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First of all, I want to apologize for my English I'll try to explain my self clearly. I want to share one question about Newtonian mechanics from my homework, of which I can't understand the logic behind it.

enter image description here

As you can see in the picture, a body of mass $m = 1$ kg moves straight along the x-axis, with a velocity of $v_i = 20$ m/s. Another object with mass $M = 10$ kg is resting without friction.

  • What is the maximal height that mass $m$ will rise (considering gravitational force), after the collision between two masses?

In the exercise it is stated that $mv=MU$ on the x-axis and the collision is absolutely elastic (by conservation of momentum law).

Now my question is how is it possible that mass $m$ still has velocity on y-axis when all the momentum after the collision moved to mass $M$?

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closed as off-topic by ZeroTheHero, G. Smith, Aaron Stevens, Jon Custer, Buzz Jan 15 at 0:57

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  • $\begingroup$ Does the problem include a gravitational force? $\endgroup$ – Alex S Jan 14 at 0:37
  • $\begingroup$ Yes, I will add it right know. $\endgroup$ – Rasim Muradov Jan 14 at 0:43
  • $\begingroup$ Okay, be sure to explain when gravity begins because if it starts before the collision, the trajectory of mass $\mathbf{m}$ will change. $\endgroup$ – Alex S Jan 14 at 0:45
  • $\begingroup$ The trajectory of mass m is paralel to x-axis until the collision. $\endgroup$ – Rasim Muradov Jan 14 at 0:52
  • $\begingroup$ @RasimMuradov Does the diagram and/or the text indicate that after collision the body is moving vertically upwards. $\endgroup$ – Farcher Jan 14 at 12:30
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You can apply conservation of momentum in the horizontal direction because the only horizontal forces acting on the system are the internal forces between the two masses when they collide.

But you can't easily apply conservation of momentum in the vertical direction, because when the collision occurs there is also a vertical impulsive force between M and whatever it is resting on. After the collision, the momentum of the whole earth has increased in the downward direction, and m has an equal and opposite momentum upwards.

Of course the mass of the earth is so big that the velocity change caused by the increase in momentum is too small to measure, and can be ignored.

If you use conservation of horizontal momentum, and conservation of energy (because the collision is perfectly elastic), you have enough equations to solve the problem without trying to use conservation of vertical momentum as well.

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  • $\begingroup$ but does its a right to say that on x-axis mV=MU ? $\endgroup$ – Rasim Muradov Jan 14 at 1:25
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After the collision the mass $m$ has no horizontal momentum so all the horizontal momentum that it did have before the collision is transferred to the momentum of the wedge.
This is justified because there are no horizontal external forces acting on the mass & wedge system.

Vertical momentum is not conserved by the mass & wedge system as the ground exerts a vertical force on the system.

The only other thing to note is that because the collision is elastic (kinetic energy conserved before and after the collision) the law of conservation of (mechanical) energy can be used.

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  • $\begingroup$ But does not that mean there is a vertical component also for MU? if mass m changes his lane from horizontal to vertical some component must to on the same force with an inverse side? $\endgroup$ – Rasim Muradov Jan 14 at 14:00
  • $\begingroup$ @RasimMuradov $MU$ is the horizontal momentum of the wedge after the collision. After the collision only the vertical momentum (velocity) of the mass needs to be considered. $\endgroup$ – Farcher Jan 14 at 14:05
  • $\begingroup$ link can you please watch that and tell me if it is true to calculate this way? $\endgroup$ – Rasim Muradov Jan 14 at 14:08
  • $\begingroup$ @RasimMuradov The wedge cannot have a vertical momentum because it cannot move downwards as it is resting on the ground. A better way of putting this is to say that the mass of the Earth and wedge is much greater than the mass hitting the wedge. $\endgroup$ – Farcher Jan 14 at 14:14
  • $\begingroup$ Ok, Thank you @Fracher $\endgroup$ – Rasim Muradov Jan 14 at 14:16
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We will start by resolving the velocity of mass 'm' just before collision in two directions. One along the plane of the mass 'M' and one perpendicular to it.

I'll call the velocity component parallel to the plane of 'M' as $V_a$ and the velocity component perpendicular (into the surface as $V_b$).

Fig shows what I'm talking about

Now.. you see.. during the collision.. the mass 'M' cannot alter the velocity component $V_a$ in any way ((unless there's friction between them. But there isn't in this case)).

So the instant after collision, mass 'm' indeed has an unchanged velocity component!!

Also if you're wondering about gravity taking away that momentum, then you're right.. kinda.. During the instant of collision, the impulse being so high plus the time interval being soo small, other forces (in this case gravity) can be neglected.

But mind you! After the collision event, that is, after the masses have separated gravity starts taking away all the vertical momentum!

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