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## Physics library

### Course: Physics library > Unit 6

Lesson 2: Elastic and inelastic collisions# Solving elastic collision problems the hard way

In this video, David shows how to solve elastic problems the hard way. In other words, using conservation of momentum and conservation of kinetic energy, David substitutes one equation into the other and solves for the final velocities. Created by David SantoPietro.

## Want to join the conversation?

- Is it impossible for an object to come to a complete stop after an elastic collision?(5 votes)
- If a ball of mass m and velocity v hits another ball of mass m but without any speed, the first ball will come to a complete stop due to the elastic collision, while the second ball acquires speed v ( assuming they collided with their centres aligned with the velocity).(12 votes)

- I know that it's not possible for an elastic (or "perfectly elastic") collision in nature. However, is it possible for a perfectly inelastic collision to occur?

Why, or why not?(3 votes)- Yes, when objects stick together after the collision, that's perfectly inelastic. Clay balls can collide and stick together, train cars link together, paint balls go splat, etc.(6 votes)

- When we rewrote Vg in terms of Vt, why didn't we substitute it directly in the momentum equation instead of the kinetic energy equation?(2 votes)
- Let's try(omitting the units):

(0.07 - 0.058*V_t)/0.045 = V_g

1.56 - 1.29 * V_t = V_g

Insert it:

0.07 = 0.058 * V_t + 0.045 * V_g

0.07 = 0.058 * V_t + 0.045 * (1.56 - 1.29 * V_t)

0.07 = 0.058 * V_t + 0.07 - 0.058 * V_t

0.07 = 0.058 * V_t - 0.058 * V_t + 0.07

0.07 = 0.07(5 votes)

- Could you have found an expression for Vg using the KE formula, and then used it to solve the momentum equation, rather than the other way around?(1 vote)
- I was given the formula at school as (m1*v1)+(m2+v2)=(m1*f1)(m2*f2) how do I use this?(2 votes)
- well, first step: you should really ask your teacher and explain that you are not sure what it is for.

It looks like the equation for conservation of momentum. You can use it to work out the speed of objects after they collide or explode for example... but speak to your teacher first and see if she can give you some exercises to try.

OK??(0 votes)

- What if you had two balls with the same mass, but one ball bounces off and one ball sticks on a block. How can you tell which direction the block moves after the collision?(1 vote)
- how can we know whether the two balls after collision will move in same direction or opposite? Both in Elastic & Inelastic collision?(1 vote)
- You'd have to work out the momentum before and after the collision. Without knowing the masses of the two balls and their velocity before the collision, plus the trigonometry of the collision, it's impossible to say what the outcome of the collision will be.(2 votes)

- For a collision to be accepted as elastic, how close do the kinetic energies have to come to become equal to one another? I understand that nothing is perfect, so to be known as 'elastic', do the kinetic energies differ by 0.01 or 0.1 J, etc.?(1 vote)
- That’s arbitrary. An elastic collision is defined as one in which kinetic energies(initial and final) are equal. However, if the difference in energy is insignificant compared to the total final and initial energies, we can say that the collision is elastic for the sake of the experiment.(1 vote)

- If it weren't an elastic collision could we do it? (Without knowing anything but what it is know in this one)(1 vote)
- Yes we could, in case we were given the fact that it is perfectly inelastic collision. Because if it is inelastic collision then we know that final velocities are same and it would be pretty straightforward.(1 vote)

- you assumed almost no time while collision as a provision for momentum conservation by avoiding external impulse due to external forces, and said "like gravity". why do we concern about the impulse due to gravitational force while it acts perpendicular to the velocity and momentum direction which implies that it will not affect our velocities and momentum in the x direction. even the kinetic energy in the x direction will not be affected. actually i think that this duration of time is required for momentum conservation, it is the time where momentum redistributed or partially interchanged between the collided objects. and eventually we are concerned with the impulses due to forces in the same direction of our interest, you made emphasis on the one direction in our example here, again, is gravitational force live in our dimension?(1 vote)

