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Two balls, with masses $m_1$ and $m_2$ are falling from the heigh $h$. All the collisions are perfectly elastic. Do not consider the size of the balls. $m_1 < m_2$ and lighter ball is on the top.

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What heights $h_1$ and $h_2$ will the balls jump to?


And so I have many equation like: $V(-m_1+m_2)=m_1v_1+m_2v_2$, $(m_1+m_2)gh=m_1gh_1+m_2gh_2$ and $\frac12 V^2(m_1+m_2)= \frac12 m_1 v_1^2 + \frac12 m_2 v_2^2$ (Where $V$ is velocity of both balls right before collision. I consider lighter ball is falling down the same velocity as heavier ball is flying up, so I put opposite sign in the equation of momentum. $v_1$ is a velocity of lighter ball after collision and $v_2$ velocity of heavier ball). But I have no idea how to get $h_1$ and $h_2$ independent of each other.

Please don't describe me "you have to do this and this" and "think about what happen when...". I need to see the progress. I need to see these equations and what's happening.

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  • $\begingroup$ You can't solve two variables from a single equation(must need two). If you want the answer using your equations, just add another equation containing h1 and h2. $\endgroup$ – blackSmith Aug 27 '13 at 13:12
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    $\begingroup$ possible duplicate of Two balls falling one above the other $\endgroup$ – Kyle Oman Aug 27 '13 at 20:46
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When the two balls arrive at the ground, they both have speed $V=\sqrt{gh}$ and are travelling downward. Break the interaction down into two stages. First, $m_2$ collides with the ground. Then, $m_1$ and $m_2$ collide (they cannot collide before $m_2$ collides with the ground because they have the same velocity). After this, $m_1$ necessarily has positive velocity (which can be shown). $m_2$ can have positive or negative velocity, depending on the ratio $m_1/m_2$. If it has negative velocity, it will collide with the ground again.

First, let's deal with $m_2$ colliding with the ground. This can be modelled as an elastic collision by thinking of the ground as a ball with mass that tends to $\infty$. Carrying out the math, it's pretty easy to show that the ground doesn't move and $m_2$ just has the sign of its velocity switched.

Now the second collision is between $m_1$ moving downward with speed $V$ and $m_2$ moving upward, also with speed $V$. The equations to solve are conservation of momentum:

$$m_1v_{1,i} + m_2v_{2,i} = m_1v_{1,f} + m_2v_{2,f}$$

putting in the known velocities gives:

$$m_1(-V) + m_2V = m_1v_{1,f} + m_2v_{2,f}~~~~~~~~~~~~~~(1)$$

The second equation is for conservation of energy. I'll consider the change in energy between just before and just after the collision, so there is no change in heights and therefore no change in potential energy. This gives:

$$\frac{1}{2}m_1v_{1,i}^2 + \frac{1}{2}m_2v_{2,i}^2 = \frac{1}{2}m_1v_{1,f}^2 + \frac{1}{2}m_2v_{2,f}^2$$

(the $\frac{1}{2}$ aren't necessary since they cancel out, but you may be used to seeing them there). Again, putting in the known velocities:

$$\frac{1}{2}m_1(-V)^2 + \frac{1}{2}m_2V^2 = \frac{1}{2}m_1v_{1,f}^2 + \frac{1}{2}m_2v_{2,f}^2~~~~~~~~~~~~~~(2)$$

Now we have 2 equations ($(1)$ and $(2)$) and 2 unknown quantities $v_{1,f}$ and $v_{2,f}$. They can be solved (I won't go through the details here), and you should get:

$$v_{1,f} = V\frac{3\frac{m_2}{m_1}-1}{\frac{m_2}{m_1}+1}$$ $$v_{2,f} = V\frac{\frac{m_2}{m_1}-3}{\frac{m_2}{m_1}+1}$$

A nice thing to check right now is what happens if $m_1=m_2$ (in this special case the two balls should keep the same speed, but just change directions).

Now you can check that $v_{1,f}$ is actually moving upward (we know that $m_1<m_2$, so this should be pretty obvious). Without knowing the ratio $m_1/m_2$, it isn't clear whether $v_{2,f}$ is positive or negative (but you can, and should, show that $|v_{2,f}|<|v_{1,f}|$, so $m_2$ will not "catch up" to $m_1$ and collide again). If $m_2$ happens to have negative velocity after this collision, it will hit the ground again. Doing the same thing as before, its velocity will just reverse direction and it will immediately be moving upward anyway, so we can just assume it's positive.

Now the only thing left to do is solve for the heights. This is done again using conservation of energy, this time between right after the collision and the instant each ball reaches its maximum height (treat them separately). The relevant equations:

$$\frac{1}{2}m_1v_{1,f}^2 = m_1gh_1$$ $$\frac{1}{2}m_2v_{2,f}^2 = m_2gh_2$$

Solve those for $h_1$ and $h_2$ and you're done!

