Collision of two balls [duplicate]

Question

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.

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

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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.

Answer 1

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!

Answer 2

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.

Answer 3

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.