## Video transcript

- [Narrator] So, I looked up
the mass of a tennis ball. And it turns out a tennis
ball is about 58 grams or point o five eight kilograms. And I wondered, if we
shot that tennis ball to the right, straight toward a golf ball, and I looked up the mass of a golf ball. A golf ball's about 45 grams or point o four five kilograms. So if we shot these balls straight toward each other, at a certain speed, let's say the golf ball's moving around 50 meters per second. That's pretty fast. That's over 100 miles an hour. And we shoot the tennis ball to the right, at a speed of 40 meters per second, so that these balls collide. And they collide head on. In other words, I want them to collide and stay in this single direction. I don't want a glancing collision, where the golf ball goes flying up this way, or something like that. Let's not do that. So it's all gonna happen in one dimension. And my question is this, just given the initial velocities and the masses, can we figure out the final velocities of the golf ball and the tennis ball? Let's try it. So how can we start? When I'm doing a collision problem, I typically just start
with conservation momentum. I just know, if it's
gonna be a quick collision the momentum right before the collision should equal the momentum right after the collision. At least the total amounts. Why is that true? It's because this golf ball, the time that it's actually in contact with the tennis ball,
it's gonna be so small that any external forces
that might be there, like gravity, are gonna
have so little time to act on the system, the external forces can't really impart a large amount of external impulse. And if there's no external impulse, the total momentum of our system, golf ball and tennis ball, has to stay constant. So, because these collisions happen, typically, over a very
short time interval, we're just gonna say, the
momentum right before total and the momentum right after total is gonna be the same. And that goes for basically any collision between two freely moving objects. You can just assume the total momentum's gonna be conserved. So what will that mean mathematically? Well it's gonna be that
the total initial momentum, p is the letter we use for momentum, and the total, I'm gonna use Sigma to represent the total. This just means add up
all the initial momentum, not just the momentum
of one of the objects, but all the momentum of all the objects. And I'm even gonna put
a vector sign up here because momentum's a vector. That's important, because
momentum can be negative. So you can't forget the
negative signs in here. And if momentum's conserved, then this, initial, total momentum should equal the final total momentum. This is what we mean when we say, "Momentum is conserved." so remember, the formula for momentum is mass times velocity. So the initial momentum of the tennis ball would be mass times velocity. So that would be zero point zero five eight kilograms. Times the velocity,
initially, of the tennis ball is positive 40. And I'm gonna put a positive here to remind me that this is to the right. That's why I'm making it positive. The initial momentum of the golf ball would be also mass times velocity. So it'd be plus the mass of the golf ball is point o four five kilograms. And the initial velocity of the golf ball would be negative 50 meters per second. Because the golf ball
is moving to the left. And we're gonna assume
leftward is negative and rightward is positive. So if this is the total, initial momentum, and momentum's conserved,
this should equal the total final momentum. So the final total
momentum of the tennis ball is gonna be zero point
zero five eight kilograms times v final of the tennis ball. I'm just gonna call that V-T, for v of the tennis ball, plus the final momentum of the golf ball's gonna be plus zero point zero four five kilograms times the final velocity
of the golf ball's gonna be v, I'm gonna put V-G, for v of the golf ball. And that's gonna be the
velocity after the collision. So can I solve now for the final velocity of the tennis ball and the golf ball? No, I can't. I've got one equation
and I've got two unknowns sitting over here. So I'm not gonna be able
to solve for either of them if I've got two variables
in my single equation. In other words, I can add up this whole left hand side if I wanted to. If you add all this up you're gonna get zero point zero seven
kilogram meters per second, is your total, initial momentum. Then if I solve this
over here I'm gonna have equals two unknowns. I don't know V-T and I don't know the velocity of the golf ball either. So I need at least one
more piece of information. I need to know, for instance, I knew one of these final velocities. I could easily solve for the other. So if the problem gave
me the final velocity of the tennis ball. Well, I can plug that number into here and just solve, then for my final velocity of the golf ball. Or the problem could tell
you that this collision, what type of collision is it? If it tells us that they stick together. Then I can assume that they both move off at the same velocity. That would be a perfectly
inelastic collision. So if it was a perfectly
inelastic collision, I'd just have equals
one big mass over here. In other words, point
o five eight kilograms. Plus the mass of the golf ball. Point o four five kilograms. It'd be one big mass because
