# Ugh...Physics

Could someone explain how to solve 2D collisions using conservation of momentum? The textbook doesn’t explain it, my teacher can’t teach, and I can’t find anything online that shows a problem worked out and explained. It’s “simple” stuff; Xkg cue ball hits another ball of the same mass at an angle of X degrees, find the final velocity of the two balls if the cue ball has an initial momentum of X. I’d appreciate it if someone could either explain it themselves or point me in the right direction.

Give us a problem, it 'll be easier to explain it to you through that.

Essentially conservation of momentum is M1V1i +M2V2i = M1V1f +M2V2f where M is mass and V is velocity and the subscript i signifies before and f after the event. Everything that you’re concerned with is algebra tricks around that theme.

In an inelastic collision, you’ll have the 2 objects moving together in the same direction. So if you’re wondering how fast the object is going, you take into consideration that V1f and V2f are the same because the objects move together as one object. So

M1V1i +M2V2i = M1V1f +M2V2f

becomes

M1V1i +M2V2i = ( M1 +M2 )Vf

In elastic collisions, the objects will not move as 1. They will move as 2 objects, they bounce off each other therefore the initial equation applies.

M1V1i +M2V2i = M1V1f +M2V2f

What is worth noting however is that since the collision is elastic, there is no loss of energy, therefore the kinetic energy that was there initially is the same as the kinetic energy that is there after the collision. The energy is simply transfered from 1 mass to the other mass, to make 1 mass move faster and 1 slower.

The formula for kinetic energy is 1/2mv^2 therefore if we apply this to our original equation, it becomes

(1/2)M1V1i^2 +(1/2)M2V2i^2 = (1/2)M1V1f^2 +(1/2)M2V2f^2

Ei = Ef ; I was going to write a lot more, but I’ll spare you the rest I’m not sure you need it.

Anyway, the only thing that happens when you go from 1D (what I talked about) to 2D is that you have a y component as well as an x component and not just an x component.

So if you had M1V1i +M2V2i = M1V1f +M2V2f

You now have M1V1ix +M2V2ix = + M1V1fx + M2V2fx and M1V1iy +M2V2iy = M1V1fy +M2V2fy

All you need to take into consideration when it comes to x and y is the same thing as with forces: you have an x component and a y component and you can treat each independently. You just have to draw out your diagrams, establish your angles and use the sin and cos.

Let’s say you have a mass 1 and a mass 2. Mass 1 is in motion towards mass 2 and collides with it elastically. This will mean each mass will go in its own direction at an angle theta and phi with a certain velocity f1 and f2 respectively. Assume the angle theta is going up and phi down so tha the ball m1 goes up and ball m2 goes down.

So you have the x component where

M1Vi = M1V1fcostheta + M2V2fcosphi

And a y component where

0 + 0 = m1v1fsintheta + m2v2fsinphi

If the collision is elastic, the energy is conserved therefore

(1/2)m1vi^2 = (1/2)m1v1f^2 + (1/2)m2v2f^2.

You now have 3 equations to figure stuff from. All you need to do is pull some algebra tricks to isolate the variables you need from the ones you’re given and plug.

Woo. I love you Sin. =D

I don’t remember the problem exactly but it went something like this:

A 0.2kg cue ball hits another ball with the same mass at a 60 degree angle with an initial velocity. Then one ball when up at a 60 degree angle (relative to the straight line the cue ball was traveling prior to the collision) and the other ball went down at like a 60 or 30 degree angle, I think.

Mini-rant: Whoever wrote my Physics textbook just arbitrarily threw numbers in the answer section of the book. I’ve done a number of exercises in it and had them checked and everything to be right, but the answer given in the book always ends up being some completely odd ass number that didn’t even return the given values when I plugged them into the equations. -_-’

sig figs?

Remember to choose a consistent coordinate axis and be careful of positives and negatives, and of using sine/cosine correctly.

If you are learning about kinetic energy at the same time, remember that kinetic energy is only conserved in <i>perfectly elastic</i> collisions.

I’ll post a problem or 2 later this afternoon when I come back from class.

Thanks Sin, that’d be alot of help.

My Physics book is…uhm, I don’t remember. I think it’s a Houghton Muffin (or whatever the publisher’s name is) textbook, but I’m not sure; I usually don’t bring it home.

I use Serway and it gets the job done.

Here’s a simple 1D example:

Some baseball meatead wishes to maintain his physical condition during the winter. Evidently, he is too stupid to realize it is too fucking cold. He uses a 50.0kg pitching machine to help him, placing the machine on a thin layer of ice so that friction between the ground and the machine is negligible. The machine fires a .15kg baseball horizontally with a speed of 36m/s, straight towards his head. What’s the recoil speed of the machine from the throw (not the speed of the ball after it collides with his head):

m1v1i + m2v2i = m1v1f + m2v2f

v1i and v2i because the machine and ball are stationary before the pitch. Therefore

0 = m1v1f + m2v2f.

If the ball is m1 and the machine is m2 and the ball’s speed v1f is 36m/s, then

v2f = -(m1v1f)/m2
v2f = -(.15kg x 36m/s) / 50.0kg
v2f = -.011m/s.

