Elastic Collision: Example

by Jared Rovny

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    00:01 Let's do an example of an elastic collision, where these two objects bounce off of each other, where we have a 200-kg bumper car -- that's our elastic collision for you -- moving at 30 meters per second and colliding with a stationary 400-kg bumper car, and we would like to know the velocity of these two cars after the collision.

    00:21 So this is a classic case where we look at the variables beforehand and try to use physical intuition to predict what will happen in the future if these two things collided.

    00:29 I'd like you to give this problem a shot, but be careful, it could be a long problem -- an involved problem, once we get into how to solve the equations that results from our analysis of the collision.

    00:41 So give it a try, and then we're going to do it here as well.

    00:45 So were going to try this problem first by writing out the entire scenario.

    00:49 What I recommend before we jump into this, since this might be one of the longest examples we actually do, is to grab yourself a cup of coffee, sit down and be prepared, and just watch how we both derive what the equations are that we need to solve and then how we solve those equations.

    01:03 We'll think about those as two sort of separate processes.

    01:06 So, first, let's write out what happens in this problem.

    01:10 We have an object.

    01:11 It's moving along with some velocity, in this case 30 meters per second, and it collides with a stationary object.

    01:17 So this first object has a mass of 200 while the stationary object has a mass of 400.

    01:24 I'm writing everything in SI units -- kilograms and meters per second.

    01:30 So these are our objects and this is our situation here.

    01:32 We have two masses with different masses.

    01:38 And then the velocity of the first one, which is non-zero, 30 meters per second but the velocity of our second object initially is 0.

    01:46 So this is going to be important.

    01:48 So let's first write out our momentum and our energy.

    01:52 So initially, our momentum -- and this is all just in the X directions, so let's just write the X component of our momentum, since we're just doing one dimension is equal to M1V1, so the momentum of the first car plus M2V2, the momentum of the second car.

    02:09 So this is our initial momentum.

    02:11 So I'll just write initial "i" right here.

    02:14 So then, what we could do is write the final momentum: Px final is equal to-- -- and let's write a final version of this scenario.

    02:24 You have these two objects after the collision and they're moving with their own velocities which we don't know.

    02:29 We're going to call these velocities, V1 prime and V2 prime, since we don't know what these velocities are and they are our final velocities.

    02:37 So writing the final momentum of this situation on the right, what we have is: M1V1 prime + M2V2 prime.

    02:48 And we know that these must be equal to each other by conservation of momentum.

    02:52 The first thing we should do here is say that the second object initially, this is not moving.

    02:57 So its velocity is zero.

    02:58 So we can just get rid of that immediately.

    03:00 The second thing we'll do is in an elastic collision, as we have here, is we'll see the initial energy (Ei) is equal to the final energy (Ef).

    03:09 So we use conservation of energy and momentum in an elastic collision problem.

    03:14 So the initial energy is just the kinetic energy of your two objects: 1/2 [(M1V1 squared) + 1/2 (M2 x V squared)] and the final momentum will look exactly the same except we're using our prime coordinates, V1 prime squared, because these are our new velocities.

    03:35 The next thing we can do with this energy equation that we've written, is because the velocity of the second object is zero, we can also get rid of this term as well because that object isn't moving.

    03:46 So the next thing we'll do is just apply our conservation of momentum and conservation of energy.

    03:51 We'll say that these two are equal, initial and final momentum.

    03:54 We'll say that these two are also equal the initial and final energy.

    03:58 So doing that down here, we have: M1V1 = M1V1 prime + M2V2 prime And writing conservation of energy, we can equate this term with a slower term.

    04:13 We now have: 1/2 M1V1 squared = 1/2 M1V1 prime squared + 1/2 M2V2 prime squared And these are the equations that we need to solve.

    04:29 We have two equations and two unknowns, the unknowns being the new velocities, V1 prime and V2 prime.

    04:35 So its important for you to understand at this point, that we've already done most of the physics for this problem.

    04:40 We've written down momentum initially.

