Derive trig equations 4, 5, and 6 from 1, 2, and 3

[jfnn]

Homework Statement

Hi again, I will re-write this on behalf of a request. I am struggling trying to derive these equations for a physics lab. The equations asked to derive are the equations that describe a perfectly elastic collision between objects of identical mass.

These are given to me:

Equation 1: v1 = v'1 cos (theta1) + v'2 cos (theta 2)
Equation 2: 0 = v'1 sin (theta1) + v'2 sin (theta 2)
Equation 3: v1^2 = v'1^2 + v'2^2

These equations are given to but (but they come from the conservation of momentum equation in the horizontal and vertical direction and the conservation of kinetic energy for two equal masses undergoing an identical collision).

The equations I am asked to derive from equation 1, 2, and 3 are as follows:

Equation 4: v'1 = v1cos(theta1)
Equation 5: v'2 = v1sin(theta1)
Equation 6: theta1 + theta2 = 90 degrees

Homework Equations

Listed above. Equations 1-6.

The Attempt at a Solution

[/B]
I solved equation 2 for v'1 and then put this expression into equation 1 first as shown below:

0 = v'1sin(theta1) + v'2sin(theta2)
v'1 = - (v'2sin(theta2)/sin(theta1))

Plug into equation 2 and solved for v'2 I got equation 5.

I then plugged equation 5 into v'1 = - (v'2sin(theta2)/sin(theta1)) to get equation 4.

After this, I am lost and have no idea how to derive equation 6 (the one for the angles)...

Please let me know and offer any assistance, this is for my report due tomorrow.

Thank you,
J

 

[haruspex]

Plug into equation 2 and solved for v'2 I got equation 5.

You couldn't have got that by plugging into eqn 2. You used eqn 2 to get to this. But I don't see how you got 5 merely by plugging it into 1 or 3 either. Seems to me you need to use both. Please post your working for this step.

 

[lsie]

Did anyone manage to figure this out? I'm hitting a brick wall.

 

[lsie]

I think, for equation 6, I need to end up with something like sin (theta) - cos (theta) = 0. But I can't seem to get there. Ditto for equations 4 and 5. I've tried all kinds of substitutions and the equations seem to get longer and longer without resolving.

 

[PeroK]

Did anyone manage to figure this out? I'm hitting a brick wall.

You'll need to post what you've tried - and possibly better in a new homework thread.

One idea is to use the Centre of Mass frame. That's probably what I'd do.

 

[PeroK]

I think, for equation 6, I need to end up with something like sin (theta) - cos (theta) = 0. But I can't seem to get there. Ditto for equations 4 and 5. I've tried all kinds of substitutions and the equations seem to get longer and longer without resolving.

There might be a problem with this equation:

Equation 2: 0 = v'1 sin (theta1) + v'2 sin (theta 2)

If you draw a diagram where one mass is moving to the right, then both masses must be moving to the right after the collision. In this case, if we draw $\theta_1$ above the initial direction of motion, and $\theta_2$ below the direction of motion, then we have:
$$v'_1\sin \theta_1 = v'_2\sin \theta_2$$And, you need to show that $\theta_1 + \theta_2 = \frac \pi 2$.

If, however, we take the equation given, then $\theta_2$ is a negative angle. And $\theta_1 + \theta_2 \ne \frac \pi 2$. In fact, in this case, we should get:
$$\theta_1 + \theta_2 = \theta_1 - (\frac \pi 2 - \theta_1) = 2\theta_1 - \frac \pi 2$$And, it should be:$$\theta_1 - \theta_2 = \frac \pi 2$$
Note that for a head-on collision, we have $\theta_1 = \theta_2 = 0$ and the first mass has zero velocity after the collision. That's an exception that should emerge if you are being careful with the maths.

 

[lsie]

You're absolutely right that the OP wrote equation 2 down wrong. I'll be back to look over what you've written tomorrow, when I have more energy. Thank you for the help!

 

[PeroK]

You're absolutely right that the OP wrote equation 2 down wrong. I'll be back to look over what you've written tomorrow, when I have more energy. Thank you for the help!

Ignore my suggestion about the CoM frame. It's not any easier that way.

 

[lsie]

I've worked on this and still haven't figured it out. I'm wondering if it's basically a matter of substitution tetris or if there's something I'm missing.

