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### Course: Mechanics (Essentials) - Class 11th > Unit 8

Lesson 3: Can you push a car from inside?- Conservation of momentum
- Momentum conservation derivation
- Conservation of linear momentum (basic)
- Conservation of momentum calculations
- Bouncing fruit collision example
- Momentum: Ice skater throws a ball
- Apply: conservation of momentum calculations

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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.

## Want to join the conversation?

- Why does the start momentum equal zero?(41 votes)
- Momentum equals the mass times velocity: at the beginning, the skater isn't moving, so no matter what the combined mass of her+ball is, their velocity is 0. 50 * 0 = 0.(99 votes)

- 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)
- No, the explosion is not an external force, it just pushes the bullet and gun away from each other.(6 votes)

- So basically, we assume numerous items to simplify concepts in physics? When do we stop assuming? At what stage in learning physics?(5 votes)
- Everything in physics is based on underlying assumptions. When the assumptions are shown to be false from experiment, they get thrown out and a new theory with new assumptions replaces it(2 votes)

- 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)- Speed is low enough to ignore relativistic effects, so all you need to do is apply conservation of momentum

Initial momentum = 495*222u

Final momentum = 448*(222u - 4u) + v*4u

Set those two momentum equal to each other and solve for v.(9 votes)

- 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)- 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)

- 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)
- both momentum and kinetic energy is conserved so relate the two equations and u will get final velocity of both first and second body(3 votes)

- Who discovered Momentum ? And What is unit of Momentum ?(2 votes)
- Momentum does not have its own units, it is just mass*velocity, so in SI units it would be kg*m/s

I am not sure how we would say who "discovered" it. Isaac Newton used it in his equations of motion.(4 votes)

- Why it is .105 not 0.105 what is the notation of .105?(2 votes)
- .105 and 0.105 mean the same thing, .105 is just shorter and easier to write than the full 0.105. Both forms are common enough that you'll see them both used very frequently and should keep in mind that they're just different ways of writing the same number.(4 votes)

- 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)- 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)

- 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)

## 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.