    04:42 We've written down momentum finally, and done the same for energy.

    04:45 And then applied conservation on momentum and conservation of energy in order to try to solve and come up with equations that we can solve for the two variables that we're trying to find in this problem.

    04:55 So for the sake of completeness, I think we should do this next because this is exactly the kind of situation we might need to do a complicated equation to find a variable where you have squared quantities.

    05:06 So let's do that.

    05:08 Again, what we're looking for is V1 prime and V2 prime, where we have, again, two equations and two unknowns.

    05:15 There are a number of ways to solve a set of equations when you have them like this.

    05:19 One way is that we find or we solve each equation for V1.

    05:23 So we just take this V1 prime term and this V1 prime squared term.

    05:27 We rewrite each equation just in terms of this quantity.

    05:31 If we do that, we know that the two quantities that we come up with must be equal to each other.

    05:35 So, for example, let's take this first equation: M1V1 = M1 prime + M1V1 prime + M2V2 prime and solve it for V1 prime.

    05:47 We see that V1 prime is: 1/M1 [M1V1 - M2V2 prime] We can do the same thing with this energy equation, which we'll do here.

    06:01 We see that, V1 prime squared equals-- and then we should have -- first of all, multiplying both sides by 2, just to simplify this.

    06:12 We then have: V1 prime squared = M1V1 squared - M2V2 prime squared, all divided by M1.

    06:25 So make sure this next step -- just rewriting these two equations -- is something that you can do.

    06:30 So we've rewritten our equations here, and the reason we've done this is because these quantities, V1 prime, we can equate them.

    06:39 We need to square this first equation in order to do that.

    06:41 So this equation we're going to square and then these will be equal to each other and we can write them just like that.

    06:47 So let's do that with square of this first equation and then recognize that it will be equal to the second equation since they'll be both be equal to V1 prime squared.

    06:54 In that case, we have: (M1V1 - M2V2 prime) / M1 -- and we're going to square both of these.

    07:07 Square the entire quantity, which looks like this: So, this is equal to V1 prime squared.

    07:13 But this is going to be equal to this expression down here, because we know what V1 prime squared is equal to.

    07:19 So this is: M1V1 squared - M2V2 prime squared all divided by M1.

    07:29 So now we have to get into the final messy details of this problem.

    07:33 We just need to rearrange this equation until we have V2 prime.

    07:37 The first thing that I can do is to multiply both sides by this M1 prime squared, and then I'm going to write out these full squared terms.

    07:43 This will be the only really bad part of this problem, so stick with me for this last bit, and then we'll watch it simplify right in front of our eyes.

    07:50 So that will be the more gratifying part.

    07:52 So let's first do this.

    07:53 We have: M1V1 -- and we have to square this.

    07:58 So, this is: M1 squared x V1 squared.

    08:01 And then were gonna to do the cross term.

    08:03 This is going to come from algebra: -2 M1M2 V1V2 prime + and then our last term is squared and that term is: M2 squared V2 prime squared equals and then we multiply both sides by M1 squared, so let's do that.

    08:26 That will cancel out one of these and now this is equal to: M1 ( M1V1 squared - M2V2 prime squared) So, we're almost done here.

    08:38 I did one thing that you may or may not remember from algebra, which is that I took the squared term, and anytime you have something that looks like A - B and you're squaring it, you should remember that you always get: A squared - 2AB + B squared And there are few methods that we have in algebra for coming up with these results So I recommend you review all the basic algebra because those skills will be useful, when we're trying to solve variables, especially in long problems like this one.

    09:05 So now, we're nearly done.

    09:07 We get to do the fun part now which is the simplifying where we get to get rid of many terms.

    09:11 First, notice that this first term is M1 squared V1 squared.

    09:15 That's the same as this first term: M1 x M1 = M1 squared M1 x M1 is M1 squared.

    09:19 So M1 squared, V1 squared so these two terms go away by subtracting them from both sides.

    09:25 And now, we get to just rewrite this whole equation, where, again, we're trying to solve for V2 prime.

