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In an elastic collision, I understand that momentum is conserved and kinetic energy is conserved. If billiard ball of silver (with velocity $v_{(Ag)}$ impacts a stationary billiard ball of aluminum, I am trying to calculate the velocity of the aluminum ball after the collision, $v_{(Al)}$. After an elastic collision, the impactor is at rest and the impactee has the motion.

Using momentum, $= m \cdot v$

$$m_{(Ag)} \cdot v_{(Ag)} = m_{(Al)} \cdot v_{(Al)}$$

Assuming silver is 4x denser than aluminium, then using momentum, the aluminium ball should have velocity

$$v_{(Al)} = 4\cdot v_{(Ag)}$$

But if we use kinetic energy, $1/2 m \cdot v^2$

$$\frac12m_{(Ag)}\cdot v_{(ag)}^2=\frac12m_{(Al)}\cdot v_{(Al)}^2$$

$$v_{(Al)}^2=\frac{m_{(Ag)}}{m_{(Al)}}\cdot v_{(Ag)}^2$$

$$v_{(Al)}=\left(\frac{m_{(Ag)}}{m_{(Al)}}\right)^{\frac12}\cdot v_{(Ag)}$$

$$v_{(Al)}=2\cdot v_{(Ag)}$$

Somewhere I have lost some neuron connections in my brain because I cannot resolve this conflict. This is a perfectly elastic collision so both momentum and kinetic energy should be conserved.


I have read multiple threads including:

When is energy conserved in a collision and not momentum?

How to calculate velocities after collision?

How can I calculate the final velocities of two spheres after an elastic collision?

Calculating new velocities of $n$-dimensional particles after collision

Velocities in an elastic collision

Summation of the velocities before and after an elastic collision

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  • $\begingroup$ If you throw a ping pong ball (head-on) against a heavy led ball (with the same radius), what will happen? $\endgroup$ Commented May 13, 2021 at 12:42
  • $\begingroup$ Hello! I have edited your question using MathJax (LaTeX) math typesetting. For future questions, you can refer to MathJax basic tutorial and quick reference. Thanks! $\endgroup$
    – jng224
    Commented May 13, 2021 at 13:17
  • $\begingroup$ I have edited the subscripts to make them a look a bit more nicer and upvoted this question. Hope your neurons get reconnected with the answer(s) to this question :D $\endgroup$
    – Babu
    Commented May 13, 2021 at 13:32

3 Answers 3

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If the objects have different masses, then there isn't a way to start the collision with object 1 moving and object 2 at rest and then end the collision with object 1 at rest and object 2 moving while also having the collision be elastic. You have over-constrained your system, and so you will find contradictions like the one you found here.

Using the equations from this answer in one of your linked questions, if we are setting $v_{A,f}=v_{B,i}=0$, then we end up with the system of equations $$0 = \dfrac{m_A - m_B}{m_A+m_B} v_{A,i}$$ $$v_{B,f} = \dfrac{2m_A}{m_A+m_B} v_{A,i}$$

Which you can see is only consistent if $v_{A,i}=v_{B,f}=0$ for $m_A\neq m_B$ (which is the case of no collision), or if $m_A=m_B$.

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  • $\begingroup$ thank you for taking the time to flesh out your answer. I actually preferred your initial answer which was more hint-like. It forced me to go back and do the formal analysis setting the initial momentum of the whole system, m1v1, equal to the momentum of the system after the collision, m1v1' + m2v2'. Doing the same for kinetic energy and combining equations gives the final equations in the linked answer. $\endgroup$ Commented May 14, 2021 at 4:38
  • $\begingroup$ it gets really interesting at 1)relativistic velocities when the mass is a function of velocity or 2) when the collision has elastic and inelastic character. $\endgroup$ Commented May 14, 2021 at 4:46
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If a small mass collides elastically with a larger one which was at rest, the smaller one will bounce back, not stop.

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You say...

 After an elastic collision, the impactor is at rest and the impactee has the motion.

That is only true if the masses of the two balls are equal. Clearly they will not be equal if one is made of silver and the other aluminium.

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