If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

Impulse and momentum dodgeball example

In this video, David shows how to solve for the impulse and force applied during a dodgeball collision using the impulse momentum relationship. Created by David SantoPietro.

Want to join the conversation?

  • leafers ultimate style avatar for user Cooper Lang
    Is this classified as an elastic collision?
    (17 votes)
    Default Khan Academy avatar avatar for user
    • purple pi purple style avatar for user Adi Pasic
      An elastic collision is one in which kinetic energy remains the same before and after the impact. The only way to determine whether this specific collision is elastic is the find the final and initial kinetic energy of the ball-face system. Since the formula for finding kinetic energy, KE = 1/2(m)(v^2), requires both mass and velocity, and neither the mass nor the initial velocity of the face are provided, it cannot be determined if this collision is elastic.
      (26 votes)
  • male robot hal style avatar for user Sualeha
    What is basically an impulse? (Apart from its mathematical definition).
    (14 votes)
    Default Khan Academy avatar avatar for user
    • starky seed style avatar for user Dishita
      Love the question dude!
      anyways, this is just how I've understood impulse intuitively, and I'm just an 11th grader studying in India,

      J is like the total increase/decrease in momentum over a given period of time (I know it's synonymous with J=change P, but read this through)
      basically, what do we mean by a force of, say, 5N?
      = The momentum is increasing by 5Ns per s, so the total increase in p over 2s = 5n*2 = 10N right?!
      so over t seconds, it would be 5t Ns.
      J = force acting over a period of time, what does this lead to, a change in momentum (over that period of time) hence the 2 have an equivalent formula.
      so for the above case, we can say a force of 5 N exerted over a time period of 2s causes an impulse of 10Ns which is basically the total change in momentum
      I know this is basic, but simplicity is (usually) the key,
      I hope this appeals to your imagination as well,
      if wrong, do let me know,
      hope this helps :D
      Onward!
      (3 votes)
  • female robot amelia style avatar for user Baldeep Arora
    What about the air resistance?Doesn't air resistance affects the impulse?
    (8 votes)
    Default Khan Academy avatar avatar for user
    • duskpin tree style avatar for user M.Sanathan Sai
      In most of the cases, while calculating such questions , Air resistance is usually neglected. If we consider air resistance, friction and such other non-conservative forces, we wont be able to apply conservation of energy / momentum in the problems as they make the system more complicated and prone to errors. If we are supposed to take air resistance under consideration, the value must be given in the question. Then we can calculate the force exerted by it and subtract it from the initial momentum. But still it will make the concept more complicated.
      (7 votes)
  • primosaur ultimate style avatar for user Raymond Muller
    Doesn't J stand for joules? What's the difference between J in impulse and J in joules?
    (4 votes)
    Default Khan Academy avatar avatar for user
  • hopper jumping style avatar for user Kartik Nagpure
    So the final answer of the Second question (at the bottom) is +150N or -150N ?
    (3 votes)
    Default Khan Academy avatar avatar for user
    • female robot grace style avatar for user Paulina Enck
      The answer would be -150 N because of the way the speaker formulated negatives and positives. The choice between which way is negative and which is positive is up to you, but just remember to be consistent. However, if the problem had asked for the magnitude impulse (hypothetically), the answer would always be 150 N, as magnitude means asking for the absolute value.
      (5 votes)
  • blobby green style avatar for user azhai21
    What happens if the ball impacts at an angle and leaves at the same angle, do we need to consider that since Impulse is a vector?
    (5 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user nateeshamccready
    Hi I thought you multiply the net impulse force 3kg m/s by the time interval 0.02s but it is divided to give 150N, am I missing something?
    (3 votes)
    Default Khan Academy avatar avatar for user
    • starky seedling style avatar for user deka
      if you want to get an impulse given you know the net force and time interval, you can multiply them
      : impulse = net_force * change_time

      but here we know the net impulse (impulse is not a force, by the way) and time interval. thus we use the same formula above but with a bit of modification
      : impulse = net_force * change_time
      dividing both sides with change_time
      impluse/change_time = net_force * change_time/change_time
      which is same as
      impluse/change_time = net_force
      #change_time/change_time = 1

      this is why we divide impulse (not a force, once again) by change_time to get a net force
      (1 vote)
  • cacteye blue style avatar for user RealPiPiper78
    At shouldn't it have been -5 instead of 5, as it comes back?
    (2 votes)
    Default Khan Academy avatar avatar for user
    • male robot hal style avatar for user Charles LaCour
      No, in the problem the horizontal direction to the right was chosen as positive. In the equation the 10 m/s is negative and the 5 m/s is positive.