they'd stick together in a perfectly inelastic collision. Times just one final velocity, because they're both moving
at the same velocity. So if you take this point o seven, divide by my total
mass, that would give me the final velocity of
the two balls combined. But that's unlikely. These balls aren't gonna stick together. I mean, a golf ball and a tennis ball, unless you've got some sort of adhesive on the front of them, I don't think these are
gonna stick together, that seems unlikely. So let's assume that doesn't happen. But if you were told they stick together, in a collision, two masses, that's what you could do. It's much more likely,
that if you're dealing with a golf ball and a tennis ball, that you're gonna be told that this collision was elastic. And remember, elastic means that the total kinetic energy in this collision is gonna be constant or conserved. You're not gonna lose any
of that kinetic energy to any thermal energy or sound. And it turns out, just being told this, that the collision is elastic is enough to solve for
these final velocities. And the reason is, this is implying the kinetic energy is conserved. So we can use that to our advantage. We can just say, "All right, not only "is momentum conserved now,
but if we say it's elastic, "that means the total amount of "kinetic energy is conserved." so the initial, total, kinetic energy has to equal the final,
total kinetic energy. And remember, kinetic
energy is 1/2 M-V squared. So I can say that, all right, 1/2 point zero five eight kilograms, the mass of the tennis ball. Times it's initial velocity
was 40 meters per second. You can't forget to square it, kinetic energy's 1/2 M-V squared. Then I do plus the initial kinetic energy of the golf ball's gonna be 1/2, mass of the golf ball was
point o four five kilograms. We multiply by it's initial speed squared. And I'm just gonna do positive 50. Because we're gonna square this. This is just the speed in kinetic energy. It doesn't matter if you
make it positive or negative. Over here, it definitely
matters in momentum, whether you make it positive or negative. But since you're squaring it. And since kinetic energy's a scalar it can't be negative, doesn't matter whether you put the positive
or negative in here. It's gonna go away when you square it. We can say that this total,
initial kinetic energy should equal the total,
final kinetic energy. So I can say that, this total amount here should equal, I'm just gonna put the equals sign down here, the final kinetic energy
of the tennis ball would be 1/2 point o five eight kilograms. Times the final velocity
of the tennis ball squared. And then I have to add to that the final kinetic energy of the golf ball. Which is gonna be 1/2. Mass of the golf ball is point o four five kilograms. Times the final velocity
of the golf ball squared. And you might be like,
"How does this help us?" Look at how horrible this looks. These are squared. How's this gonna help me now. Well, now you can solve. Because I've got two equations. And I've got two unknowns. And the two unknowns over here are the same as the
two unknowns over here. So whenever you have two equations and two unknowns, you can solve for one of your unknowns. You can actually solve
for both of your unknowns. First you're gonna solve
one of the equations and then substitute into the other. And we'll get one
equation with one unknown. In other words, let me
show you how that works. Let me clean up this side over here, this left hand side, which is kind of like the upper side right here. So if I add up all this initial, kinetic energy, over here. I get 102 point 65 joules of initial, total kinetic energy. But I've still got two
unknowns in this equation. So what I'm gonna do is
I'm gonna come over here. Let's just solve this for V-G. If I solve this for V-G, I'll subtract point o five eight V-T from both sides, point o four five V-G. And now I can divide both sides from point o four five. Over here, point o seven divided by point o four five, is equal
to one point five six. And point o five eight divided by point o four five, is equal
to one point two nine. And then this is multiplied by V-T. That's what's equal to V-G. So I have an expression for V-G. The final velocity of the golf ball is equal to this quantity right here. One point five six minus
one point two nine V-T. So I'm gonna take this total expression, which is equal to V-G, and I'm gonna plug it in right over here. Which gives me 1/2 point
o four five kilograms times the quantity, one point five six minus one point two nine V-T squared. Because this V-G was squared. And I'm just substituting the expression I have over here for V-G in for this quantity V-G. And I still have to multiply by the 1/2 and the point o four five. And I still have all of this. So I still have 102 point 65 joules equals 1/2 point o five eight kilograms times V-T squared. Plus this quantity right here. Which is what I
substituted in the V-G for. And it's getting a little messy. But at least I now have one equation with just one unknown. I just have V-T in here. So if I do the math I
have 102 point 65 joules equals, if I just take point o five eight divided by two, I'm gonna
get point o two nine and V-T squared. I'm gonna leave off the units. Things are gonna get messy. And then if I take point
o four five divided by two I'll get point o two two five. And now I've gotta square this quantity. So if you remember, if you
