An object of mass $5\, \mathrm{kg}$ is projected with a velocity $20\, \mathrm{ms}^{-1}$ at an angle $60^{\circ}$, to the horizontal. At the highest point of its path, the projectile explodes and breaks up into two fragments of masses $1\, \mathrm{kg}$ and $4\, \mathrm{kg}$. The fragments separate horizontally after the explosion, which releases internal energy such that the kinetic energy ($\text{KE}$) of the system at the highest point is doubled. What is the separation of the two fragments when they reach the ground?

In this problem, the set up is quite simple. Let $v_1$ be the velocity of the $4\,\mathrm{kg}$ mass and $v_2$ be the velocity of the $1\,\mathrm{kg}$ mass. The initial momentum along the $x$ axis is $5\cdot10\,\mathrm{kg}\,\mathrm{m}\,\mathrm{s}^{-1}$

Applying conservation of momentum along the $x$ axis: $$50= 4v_1+v_2 \tag1$$ $\text{KE}_i$=$250J$. Twice of this is equal to the final $\text{KE}$. So $$2(250)=\frac{4v_1^2}{2} + \frac{v_2^2}{2} \tag2$$

Solving equations $(1)$ and $(2)$ we get two different sets of answers: $$v_1=5\, \mathrm m/\mathrm s \qquad v_2=30\,\mathrm m/\mathrm s$$ or $$v_1=15\, \mathrm m/\mathrm s \qquad v_2=-10\, \mathrm m/\mathrm s$$

My question is that out of these two possible answers which one should occur as both masses can taken any one velocity so one of the cases should arise. The separation turns out to be the same in both cases as $|v_1- v_2|$ is same for both the cases.

If somehow this can be tested physically, (under ideal conditions with no air resistance or spin of the balls) what will we observe in nature in multiple attempts? Will every time any of the two be able to happen?
How do we know for sure what exactly is going to happen here?

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    $\begingroup$ Have you considered that the fragments could fly off in the z-axis? Where the projectile is travelling in the x-axis with increasing (and decreasing) height in y-axis. If this were the situation then if I'm not mistaken there will only be one solution. $\endgroup$
    – ludz
    Commented Aug 1, 2021 at 9:39
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    $\begingroup$ actually its mentioned in the problem that fragments separate horizontally. $\endgroup$ Commented Aug 1, 2021 at 10:16
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    $\begingroup$ Yes but horziontally only means that they travel in the x-z-plane (with the coordinate system in my previous comment). $\endgroup$
    – ludz
    Commented Aug 1, 2021 at 10:30
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    $\begingroup$ If the only constraint is that the pieces move "horizontal", the direction of separation can be arbitrary. Therefore, the same relative speed of separation will result in an entire range of possible absolute speeds. But because the relative speed is fix, they'll always land separated by the same distance, which is why the cleverly worded question asks for that. $\endgroup$ Commented Aug 2, 2021 at 5:39
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    $\begingroup$ Do yourself a favor and learn to write your numbers with units all the time. It's not just that a number 50 or 250 is physically totally meaningless without a unit (is it Joules? Calories? Meters? Pounds?), it's also that treating units like parts of the equation trivially yields the correct conversion factors (every unit conversion has exactly one conversion factor that needs to be added/removed to/from the figure). Finally, it's an easy consistency check that you didn't mess up your equations when the correct unit results. Units are extremely valuable; use them. $\endgroup$ Commented Aug 2, 2021 at 21:40

9 Answers 9


The two results that you got are absolutely acceptable both mathematically and experimentally and either of them can happen if the experiment is done (but it's not random) practically with all situations similar to the one posed in the question.

Why isn't it random ?

Because the result depends (with certainty) on the arrangement of the piece of mass $4\; kg$ and that of mass $1\; kg$ when they were a single body just before exploding.