      If you consider the initial direction of the balls movement, left, as positive then the 10 m/s is positive and 5 m/s would be negative.

      Either is correct but you need to keep the directions labeled the same throughout the problem.
      (2 votes)
  • female robot grace style avatar for user Katherine Tanner
    Couldn't you, if you had the mass of that person's head, calculate how fast their head moved away? Since momentum is conserved for collisions?
    (2 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user Nahapetyan.GorY9
    Could we answer the question"What was the average force on the person's face by the ball?" by means other than Newton's Third Law? I tried several ways but none worked.
    (2 votes)
    Default Khan Academy avatar avatar for user

Video transcript

- [Instructor] This person right here is about to play dodgeball. They're just unfortunately not gonna dodge the ball. It's gonna fly in, it's gonna bounce off their head. This may or may not have happened to you. I think this probably happened to me. It's been a long time since I played dodgeball. And although, unfortunate for this person, it's a wonderful opportunity, scientifically speaking, to talk about the impulse, momentum, force, time relationship, so let's do that. Let's put some numbers on here. So, we're gonna need to know the mass of the ball. Let's say this is a .2 kilogram ball, and we're gonna need to know some other numbers. Let's say the ball comes in at a speed of about 10 meters per second. So let's say it's comin' in at 10 meters per second and let's say it leaves at a speed of five meters per second. So it's probably gonna recoil with a little less speed than it came in with. It comes in with 10, leaves with five, and let's say that time, time right here, the time period that it's actually in contact with the person's face, let's say the time, when the ball is getting kind of compressed and then recoils and expands again, let's say the time that it's actually in contact is about .02 seconds, or about 20 milliseconds. So, knowing this information, we can ask all kinds of questions. One of them is: What was the impulse on the ball from the person? Now, the definition of impulse, we use the letter J for impulse, that always seemed a little weird to me. There's no J in impulse. I end up calling it jimpulse, just so I can remember that it's impulse, and there's a J for it. So the jimpulse, or the impulse, is defined to be the force acting on the object, multiplied by the time duration during which that force is acting. In other words, the impulse, from a force, is equal to that force, multiplied by how long that force was acting on the object. So, if we knew the force on this ball, we could use this formula to get the impulse, but we don't. I don't know the force that this person's face is exerting on the ball, so I can't use this formula to solve for the impulse. But, there's an alternate formula for impulse. If you're talking about the net impulse, in other words, the impulse from all forces on an object, like this ball, that should just equal the change in momentum of that object, like the change in momentum of this ball. So if we can figure out the change in momentum of this ball, we can figure out the net impulse on this ball. And since it's the net impulse, and this formula appears also true, this is equivalent, which is saying that it's the net force, multiplied by the time duration, during which that net force is acting. This is hard for people to remember, sometimes my students like to remember it as Jape Fat. So, if you look at this, it looks like J-A-P, this kinda looks like an E, F-A-T. So if you need a way, a pneumonic device, to remember this, Jape Fat is a way to remember how impulse, change in momentum, force, and time, are all related. So let's do it. We can't use force because we don't know it yet, but I can figure out the change in momentum 'cause I know the velocities. So, we know that the change in momentum is gonna be P final, the final momentum, minus the initial momentum. What's my final momentum? My final momentum is M times V, so it's gonna be mass times V final, minus mass times V initial, and my mass is .2, so I've got a mass of 0.2 kilograms. My final velocity is five, because the ball recoiled to the right with positive five. Positive five 'cause it's moving to the right. I'm gonna assume rightward is positive. Then minus, the mass is .2 again, so 0.2 kilograms. My initial velocity is not 10. This is 10 meters per second to the left, and momentum is a vector, it has direction, so you have to be careful with negative signs here. This is the most common mistake. People just plug in positive 10, then get the wrong answer. But this ball changed directions, so the two velocities here have to have two different sides, so this has to be a negative 10 meters per second, if I'm assuming rightward is positive. This leftward velocity, and this leftward initial velocity, has to be negative 10. And, if you didn't plug that in, you'd get a different answer, so you gotta be careful. So, what do I get here if I multiply this all out? I'm gonna get zero, no, sorry, I'm gonna get one kilogram meters per second, minus a negative two kilogram meters per second, and that's gonna give me positive three kilogram meters per second is the impulse, and that should make sense. The impulse was positive. The direction of the impulse, which is a vector, is the same direction as the direction of the force. So, which way did our face exert a force on the ball? Our face exerted a force on the ball to the right. That's why the impulse on the ball is to the right. The impulse on this person's face is to the left, but the impulse on