ever have a minus b squared, the result of that is gonna be a squared, which is one
point five six squared. Minus two, times the
quantity of the first one, one point five six times the quantity of the second one, which
is one point two nine V-T. And then, plus, the final
element here squared this b squared. Which is gonna be one
point two nine squared times the velocity of
the tennis ball squared. That may have made no sense at all. If so, what I'm really doing is I'm saying that if you ever have a minus b squared, that's just equal to a squared minus two a b plus b squared. And that's what I did. Here's my a. I did a squared, one
point five six squared. And I did minus two times this first one times the second one,
with the V-T in there. Plus b squared is gonna be
plus this final term squared. Is one point two nine
squared times V-T squared. So I've got this big mess now. I just need to clean it up. The left hand side is still 102 point 65. I've still got this point o two nine V-T squared sitting here. But I need to multiply
this point o two two five throughout this whole quantity. Because it's multiplying
this whole quantity. So if I do that, I've got
plus point o five four eight. That's what? Point o two two five times one point five six squared is. Then I'll get minus
point o nine o six V-T. That's what point o two two five times this whole quantity is. And then, finally, I'll get plus point o three seven four V-T squared. That's what point o two two five is times this quantity right here. So let's identify the V-T's. I got a V-T right here, just single V-T. And then I've got a V-T
squared, right here. So I can combine this V-T squared term, with this V-T squared term. And I'll get point o six six V-T squared minus point o nine o six V-T, plus point o five four eight. Now we're getting close, I promise. If we subtract, there's 102
point 65 from both sides. We'll have zero equals
this whole quantity again. And then point o five four eight minus 102 point 65. Is gonna be negative 102
point five nine five. And what this is right here, is the Quadratic Equation. And you can't solve this by just trying to isolate V-T on one side. It's never gonna work that way. You've got to use the Quadratic Formula. So in the Quadratic Formula, this term here, the
point o six would be a. And this negative point
o nine o six would be b. And this negative 102 point five nine five would be the c. You could either do this
the long way by hand. Or you could just use a
Quadratic Formula Solver. They're available online. They might be on your calculator. That's what I'm gonna do. I'm gonna do this on my calculator. So the two answers I'm getting out of this would be, V-T either equals, I'm
getting 40 as one answer, meters per second. Or I'm getting negative
39 meters per second. And so which one is it? Is it gonna be 40 or negative 39? Well, we can figure out which one it is. Look at this V-T here, 40? That's the initial
velocity it had already. We want the final velocity. Why is it giving us the
initial velocity again? Because it turns out, one way to conserve momentum and energy, is for these objects to just miss each other. And fly right past each other. If the golf ball doesn't actually collide with the tennis ball. And the tennis ball just
keeps going forward, they just both maintain whatever velocity they had initially. And that would correspond to this. This is a collision that missed. So we know that this collision was not the one we're looking for. That's if they didn't collide. We're looking for this
velocity right here. So, after the collision, this tennis ball gets knocked backward, with negative 39 meters per second of velocity. So how do we find the velocity of the golf ball after the collision? Well I've got the velocity
of the tennis ball. Now all I have to do is bring
that right back into here. And I can get what the
velocity of the golf ball was. The velocity of the golf ball's now just gonna be one point five six minus one point two nine. Times this quantity, negative 39. And if I'm gonna multiply this out, I'm getting about 52 meters per second. Positive 52 meters per second, for the velocity of the golf ball. That means this golf ball got knocked back to the right, Because it's a positive velocity. And it got knocked out at a speed of 52 meters per second. So, recapping what we did, we were given the initial
velocities and the masses. We tried to use Conservation of Momentum and that was fine. Except we had two unknowns. So we had to write down another equation. If we're told this collision is elastic, we know that total kinetic
energy's conserved. We wrote down that equation but it also has two unknowns. So we solved the momentum equation for one of the variables, V-G. We substituted that
expression into over here, for the V-G in this kinetic energy. We squared it, we had only one equation, with one unknown. But unfortunately, it gave
us a Quadratic Equation. So we used the Quadratic Formula to solve. One of the velocities corresponded to the same as the initial velocity the object had in the first place. We don't want that one. Because that would mean that they didn't collide at all. We take the second one, if we wanna find the
velocity of the first object. Then we take that, plug that back into this expression here. We get the velocity of the other object.