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  • $\begingroup$ +1 I'm quite sure our answers are right. Your answer is definitely more straightforward simple to solve, though. $\endgroup$ – resgh Aug 27 '13 at 15:02
  • $\begingroup$ What have you done with (1) and (2) that you got those velocities? Whatever I do, I end up with $v_2=-V$ $\endgroup$ – user50222 Aug 27 '13 at 15:28
  • $\begingroup$ @user50222 When you work it out you should get a quadratic for each of $v_{1,f}$ and $v_{2,f}$. There are therefore two solutions; one will give you the velocities before the collision (so you'd just get -V and V), and the other gives the velocities after the collision, which is what you want. Maybe that's what's happening? The solution to conservation of momentum and energy for velocity in a 2-particle elastic collision is a standard exercise, so you could just look up the derivation (like maybe here: blog.wsd.net/jrhoades/files/2010/11/…). $\endgroup$ – Kyle Oman Aug 27 '13 at 15:39
  • $\begingroup$ @user50222 Rearrange the momentum conservation equation, expressing one velocity in terms of the other. Substitute into energy conservation equation. Use quadratic formula (remember to discard the answer that corresponds to the initial situation). Use momentum conservation equation to get the remaining velocity. $\endgroup$ – resgh Aug 27 '13 at 15:39
  • $\begingroup$ @Kyle Broken link? $\endgroup$ – resgh Aug 27 '13 at 15:43
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Solve the problem step by step:

First consider the two balls dropping to ground, $$\frac{1}{2}(m_{1}+m_{2})v^2=(m_{1}+m_{2})gh$$ $$v=-\sqrt{2gh}$$ $m_{2}$ ball collides elastically with ground, $$v_{1}=-\sqrt{2gh}, v_{2}=\sqrt{2gh}$$ Balls $m_{1}$ and $m_{2}$ ball collide elastically, $$v_{1}=\frac{(-\sqrt{2gh})(m_{1}-m_{2})+2m_{2}(\sqrt{2gh})}{m_{1}+m_{2}}=\sqrt{2gh}\frac{-m_{1}+3m_{2}}{m_{1}+m_{2}}$$ $$v_{2}=\frac{(\sqrt{2gh})(m_{2}-m_{1})+2m_{1}(-\sqrt{2gh})}{m_{1}+m_{2}}=\sqrt{2gh}\frac{-3m_{1}+m_{2}}{m_{1}+m_{2}}$$ As $m_{1}<m_{2}$, clearly $v_{1}>0$ and $v_{1}>v_{2}$, If $v_{2}<0$, $m_{2}$ ball collides elastically with ground again, so $$v_{2}=|\sqrt{2gh}\frac{-3m_{1}+m_{2}}{m_{1}+m_{2}}|$$ Clearly, $v_{1}>v_{2}$ still holds, The two balls experience identical acceleration so forth and no further collisions take place, $$\frac{1}{2}m_{1}v_{1}^2=m_{1}gh_{1}, \frac{1}{2}m_{2}v_{2}^2=m_{2}gh_{2}$$ $$h_{1}=\frac{1}{2g}(\sqrt{2gh}\frac{-m_{1}+3m_{2}}{m_{1}+m_{2}})^2=h(\frac{-m_{1}+3m_{2}}{m_{1}+m_{2}})^2$$ $$h_{2}=\frac{1}{2g}(|\sqrt{2gh}\frac{-3m_{1}+m_{2}}{m_{1}+m_{2}}|)^2=h(\frac{-3m_{1}+m_{2}}{m_{1}+m_{2}})^2$$

NOTE: In calculating the final velocities after each elastic collision between the balls, I used the solution to the one dimensional elastic collision given here. The result will be the same, albeit much more involved, if the laws of conservation of energy and momentum were directly used.

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  • $\begingroup$ Heh, beat me to an answer. At least we got the same thing, so it's probably correct :) $\endgroup$ – Kyle Oman Aug 27 '13 at 14:59
  • $\begingroup$ In the point, where you used that one dimensional elastic collision equation, I can't follow. I don't see the process that would give this form. Which equations are used to make the elastic collision equation? $\endgroup$ – user50222 Aug 27 '13 at 15:10
  • $\begingroup$ @user50222 The equations are conservation of momentum and conservation of energy. Try to set up a system of equations $m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}, \frac{1}{2}m_{1}u_{1}^2+\frac{1}{2}m_{2}u_{2}^2=\frac{1}{2}m_{1}v_{1}^2+\frac{1}{2}m_{2}v_{2}^2$. Solving this system gives you the equations. See the wiki article linked in my answer. $\endgroup$ – resgh Aug 27 '13 at 15:29
  • $\begingroup$ @user50222 How to solve the system is sketched in my comment to Kyle's answer. $\endgroup$ – resgh Aug 27 '13 at 15:43
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The height reached by a vertical projection is given by the equation :

$h=v^2/(2g)$ where $v$ is the projected velocity, ie - initial velocity.

Since $v_1$ and $v_2$ are known, you have the solutions for $h_1$ and $h_2$.

NB: If you want to calculate $v_1$,$v_2$ from $V,m_1,m_2$ itself, I have to update the answer accordingly. However I afraid that the outcome may be the same as udiboy posted.

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    $\begingroup$ $V(-m_1+m_2) = m_1v_1+m_2v_2$ is conservation of momentum, I don't immediately see anything wrong with it. KE will still be positive because it is proportional to $v^2$. $\endgroup$ – Kyle Oman Aug 27 '13 at 14:22
  • $\begingroup$ @Kyle : Sorry I missed that. Thanks for correcting. Updating the answer... $\endgroup$ – blackSmith Aug 27 '13 at 14:31

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