the ball is to the right, because the ball was initially going left and it had a force on it to the right that made it recoil and bounce back to the right. That's why this impulse has a positive direction to it. Now, if you've been paying attention, you might be like, wait a minute, hold on. What we really did was we found the change in momentum of the ball, and when we do that, what we're finding is the net impulse on the ball. In other words, the impulse from all forces on the ball. But what this question was asking for was the impulse from a single force. The impulse from just the person's face. Now, aren't there other forces on this ball? Isn't there a force of gravity? And if there is, doesn't that mean what we really found here wasn't the impulse from just our face, but the impulse from the person's face and the force of gravity during this time period? And the answer is no, not really, for a few reasons. Most important reason being that, what I gave you up here was the initial horizontal velocity. This 10 meters per second was in the X direction, and this five meters per second, I'm assuming is also in the X direction. 'Cause if I'm taking the initial velocity in the X, and the final velocity in the X, and I take the difference in momentum, what I really found was the change in momentum in the X direction. When I do that, I'm finding the net impulse in the X direction, and there was only one X directed force during this time and that was our face on the ball, pushing it to the right. There was a force of gravity. That force of gravity was downward. But what that force of gravity does, it doesn't add or subtract any impulse in the X direction. It tries to add impulse in the downward direction, in the Y direction, so it tries to add vertical component of velocity downward, and so we're not even considering that over here. We're just gonna consider that we're lookin' at the horizontal components of velocity. How much velocity does it add vertically, gravity? Typically, not much during the situation, because the time period during which this collision acted is very small and the weight of this ball, compared to the force that our face is acting on the ball with, the weight is typically much smaller than this collision force. So that's why, in these collision problems, we typically ignore the force of gravity. So, we don't have to worry about that here. That's not actually posing much of a problem. We did find the net impulse in the X direction since our face was the only X directed force, this had to be the impulse our face exerted on the ball. Now, let's solve one more problem. Let's say we wanted to know: What was the average force on this person's face from the ball? Well, we know the net impulse on the ball, that means we can figure out the net force on the ball, because I can use this relationship now. Since I know that the net impulse on the ball in the X direction should just equal the net force on the ball in the X direction, multiplied by the time interval during which the force was applied, I can say that the net impulse on the ball was three kilogram meters per second, and that should equal the net force on the ball in the X direction, which was supplied by, unfortunately, this person's face, multiplied by the time interval, which is 0.02, 20 milliseconds. So, now I can solve. The force, the net force on the ball, during this time interval in the X direction, was three, divided by .02. If I take three kilogram meters per second and I divide by .02 seconds, I'm gonna get 150 Newtons was the net force on the ball. We got a positive number, and that makes sense, because this person's face exerted a positive force on this ball, 'cause the force was exerted to the right. So, these were positive, and the impulse from the face on the ball should be going the same direction as the force from the face on the ball. So, this is the force on the ball by the person's face, but notice this question is asking: What was the average force on the person's face from the ball? Not on the ball by the face. You might think, oh no, we gotta start all over, we solved for the wrong question, but we're in luck. Newton's third law says that the force on the face from the ball should be equal and opposite. So, this force on the face from the ball has got to be equal and opposite to the force on the ball from the face, so that's what we found here, the force on the ball from the face, that means the force on the face from the ball is gonna have the same size. It's gonna be 150 Newtons. It's just gonna be directed in the leftward direction, that means it's gonna be a negative force, so technically, you could say, this would be negative 150 Newtons on the face from the ball. So, to recap, the impulse, from an individual force, is defined to be that force, multiplied by the time interval during which that force is applied. And if you're talkin' about the net force in a given direction, multiplied by the time interval, you'd be finding the net impulse in that direction, and this also happens to equal the change in momentum in that direction. So, in other words, if there is a net impulse in a given direction, there's gotta be a change in momentum in that direction by the same amount. And one convenient way to remember how are these related, is you could use the pneumonic device, Jape Fat. I have no idea what Jape Fat means, but it helped me remember that the net impulse equals the change in momentum, and that also equals the net force, multiplied by the time interval, during which that force was applied. And finally, remember that during these collisions, there's always an equal and opposite force exerted on the two objects participating in the collision.