    09:32 So this will be what we solve for: So we'll just rearrange by adding M2V2 prime squared to both sides.

    09:38 So we'll then get something that should look like: (M2 squared + M1M2 x V2 prime squared) and this came from this term, M2 squared and this term, M2M1 So we've taken care of these two terms.

    10:05 Minus from this negative sign 2 M1M2V1 x V2 prime So you see what we're doing here, and the only thing we have left there is 0 because I moved everything over to the left side.

    10:23 So again practice your algebra here.

    10:25 Try to do this on your own and see if you can get the same results.

    10:28 There's a lot going on but you can see what I've done is I've moved everything over and written it all in terms of V2 prime because that is exactly what we're looking for.

    10:36 Everything else here is just details.

    10:37 These are just numbers that we can plug in when we're done.

    10:40 So the last step then we're pretty well done here, is we factor out V2 prime and then we have an equation that is: (M2 squared + M1M2) V2 prime - (2M1M2V1) = 0 and this gives us a solution.

    11:04 Any time you have two quantities being multiplied by each other, in this case, V2 prime times this big long expression that we have next to it, you can know that one of those two numbers must be zero, because if I tell you, you have two numbers when you multiply them together, you get zero, You could tell me that: "Well, one of those two numbers must be a zero because if you multiply them together that's the only way to get zero." So, this factoring technique that I just showed you here is very, very common for any problem where you have squared quantitities.

    11:31 Like this V1 squared or V2 prime squared as we have here.

    11:35 So, let's do that really quickly.

    11:36 We see that the math is telling us that if the laws of phyics we told are true, we have one of two options.

    11:41 We can say that V2 prime is zero from this term, or where we can do the opposite thing and say that this whole term is equal to zero because again one of these two has got to be zero.

    11:52 So either the second object, -- this object here that initially wasn't moving -- is also finally not moving.

    11:59 It just stayed still through this entire thing.

    12:00 In other words, what the math is telling us here is it says: "if you want these laws of conservation of momentum and energy to be true, one of your options is that V2 prime is zero the velocity of the second object stays zero but that would correspond to the first mass going right through the second mass and never actually impacting it.

    12:18 The mathematics will sometimes deliver you these kinds of answers because the mathematics has no idea what's going on here.

    12:24 It just says: "Here's something that would conserve energy and momentum." And it is true.

    12:29 It would conserve energy and momentum if we let one object go right through the other one, but it's not a physical scenario.

    12:35 This example is one of many times where you'll use the mathematics to try to tell you something about the physics and see a non-physical result and you'll have to use your intuition always to understand the physics of the problem and what's going on, to pick the right answer.

    12:51 In this case, we can say that this is a non-physical answer.

    12:54 There's no way that the objects are going through each other and so now we have our final answer, which comes from setting this equal to zero.

    13:00 If this is equal to zero, I can add this term to both sides, and then divided by this first term and we have: V2 prime = (2 x M1 x M2) / (M2 square + M1 x M2) x V1 So, in other words, the final velocity of the second object is equal to something and has to do with the masses times the velocity of the first object originally.

    13:29 The last step in this problem would be a very easy step because we have to find the velocity of the first object after it's run into and hit the second object.

    13:39 but this is very easy, we don't have to do any of this analysis anymore because we already have a simple equation relating V1 prime and V2 prime.

    13:48 This comes from our conservation of momentum.

    13:49 So what we have to do here, is for the equations that we have on the left here.

    13:55 You see this conservation of momentum equation.

    13:58 We can very easily see, how V1 prime and V2 prime are related.

    14:02 So we're going to do our last step over here.

    14:05 Specifically, we can see just by rearranging this momentum equation here.

    14:10 The V1 prime has to be equal to M1V1 minus M2V2 prime divided by M1.

    14:23 Since we know all these quantities that I've written here, we can solve this problem just by putting in our numbers.

    14:30 So let's first do this with V2 prime.

    14:32 This will be equal to: 2 x the first mass, which is 200 times the second mass which is 400, all divided by the second mass squared, which is 400 squared plus the product of the two masses -- in this case, this is M1 x M2, which is 200 x 400.

