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I am asking about this video where The Physics Girl drops a system of balls from a height.
I am trying to estimate the velocities just after the collision using the law of conservation of momentum. And I proceed as below:

Consider the system as comprising of two balls (1: tennis ball, mass $m$, and 2: basketball, mass $M$). Now initial momentum ($p_i$) is

$$p_i = m u_1 + M u_2 = 0,$$ as the ball is dropped from rest.

The final momentum after the collision $p_f = m v_1 + M v_2.$ By law of conservation of momentum $p_i = p_f$

$$ 0 = m v_1 + M v_2 $$

or

$$v_1 = - (M/m)v_2 $$

Is this explanation correct? If so what is the meaning of negative sign?

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  • $\begingroup$ @Paul Thank you very much for editing. How exactly you do it? $\endgroup$
    – gpuguy
    Oct 23, 2019 at 14:51
  • $\begingroup$ you can display inline LaTeX style math statements by wrapping them in dollar signs $. Double dollar signs $$ for stand alone equations. So $p_i$ displays as $p_i$. $\endgroup$
    – Paul T.
    Oct 23, 2019 at 14:55

3 Answers 3

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You have not applied the conservation of momentum rule consistently. At the start the balls and the earth have zero momentum in their combined rest frame. By the time the balls reach the ground, at a velocity v say, their combined momentum is (m+M)v, not zero. For momentum to be conserved you have to take into account an equal and opposite effect on the Earth's momentum. That is, the Earth's momentum changes by -(m+M)v as a result of the reaction from attracting the balls. So if you start with the assumption that the combined momentum of the balls immediately before the collision is (m+M)v, then assume that the collision is elastic, you should be able to work out v1 and v2.

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  • $\begingroup$ Thanks. In your expressions, M is the mass of earth? Would you mind showing a few steps clearly, so I can pick up and extend the derivation? $\endgroup$
    – gpuguy
    Oct 23, 2019 at 17:54
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If you place your initial point before the ball was dropped, the momentum of the system is not conserved. The force of gravity imparts an impulse ($J$) on the system. You need to use the Impulse-Momentum Theorem:

$$ \vec{p}_i + \vec{J} = \vec{p}_f, $$

where $\vec{J} = \vec{F}_\mathrm{ave} \Delta t$; $\vec{F}_\mathrm{ave}$ is the average net force on the system; and $\Delta t$ is the time interval over which the force is applied.

To figure out the impulse, we need to know how much force was applied and for how long.

Even during the collision with the ground there is a net force pushing up on the system, so momentum won't be conserved. The Impulse-Momentum Theorem will still work, if you can determine the impulse.

In the video, they approach the problem by thinking about energy. The video states that the initial gravitational potential energy is converted to elastic potential energy when the balls compress against the ground. That elastic potential energy becomes kinetic energy after the bounce.

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  • $\begingroup$ Thanks. But what if I include the earth also in the system? Will the momentum be conserved? $\endgroup$
    – gpuguy
    Oct 23, 2019 at 17:42
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The minus sign just means different direction. What you are writing is more of a system non-elastic collision between two balls with $u_1$ and $u_2$ as the velocities in the center of mass frame.

What happened in the video is pretty different. First of all the collision happens with earth, but if you add the term $M_{earth}v_{earth}$ to your equation you notice that after the collision $V_{earth} - v_{earth} = C/M_{earth} \approx 0$, where capital letter is used to indicate value after collision.

Also, the collision between the two balls is not an elastic collision, so the equations you write are not valid.

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  • $\begingroup$ Thanks for your answer. Can we not assume the elastic collision and still see the same behavior of the balls? $\endgroup$
    – gpuguy
    Oct 23, 2019 at 17:41
  • $\begingroup$ Not at all ! Imagine replacing the basketball with a bowling ball (not elastic at all). $\endgroup$
    – Edgar
    Oct 24, 2019 at 9:14

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