We assume the machine can move backwards because it is on a frictionless surfance. If you draw a force diagram, you will not have a forward force to counter the force pushing the machine back as the ball is thrown forward. Newton’s second law is that for every action, there is an equal and opposite reaction. This is that in action.

Let’s do an inelastic collision problem:

Some dude in a fat ass 1800kg car is sitting at a traffic light drinking a beer. Some asshole in a 900kg car rams him up his fat ass at 20m/s, causing the fat ass to drop his beer all over himself and begin yelling expletives at said asshole. The cars become an entangled fucking mesh. What is the speed of the mesh after the collision?

This is inelastic because the 2 objects that collided moved as 1 mass after the collision.

m1v1i + m2v2i = (m1+m2)vf ;

(1800)(0) + (900)(20) = (1800 + 900)vf

[(900kg)(20 m/s)]/(2700 kg) = 6.67m/s

Let’s do a 2D example now that we’ve done 1D. Its the same as 1D except now you have the y and x components to deal with separately and you do so by using sin and cos.

Let’s do a nice problem:

2 cars coming towards an intersection at 90 degree angles: 1 is going east, the other is going north. Car 1 has a mass of 1500kg and is going at 25.0 m/s. Car 2 has a mass of 2500 kg and is going at 20m/s. Whm at’s their final velocities when they crash and undergo a perfectly inelastic collision and at what angle do they fly off after crashing?

m1v1x + m1v1y = 1500kg x 25.0 m/s + 0 = 3.75 x 10^4 kg m /s + 0 in the x direction.

m2v2x + m2v2y = 2500kg x 20.0 m/s + 0 = 5.00 x^5 kg m/s + 0 in the y direction

So after the crash,

m1v1x = 4000 vfcostheta

and

m2v2y = 4000vfsin theta

if we divide those 2 by each other we get

m2v2y/m1v1x = 4000vfsin theta/4000vfcostheta

4000 and vf cancel out so

m2v2y/m1v1x = tantheta.

theta = tan^-1 ( 5x10^4 / 3.75x10^4) = 53.1 degrees.

Because m2v2y = 4000vfsin theta

vf= m2v2y / 4000sin theta

vf = 5x10^4 / (4000 sin 53.1) = 15.6m/s.

Voila.

Let’s do a mean problem:

We now live in the happy world of molecules. 2 protons get really excited and decided to show each other some much needed affection. But they’re not very smart protons and proton 1 doesn’t realize that he will collide elastically with proton 2, which is initially at rest. Proton 1 is moving at 3.5 x 10^5 m/s (he is THAT excited). After the collision, one proton moves off at an angle of 37 degrees to the original direction of motion and the 2nd deflects at an angle of phi to the same axis. What is the final speed of the 2 protons and what is the angle phi?

So now we use those 3 funky equations I gave you earlier:

So you have the x component where

M1Vi = M1V1fcostheta + M2V2fcosphi

And a y component where

0 + 0 = m1v1fsintheta + m2v2fsinphi

If the collision is elastic, the energy is conserved therefore

(1/2)m1vi^2 = (1/2)m1v1f^2 + (1/2)m2v2f^2

Firstly, we can take out mass and the 1/2 because mass and the 1/2 are everywhere; that leaves us with

the x component where

V1i + 0 = V1fcostheta + V2fcosphi

And a y component where

0 + 0 = v1fsintheta + v2fsinphi

And the energy equation:

vi^2 = v1f^2 + v2f^2

Let’s plug in some values:

3.5x10^5 = v1fcos37 + v2fcosphi

0 = v1fsin37 + v2fsinphi

(3.5x10^5 m/s)^2 = v1f^2 + V2f^2
1.2x10^11 m^2/s^2 = v1f^2 + V2f^2

Now we rewrite the first 2 equations so that

v2fcosphi = 3.5x10^2 - v1fcos37
v2fsinphi = v1fsin37

Now we square each and then we add them such that

v2f^2cos^2(phi) + v2f^2sin^2(phi)= v1f^2sin^2(37) + v1f^2cos^2(37) + 1.2x10^11 m^2/s^2 - (7.0x10^5 m/s)v1fcos37

v2f^2 [( cos^2(phi) + 2sin^2(phi)] = v1f^2[ (sin^2(37) + cos^2(37) ]+ 1.2x10^11 m^2/s^2 - (7.0x10^5 m/s)v1fcos37

sin^2(phi) + cos^2(phi) = 1

v2f^2 = v1f^2 + 1.2x10^11 m^2/s^2 - (5.6x10^5 m/s)v1f

Now we can substitute this into the energy equation:

v1f^2 + (v1f^2 + 1.2x10^11 m^2/s^2 - (5.6x10^5 m/s)v1f) = 1.2x10^11 m^2/s^2

2v1f^2 - (5.6x10^5 m/s)v1f = 0

2v1f = 5.6x10^5 m/s

v1f = 2.8x10^5 m/s

If we plug v1f into the energy equation vi^2 = v1f^2 + v2f^2 and solve for v2f, we get 2.1x10^5 m/s.

To get the angle phi, we solve for phi in 0 = v1fsin37 + v2fsinphi

phi = sin^-1 (v1fsin37 / v2f) = 53 degrees.

That’s awesome, Sin. Funniest problems I’ve ever read. And they helped, too. Thanks a lot. =D