    15:04 And now what we need to do is simplify.

    15:07 One thing that you might be able to notice here is we can cancel this 400 immediately, which we could have also done immediately over here by cancelling one version of M2.

    15:17 And then we can solve this by saying: 2 x 200 is 400, divided by (400 + 200) which is 600 So, we also have to multiply, don't forget, by your velocity V1.

    15:31 So, we multipy this four-sixths by our velocity which is 30 and then we can do our final step, which is: 4 x 5, after cancelling 30 with 6, 20 meters per second.

    15:43 So this is V2 prime and it's very easy to calculate V1 prime.

    15:47 It's nowhere near as difficult since we've already done all these work, which will just be: M1 which is 200, times V1 which is 30, minus M2 which is 400, times V2 prime, which we just found to be 20, all divided by M1 which is 200.

    16:03 Cancelling this 200 with these two, so this is 400, we now see that the V1 prime, the velocity of the first object afterwards is 30 - 2 x 20 (which is 40), thus equals -10 meters per second.

    16:20 And this is the velocity of our first object after the collision.

    16:23 What you can see here, is that the second object after being hit by the first object moves off with a velocity of 20 meters per second, where the first object bounced off and is now moving backwards in the negative direction, at a speed of 10 meters per second.

    16:37 So this was a very long and involved problem, and in fact, it's one of the longest, if not the longest problem, you would ever need to see because it involves solving an equation which had squared variables in it.

    16:49 I worked through all the messy details, just this one time, because I think it's important for you to see it, at least once, to sort of follow the logic and try to deal with the tricky things that pop up here and there with maybe one of these velocities can be zero, and how do we square this and equate these two equations.

    17:05 So we actually worked through the details here.

    17:07 I've tried to separate what the physics is, which was on the first left column, from what the mathematics are when you're trying to solve a very long and complicated problem like this.

    17:16 And of course, anytime you get into the weeds as you just saw me do problems are popping up and you can see maybe something is forgotten, maybe I didn't put a variable here and there.

    17:24 So be careful and just be slow and vigilant, as you go through problems like this because they can be sticky.

    17:31 But if you're just careful and slow and follow the logic of a problem it shouldn't be too bad.

    About the Lecture

    The lecture Elastic Collision: Example by Jared Rovny is from the course Momentum.

    Included Quiz Questions

    1. vA = 0m/s and vB = 6m/s
    2. vA = 2m/s and vB = 6m/s
    3. vA = 3m/s and vB = 5m/s
    4. vA = 2m/s and vB = 4m/s
    5. vA = 10m/s and vB = 16m/s
    1. 14.4m/s
    2. 24m/s
    3. 1.44m/s
    4. 25m/s
    5. 2.4m/s
    1. -5.2m/s and 0.8m/s
    2. -5.2m/s and 8m/s
    3. -8m/s and 4m/s
    4. -0.8m/s and 5.2m/s
    5. -0.8m/s and -5.2m/s
    1. The light object bounces off the target, maintaining the same speed but with opposite direction; the heavy target remains at rest
    2. The heavy object bounces off the target, maintaining the same speed but with opposite direction; the light target remains at rest
    3. The light object bounces off the target, with increased speed but with opposite direction; the heavy target remains at rest
    4. The light object bounces off the target, with increased speed but with the same direction; the heavy target remains at rest
    5. The light object bounces off the target, maintaining the same speed but with same direction; the heavy target remains at rest
    1. Speed of the shuttle will be doubled because the mass of the racket is larger than the shuttle
    2. Speed of the shuttle will be same after collision
    3. Speed of the shuttle will be quartered because the mass of the racket and the serving force is smaller
    4. Speed of the shuttle will be halved because the mass of the racket is lower than the shuttle
    5. Speed of the shuttle will be quadrupled because the mass of the racket and the serving force is larger

    Author of lecture Elastic Collision: Example

     Jared Rovny

    Jared Rovny

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