Consider the diagram. Suppose Initially it was a sphere (and I have shown a dashed line in subsequent figures to indicate the points where they separate).

enter image description here

Now in practical experiments, you will get your first set of values if the masses just before the explosion are arranged as shown in the figure

enter image description here

And the second set of values correspond to the situation when the masses are arranged in this way :-

enter image description here

So in practical experiments, which set of values we get depend on the position of the pieces (just before the explosion) inside the main sphere . But it's not random.

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    $\begingroup$ Great man thanks. The only little addition of intuition that isn’t explicitly covered in the excellent answers is this: The two solution values for $v_1$, and the two for $v_2$, are centered around 10, ie around $v_i$. The 4kg mass goes backward relatively at 5 (at speed 10-5=5) or forward relatively at 5 (at 10+5=15). And the 1kg with +/- 20 (at 10 +/- 20). I did some simple trig and unsurprisingly.. it works in any horizontal direction not just parallel and anti-parallel to the object’s horizontal velocity. $\endgroup$
    – Al Brown
    Commented Aug 2, 2021 at 2:22
  • $\begingroup$ I added that to my new answer, the any angle thing. Your pic goes well w it $\endgroup$
    – Al Brown
    Commented Aug 2, 2021 at 3:33

What a lovely conundrum!

I'm afraid that the solution is a rather simple. You didn't specify in your maths whether the 4 kg mass was ejected in the direction of the initial velocity of the 5 kg mass and the 1 kg in the opposite direction, or vice versa. The different pairs of solutions correspond to the two different cases.

It's not your fault that you didn't specify which case you were considering. Neither did the question. The question didn't even say that the fragments were emitted parallel and/or antiparallel to the initial velocity!

Does it actually matter if the fragments are emitted with 'sideways' velocity components? As you say, the question asked only for the fragments' separation when they reached the ground. Perhaps you should investigate this.

You might also like to try working in a frame of reference moving with respect to the ground with the object's initial horizontal velocity (that is in the frame of reference in which the total momentum is zero, the so-called 'CM frame'). In this CM frame the fragments have a total energy of 250 J due only to the explosion, and their momenta are equal and opposite. Makes life very easy!

  • $\begingroup$ The question focused on final separation when they land back to ground which turns out to be same for both cases. But it's interesting how it's difficult to precisely decribe what's going on using simple equations. If anything can happen out of the two cases, what would happen if the experiment was really carried out? What is the deciding factor? $\endgroup$ Commented Aug 1, 2021 at 8:50
  • $\begingroup$ (a) "The question focused on final separation when they land back to ground ... the same for both cases." Would it be the same if there were sideways component to the fragments' velocities?" (b) But it's interesting how it's difficult to precisely decribe what's going on using simple equations." If you knew that the 4 kg fragment was going to be emitted in the 'forward' direction you'd simply need to add an inequality to your equations: namely $v_1>10$. (c) "What is the deciding factor?" Structural weaknesses in the object, determining how it will break apart (as in a hand grenade). $\endgroup$ Commented Aug 1, 2021 at 17:12
  • $\begingroup$ Take a look at my answer. The question contains some extremely useful simplifications, which make the answer for all horizontal directions, easy. $\endgroup$
    – Stilez
    Commented Aug 2, 2021 at 13:49

Before explaining the two.solutions, i would like to observe that we can solve this very quickly by making one important simplifying observation:

At the top of their curve, they separate. Their velocity at the top of the curve is horizontal. The new energy of explosion is directed horizontally too. So there is zero change to their vertical velocity due to the explosion. So they will take the same time to reach ground, among other things.

If they separated but no extra energy was added, their separation at landing would obviously be zero. So all we need to do is totally ignore everything except two factors - how long till they land, and the relative horizontal velocity of the fragments due to the separation.

Time of flight after separation: vertical velocity 20.sin(60) = 17.3m/s hence 17.3/g = 1.46 seconds up and therefore also 1.46 seconds down.

Relative velocity due to separation is said to be velocities that result in double the previous KE as at highest point. At the highest point the system has no vertical velocity,its only KE is horizontal velocity: 5kg moving at 20.cos(60) m/s = 10 m/s. KE is mv2/2 = 250 J. So KE after explosion is twice this, or 500J.