Here's what I've tried:

Substituting equation one (V1 = V1'cos theta + V2'cos phi) and equation two (0 = V1' sin theta - V2' sin phi) together, we get:

V2'= V1f (sin theta) / (sin phi)

V2' = ((V1 - V2'cos phi) / (cos theta)) * (sin theta/sin phi)

V2' = (V1 sin theta - V2' cos phi sin theta) / (cos theta sin phi)

V2' = V1 sin theta ((-V2' cos phi sin theta) / (cos theta sin phi))

Here I'm pretty happy, because I have what I'm trying to derive (V2' = V1 sin theta); but there just happens to be all that other junk in front of it. I have a chain from there that I won't bother including because it goes on for quite a while. There are promising moments, but it seems to lead to a kind of infinite regression.

Am I on the right track? Do I just keep sniff-substituting my way to an answer? Or am I going about this entirely the wrong way?

Cheers!

 

[PeroK]

You need to square the momentum equations, so that you can use the energy equation. Alternatively, use the law of cosines.

 

[PeroK]

To get the ball rolling. I'm going to use $v$ for the initial velocity and $v_1, v_2$ for the velocities after the collision. Equation 1 becomes:
$$v_2\cos \theta_2 = v - v_1 \cos \theta_1$$Squaring this equation gives:
$$v_2^2\cos^2 \theta_2 = v^2 - 2vv_1\cos \theta_1 + v_1^2\cos^2 \theta_1$$Can you continue from there? The aim is to eliminate $v_2$ using this and the other equations.

 

[lsie]

Ok, I've worked out equation number 4.

My solution was to square equations one and two, add them together, and reduce them down with the help of equation three. This worked well, because squaring Equation One produces a few squared cos and sin values that can be converted to '1,' some x and y and z squared values I could pare down with Equation Three, and a cosθ on its own.

I tried doing something similar to solve Equation Five, but I end up with a sinθsinΦ I don't know what to do with. I don't see how to get sinθ on its own, as I did previously with cosθ.

I'm even more lost about how to get Equation Six.

 

[PeroK]

We started by squaring equations 1 and 2, and adding them:
$$v_2^2\cos^2 \theta_2 = v^2 - 2vv_1\cos \theta_1 + v_1^2\cos^2 \theta_1$$$$v_2^2\sin^2 \theta_2 = v_1^2\sin^2\theta_1$$$$v_2^2 = v^2 +v_1^2 - 2vv_1\cos \theta_1$$This last equation you should recognise as the law of cosines. You could get that more directly, by drawing a triangle with $\vec v$ as the vector sum of $\vec v_1 + \vec v_2$.

Next, we use conservation of energy (equation 3) to eliminate $v_2$ from our last equation:
$$v^2 - v_1^2 = v^2 +v_1^2 - 2vv_1\cos \theta_1$$$$\implies 2v_1^2 = 2vv_1\cos \theta_1$$$$\implies v_1 = v\cos \theta_1$$Note that this is where we need the assumption $v_1 \ne 0$, which is the case for a head-on collision. And we have equation 4. I assume that is what you did.

To get equation 5, we can use equation 4 (squared) and conservation of energy again:
$$v_1^2 = v^2\cos^2 \theta_1$$$$v_2^2 = v^2 - v_1^2$$$$\implies v_2^2 = v^2(1 - \cos^2\theta_1) = v^2\sin^2\theta_1$$$$\implies v_2 = v\sin\theta_1$$The simplest way to get equation 6 is to combine equations 4 & 5 with equation 2:
$$v_1\sin\theta_1= v_2\sin\theta_2$$$$\implies (v\cos \theta_1)\sin \theta_1 = (v\sin \theta_1)\sin \theta_2$$$$\implies \cos \theta_1 = \sin \theta_2$$Then you need to know a little bit of trigonometry. E.g.
$$\sin\theta = \cos(\frac \pi 2 - \theta)$$

 

[Mister T]

V2'= V1f (sin theta) / (sin phi)

V2' = ((V1 - V2'cos phi) / (cos theta)) * (sin theta/sin phi)

V2' = (V1 sin theta - V2' cos phi sin theta) / (cos theta sin phi)

V2' = V1 sin theta ((-V2' cos phi sin theta) / (cos theta sin phi))

Learn LaTex so you can display the equations in a readable format. I find myself reaching for paper and pen so I can rewrite your equations in a way that's not tedious to read.