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Momentum: Ice skater throws a ball

A simple conservation of momentum problem involving an ice skater and a ball. Created by Sal Khan.

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  • blobby green style avatar for user Parth Thakurdesai
    Why does the start momentum equal zero?
    (41 votes)
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  • piceratops tree style avatar for user Bruno Andreghetti
    How about the gun situation? I mean, of course the momentum should be conserved, but there's an explosion going on inside the gun so that should be considered as an "external" force, right? How does that go into the equation? I'm not sure I'm right with my assumption, so if someone could please explain to me how does the explosion affects the situation, I'd be very thankful! ^^
    (5 votes)
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  • leaf green style avatar for user colin
    So basically, we assume numerous items to simplify concepts in physics? When do we stop assuming? At what stage in learning physics?
    (5 votes)
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  • leaf orange style avatar for user Liz Finlayson
    Look... I am really confused. Can you help me step by step go through this problem?
    An atomic nucleus of radon initially moving at 495 m/s emits an alpha particle in the direction of its velocity, and the new nucleus slows to 448 m/s. If the alpha particle has a mass of 4u, and the original nucleus of radon had a mass of 222u, what speed does the alpha particle have when it is emitted?
    (1 vote)
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  • male robot hal style avatar for user klimax
    Let's say there is a 500kg-car going to the east with 10 m/s. And there is a wall standing on its way with 5000kg of mass.

    Initial momentum is 5000P. That means final momentum also has to be equal to 5000P but how would this happen when car hits the wall and stops?
    (4 votes)
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    • primosaur ultimate style avatar for user saml.boyajian
      Although it is not clear that the wall nor the car moves in this case, something has to be impacted from the collision. For this particular example, it would be the earth that will be moving a little bit (which us humans wouldn't really feel). The energy of this collision turns mostly into heat and sound, so in theory, it may be said that this is not conservation of momentum.
      (2 votes)
  • leaf blue style avatar for user Brynn Tiner
    In an elastic equation where you are given the mass of both objects and initial velocity of both objects, how do you find the final velocity of the second object?
    (3 votes)
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  • duskpin ultimate style avatar for user 🚀The knowledge Hunter🔭
    Who discovered Momentum ? And What is unit of Momentum ?
    (2 votes)
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  • blobby green style avatar for user vivian Xia
    Why it is .105 not 0.105 what is the notation of .105?
    (2 votes)
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  • blobby green style avatar for user Lala Warpu
    Hi, I'm just wondering- how does this apply to Newton's three laws? Is it correct in saying that for every action there's always an equal and opposite reaction, and in this situation,
    action= throwing the ball & reaction = moving back? But the ball moved a greater distance as it has a smaller mass.
    (3 votes)
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    • piceratops tree style avatar for user Roman  Vaivod
      I wondered this question too. But what about acceleration in this case? Where did it go? seems like ball and girl are moving at the beginning with constant speed... why?

      Thinking about it - F * t = momentum - so that seems like the way we can argue about time during which force was applied. and that means that at the beginning for some time t objects have a force F acting on them. Thinking further this force can be stopped only by some air resistance friction or whatever but that means that momentum will be decreasing.

      So the momentum seems like a tool to argue about particular moment in time like collisions or separations. But what moment in this force and time formula? if there was some little time which lead to momentum - why do we choose this particular time and not the other - two seconds after for example.

      And i still don't understand how we can argue with this thing how objects will move when collide - together or separate in different or same direction.
      (1 vote)
  • blobby green style avatar for user henryh161
    When should we use momentum formulas to solve a problem and when do we use Kinetic energy formulas to solve a problem

    are there hallmarks of a question that indicate which is better to use?
    (3 votes)
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Video transcript

Welcome back. I'll now do a couple of more momentum problems. So this first problem, I have this ice skater and she's on an ice skating rink. And what she's doing is she's holding a ball. And this ball-- let me draw the ball-- this is a 0.15 kilogram ball. And she throws it. Let's just say she throws it directly straight forward in front of her, although she's staring at us. She's actually forward for her body. So she throws it exactly straight forward. And I understand it is hard to throw something straight forward, but let's assume that she can. So she throws it exactly straight forward with a speed-- or since we're going to give the direction as well, it's a velocity, right, cause speed is just a magnitude while a velocity is a magnitude and a direction-- so she throws the ball at 35 meters per second, and this ball is 0.15 kilograms. Now, what the problem says is that their combined mass, her plus the ball, is 50 kilograms. So they're both stationary before she does anything, and then she throws this ball, and the question is, after throwing this ball, what is her recoil velocity? Or essentially, well how much, by throwing the ball, does she push herself backwards? So what is her velocity in the backward direction? And if you're not familiar with the term recoil, it's often applied to when someone, I guess, not that we want to think about violent things, but if you shoot a gun, your shoulder recoils back, because once again momentum is conserved. So there's a certain amount of momentum going into that bullet, which is very light and fast going forward. But since momentum is conserved, your shoulder has velocity backwards. But we'll do another problem with that. So let's get back to this problem. So like I just said, momentum is conserved. So what's the momentum at the start of the problem, the initial momentum? Let me do a different color. So this is the initial momentum. Initially, the mass is 50 kilograms, right, cause her and the ball combined are 50 kilograms, times the velocity. Well the velocity is 0. So initially, there is 0 velocity in the system. So the momentum is 0. The P initial is equal to 0. And since we start with a net 0 momentum, we have to finish with a net 0 momentum. So what's momentum later? Well we have a ball moving at 35 meters per second and the ball has a mass of 0.15 kilograms. I'll ignore the units for now just to save space. Times the velocity of the ball. Times 35 meters per second. So this is the momentum of the ball plus the new momentum of the figure skater. So what's her mass? Well her mass is going to be 50 minus this. It actually won't matter a ton, but let's say it's 49-- what is that-- 49.85 kilograms, times her new velocity. Times velocity. Let's call that the velocity of the skater. So let me get my trusty calculator out. OK, so let's see. 0.15 times 35 is equal to 5.25. So that equals 5.25. plus 49.85 times the skater's velocity, the final velocity. And of course, this equals 0 because the initial velocity was 0. So let's, I don't know, subtract 5.25 from both sides and then the equation becomes minus 5.25 is equal to 49.85 times the velocity of the skater. So we're essentially saying that the momentum of just the ball is 5.25. And since the combined system has to have 0 net momentum, we're saying that the momentum of the skater has to be 5.25 in the other direction, going backwards, or has a momentum of minus 5.25. And to figure out the velocity, we just divide her momentum by her mass. And so divide both sides by 49.85 and you get the velocity of the skater. So let's see. Let's make this a negative number divided by 49.85 equals minus 0.105. So minus 0.105 meters per second. So that's interesting. When she throws this ball out at 35 meters per second, which is pretty fast, she will recoil back at about 10 centimeters, yeah, roughly 10 centimeters per second. So she will recoil a lot slower, although she will move back. And if you think about it, this is a form of propulsion. This is how rockets work. They eject something that maybe has less mass, but super fast. And that, since we have a conservation of momentum, it makes the rocket move in the other direction. Well anyway, let's see if we could fit another problem in. Actually, it's probably better to leave this problem done and then I'll have more time for the next problem, which will be slightly more difficult. See you soon.