for the cosine of the sum of angles,

cos(α+β) =cos(α)cos(β)-sin(α)sin(β)?

You need the cosine of (θ1+θ2). And you also know that cos(90°)=0.

ehild

 

[davidpotts]

It is the same as cosθ₁cosθ₂-sinθ₁sinθ₂=0

You know the formula for the cosine of the sum of angles,

cos(α+β) =cos(α)cos(β)-sin(α)sin(β)?

You need the cosine of (θ1+θ2). And you also know that cos(90°)=0.

ehild

No, I didn't know that formula. Though I might have looked it up and spared myself a flight of logic! So from the formula and the fact that cosθ₁cosθ₂-sinθ₁sinθ₂=0, we have cos(θ₁+θ₂)=0, so θ₁+θ₂=pi/2, which is our goal.

Nice!

 

[ehild]

Just keep in mind that such formulae exist also for sin(α+β)=sin(α)cos(β)+cos(α)sin(β);
and tan(α+β)=(tan(α)+tan(β))/(1-tan(α)tan(β)).
The addition laws are not hard to prove geometrically. For the cosine, find the length AB in the picture.


ehild

 

[davidpotts]

Just keep in mind that such formulae exist also for sin(α+β)=sin(α)cos(β)+cos(α)sin(β);
and tan(α+β)=(tan(α)+tan(β))/(1-tan(α)tan(β)).
The addition laws are not hard to prove geometrically. For the cosine, find the length AB in the picture.


ehild

Yes, I did look these up in one of my math books and read the proof of the formula for cos(α+β). Thanks for the tip!

 

[logics]

in an elastic collision between two identical bodies, one of which is initially at rest, the angle between their velocities after collision is pi/2, except for the case of central

that is not true, if you just consider that in that case direction of v1 is y-axis and v = 0

Conservation of momentum V = v1 + v2 You could draw the triangle that shows that vector sum
Conservation of kinetic energy V^2 = v1^2 + v2^2 so here is an interesting connection between the sides of your triangle

"reductio ad absurdum" [[itex]\rightarrow\leftarrow[/itex]] and these precious hints give you an instant proof by contradiction that requires no math: angle of deflection cannot be [>/< than] different from 90°, as only π/ 2 meets those conditions

 

[PeterO]

that is not true, if you just consider that in that case direction of v1 is y-axis and v = 0

"reductio ad absurdum" [[itex]\rightarrow\leftarrow[/itex]] and these precious hints give you an instant proof by contradiction that requires no math: angle of deflection cannot be [>/< than] different from 90°, as only π/ 2 meets those conditions

I am certainly happy that the two statements prove a right angle. No need to get too involved in trigonometry when a simple Pythagorus is offered.