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# 2-dimensional momentum problem

An example of conservation of momentum in two dimensions. Created by Sal Khan.

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• at , how is the momentum in the y direction for B = 10? If you subtract 10 from both sides, you get -10, but then thats in the opposite direction. • I don't understand how there was a velocity for the X and Y directions...does ball B move in one direction and then the other? • If you throw a ball straight up in the air it only has vertical or y velocity.
If you throw a ball horizontally, it only has horizontal or x velocity initially ( though gravity will rapidly change that)
in general I might pitch a ball which is neither horizontal nor vertical - I can represent the balls motion with horizontal and vertical components e.g. if i throw it up at 10 m/s at 30 degrees from the horizontal it has a horizontal component of 10 cos 30 and a vertical component of 10 sin 30.
Why can't I just represent it as 10 m/s at an angle of 30 degrees?
the answer to this is that on the good earth, gravity acts vertically in a well known and easy to calculate manner. Nothing changes to the horizontal component if I ignore air resistance while the vertical component is first retarded to zero then gravity has the nerve to pull the ball back down!
So we divide the motion into vertical and horizontal components, calculate what happens to each using what my students fondly call the SUVAT equations of motion under constant acceleration. then to find out where the ball ends up after a give time, use vector maths to add the componenets back together.
• May I please ask what is the equation for 3D collision of any two objects • Why do you write to Pay + Pby = 0? • The initial momentums (momenta) in the y direction is zero, which is the sum of the initial momentum in y direction for A and the initial momentum in y direction of B. We know that momentum will be conserved therefore the momenta after the collision will add up to equal zero. Really there could be another subscript on those... (Pay + Pby)i = 0 = (Pay + Pab)f. I --> initial and f ---> final.
• Does anyone know the difference between a conservation of momentum problem and a change of momentum problem? I want to look at every problem and say momentum is conserved before and after a collision. When do I know I've come across a problem where momentum is not conserved and have to use the Impulse equation? Do I still use conservation of momentum equations in an Impulse problem?

For example if a car collides with a car at rest I can use conservation of momentum to find the speed of the cars after they have collided.

But then I have to use Impulse to find the change in momentum. I'm not sure when I'm supposed to switch my logic from saying momentum is conserved in this case but is changing in this other case.
(1 vote) • Linear momentum is a conserved quantity. In science, laws are laws, because they are obeyed in every single case, we know of. The linear momentum of any (closed) system is conserved - always!

What could change, in case you're looking at an inelastic collision, is the kinetic energy. But the total energy of the system is again conserved. So the kinetic energy would just get transformed into another kind of energy (i.e. heat) - or vice versa.
• What is the formula for two dimensional momentum problems or a step by step method to follow? • What would happen if we were given a problem without the mass of each ball, but are told they are equal? Would we just assign a mass to them (like 1 kg)? • Thanks for the video. Something nonintuitive here is: what makes the ball A and ball B go in the direction they do? Why does ball A move with the Y component upwards and ball B with Y component downwards?Any physical reason for that? • I'm a little confused about the difference between energy and momentum. For example, if a truck has a large momentum, does it mean that it takes greater energy to stop it?
Thanks
(1 vote) • Yes, a truck with more momentum does require more work (energy) to stop it, but that relationship is not linear, because momentum is m*v and kinetic energy (the amount of energy that would need to be dissipated to stop the truck) is 1/2*m*v^2.

The reason we are interested in this quantity m*v that we call momentum is that it has an interesting and useful property that it is always conserved in a collision. Note that total energy is also conserved but there is no such rule as conservation of kinetic energy. When trucks collide, momentum is conserved no matter what. Some or even all of the kinetic energy is converted to thermal energy. 