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The equation for conservation of momentum: $$m_1\vec{v}_1 + m_2\vec{v}_2 = m_1\vec{u}_1 + m_2\vec{u}_2$$

The equation for conservation of energy: $$\frac 12m_1v_1^2 + \frac 12m_2v^2_2 = \frac 12m_1u_1^2 + \frac 12m_2u^2_2$$

Then if it's an elastic collision on one dimension (conservation of energy and also conservation of momentum): $$v_1 - v_2 = -(u_1 - u_2)$$

My question is:

If the case is that there is conservation of energy, will there be conservation of momentum?

In both cases, can you explain why?

Also, let's say they give you a question where there is a collision.

How can you prove there is conservation of energy?

Sometimes the questions tell you that a collision was elastic. But if not, how do you prove it?

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closed as off-topic by Bill N, ACuriousMind, Kyle Kanos, Norbert Schuch, Gert Mar 14 '16 at 1:20

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  • $\begingroup$ Hint: Think about a moving particle reversing direction. $\endgroup$ – WillO Mar 12 '16 at 18:09
  • $\begingroup$ @WillO You are talking about the prove right? So I guess it will be proving that $\vec{v} = -\vec{u}$ $\endgroup$ – Pichi Wuana Mar 12 '16 at 18:21
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    $\begingroup$ Suggestion: You might want to compare the harmonic oscillator with the title question (v1). $\endgroup$ – Qmechanic Mar 12 '16 at 19:05
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    $\begingroup$ Momentum is always conserved. And for a closed system with no outside forces, the momentum will be constant for a given reference frame. $\endgroup$ – Bill N Mar 13 '16 at 1:27
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Momentum is always conserved in a collision, due to Newton's 2nd and 3rd laws. Prove this to yourself by setting up a problem where there is an interaction between two colliding objects. Since the forces between these objects are equal in magnitude and opposite in direction, it is easily seen that the accelerations experienced by each object during the collision are inversely proportional to the objects' masses. When the velocities are calculated after the collision based on the reasons given above, the total momentum before the collision is seen to be equal to the total momentum after the collision, for all collisions.

To determine if a collision is elastic, you necessarily have to calculate the total kinetic energy before the collision and the total kinetic energy after the collision. If and ONLY if the kinetic energy after the collision is equal to the kinetic energy before the collision, can you say that the collision was elastic.

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  • $\begingroup$ Not all the collisions are with momentum conserved. For example, in a free fall. My condition for conservation of momentum is that all external forces are equal to $0$ or that the net force of external forces is equal to $0$. ($W$ is an external force). $\endgroup$ – Pichi Wuana Mar 12 '16 at 19:57
  • $\begingroup$ If you already knew the answer to your question, why did you post? $\endgroup$ – David White Mar 12 '16 at 23:12
  • $\begingroup$ I didn't knew the answer for my question, I just cleared something you said in your answer. Either way thanks for it $\endgroup$ – Pichi Wuana Mar 12 '16 at 23:18
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    $\begingroup$ @PichiWuana , you're welcomed, and thanks for the up-vote. $\endgroup$ – David White Mar 13 '16 at 3:02

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