Two carts on a frictionless rail collide [closed]

There're 2 carts on a frictionless rail. Their masses are $4m$ and $2m$. (The carts are on the rail in this order). The first cart has a velocity of $v_1$. The first cart hits the second. Knowing that the final velocity of the first cart is $v_1\over4$, says if the collision is elastic, inelastic or perfectly inelastic.

I forgot to mention that $v_2=0$, that's why I've set up the first equation the way it is.

I set up the conservation of momentum

$\begin{cases} m_1v_1=m_1{v_1}'+m_2{v_2}' \\ v_1'=\frac{v_1}{4} \end{cases}$

Then I solve it:

$m_1v_1=m_1\frac{v_1}{4}+m_2{v_2}'$

$4mv_1=4m\frac{v_1}{4}+2m{v_2}'$

$3v_1=2{v_2}'$

I tried calculating the initial $KE$:

$KE_i=\frac{1}{2}4m\frac{4}{9}{v_2}'^2$

which is $KE_i=m\frac{8}{9}{v_2}'^2$

As for the final $KE_f$:

First body: $m\frac{1}{18}{v_2}'^2$

Second body: $m{v_2}'^2$

Total $KE_f=\frac{19}{18}m{v_2}'^2$

Now, how is it possible that the initial KE is less than the final one? a) I calculated wrong b) dark energy(improbable)

closed as off-topic by CuriousOne, Gert, user36790, Qmechanic♦Jul 1 '16 at 4:28

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• What if a spring is released when they collide? – Bill N Jun 30 '16 at 22:46
• There's no spring in the problem picture, which is simply 2 blocks as I described. I think the problem would have said so. – Christian Jun 30 '16 at 23:02

You have proved with your analysis that $v_2$ cannot be zero. $m_2$ must be moving with some non-zero velocity, either in the same direction as $m_1$, in which case $v_2$ must be smaller than $v_1$ (or $m_1$ will never catch up to $m_2$) or $m_2$ must be moving in the opposite direction to $m_1$.
With the information given, there are 4 unknowns: $v_1, v_2, v_1'$ and $v_2'$ (five if you count the mass $m$). One equation is given: $v_1' = \frac{v_1}{4}$. The best you can hope for is to find 3 of the other velocity variables in terms of the fourth.
In inelastic collisions the Kinetic energy in the final state is less than in the initial state. For the completely inelastic case, $v_1' =v_2'$ and both masses move together as one combined mass.
I did a calculation for the completely inelastic case and came up with $v_1' = v_2' = \frac{v_1}{4}$ and $v_2 = -\frac{v_1}{4}$. The final state KE is $\frac{6}{8}m{v_1}^2$ and the initial state KE is $\frac{33}{16}m{v_1}^2$ which is larger as required.
• I wrote the steps I followed. And I forgot to mention that $v_2=0$, even though was intrinsic in my system. – Christian Jul 2 '16 at 14:01