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

In summary, the conversation is about deriving equations for a perfectly elastic collision between objects of identical mass. The equations provided are used to derive equations for velocity and angle. The attempt at a solution involves using equations 1-3 to solve for v'1 and v'2, and then plugging them into equations 4 and 5. However, the person is still struggling with deriving equation 6 for the angles.
  • #1
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
 
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  • #2
jfnn said:
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.
 
  • #3
Did anyone manage to figure this out? I'm hitting a brick wall.
 
  • #4
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.
 
  • #5
lsie said:
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.
 
  • #6
lsie said:
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:
jfnn said:
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.
 
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  • #7
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!
 
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Likes PeroK
  • #8
lsie said:
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.
 
  • #9
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!
 
  • #10
You need to square the momentum equations, so that you can use the energy equation. Alternatively, use the law of cosines.
 
  • #11
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.
 

What are the equations used to derive equations 4, 5, and 6 from 1, 2, and 3?

The equations used to derive equations 4, 5, and 6 from 1, 2, and 3 are typically referred to as mathematical transformations. These transformations involve manipulating the original equations using various algebraic and arithmetic operations to obtain the desired equations.

What is the purpose of deriving equations 4, 5, and 6 from 1, 2, and 3?

The purpose of deriving equations 4, 5, and 6 from 1, 2, and 3 is to simplify the original equations or to obtain new equations that may be more useful for solving a particular problem or understanding a phenomenon. This process allows scientists to better analyze and interpret their data and make more accurate predictions.

What are some common methods used to derive equations?

There are several common methods used to derive equations, such as substitution, elimination, and factorization. These methods involve replacing variables with equivalent expressions, canceling out terms, and breaking equations into smaller, more manageable parts.

Can equations be derived in multiple ways?

Yes, equations can be derived in multiple ways. Depending on the problem or the desired outcome, different methods or approaches may be used to derive the equations. Scientists often explore different pathways to derive equations and choose the most efficient or effective method for their specific needs.

How do derived equations differ from the original equations?

Derived equations may differ from the original equations in terms of complexity, form, and number of variables. The goal of deriving equations is to obtain a simpler, more concise expression that retains the same underlying relationships and principles as the original equations. However, derived equations may also introduce new variables or terms to better describe a system or process.

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