This is easy now, we solve for no net momentum change, final KE of 500J and fragments of mass 1kg+4kg:

  1. m1 = 4 and m2 = 1 (labelling larger fragment m1)
  2. m1v1 + m2v2 = (m1 + m2).10 (no net change to momentum), therefore substituting and rearranging: v2 = 50 - 4.v1
  3. m1v12/2 + m2v22/2 = 500 (system KE is doubled from 250J to 500J). Rearranging and substituting we get 4.v12 + (50 - 4.v1)2 = 1000 => 20.v12 - 400.v1 + 1500 = 0. Solving the quadratic, we find v1 = 15 or 5. Since v2 = 50 - 4.v1 we have two pairs of solutions:
  • (v1, v2) = (15, -10) or (5, 30).

Checking KE and momentum, we find indeed both solutions have KE = 500J and momentum unchanged (50 kg.m/s). And totally unsurprisingly in both cases the relative velocity of the fragments is identical, as we would expect: | 15 - (-10) | = | 30 - 5 | = 25. The relative velocity after separation is 25 m/s in both cases.

So after separation we have particles in flight for 1.46 seconds separating at 25 m/s.

(The fact they accelerate downwards doesnt matter because the downwards element of the fragments' motion have identical velocity at all times, so vertical motion and gravity wont make any relative change to separation or flight time of either fragment.)

Separation at landing is 1.46 x 25 = 36.5 m

Why two solutions?

Because there are 2 ways to double the systems KE. There are 2 ways to describe whats going on.....

  • Remember the mass reaches separation with positive velocity, 10 m/s. You could add a small amount of energy, both fragments still end up with positive (horizontal) velocities immediately after separation. Add a bit more, one of them ends up with zero horizontal velocity however, and in this case total system KE is less than doubled. Add even more energy and that fragment is sent into reverse, and has positive KE again. Thats why there are 2 solutions.

  • Another way to look at it is, one fragment is projected forward and one backward. Because they have different masses and they have a non-zero existing velocity (just before separation), there will be 2 sets of solutions, one for when the heavy mass is sent forwards (and the light mass is very greatly sent backwards, into reverse), the other when the heavy mass is sent backwards (but not enough to send it into reverse) and the light mass is very greatly sent forwards. So again, we get 2 solutions.

These two explanations are flip sides of the same coin. If you want a separation that exactly doubles the system KE, you have two choices: a smaller explosion that separates them with 4kg going backward and 1kg going forward, or a larger explosion sufficient to sent 4kg forward and 1kg backwards.

Anything in between these, ends up with less than doubled system KE. Anything beyond these, more than doubles it.

What if they split apart in a different horizontal direction?

Note that for simplicity I've assumed the fragments fly off in line with their existing flight. The question says they fly off horizontally but not that they fly off in the same direction horizontally. They could both fly off sideways, in opposite directions. In that case you'd have an extra variable Θ (the angle they fly off) and an extra equation ( total momentum conserved = 0 "sideways). Some of these equations would have terms in sin Θ and cos Θ, but otherwise nothing would really change. You'd still solve it the exact same way.

What would we expect and why?

We would expect exactly the same distance.

Remember that all we cared about is their relative velocity - not their absolute velocity. Yes the absolute velocity matters for final KE, but i think this logic is correct, and we can bypass that aspect by changing reference frame in a way that should not change the result. Read on......

If we used fragment 1's horizontal position as our frame of reference (we dont care about any velocity in common, nor the actual KE, only the separation at landing), v1 would have been zero, and we know from above that v2 after impact would have been 25 or -25 (still 2 solutions, but now they are clearer as to meaning, the relative direction of the smaller mass).

So intuitively, that itself has a physical interpretation: if they were launched straight up, and split apart, a relative velocity of 25 m/s would be the correct answer - you'd get two solutions again (+5, -20) and (-5, +20), obtained by subtracting 10 m/s from the above solution. This would still, give conservation of momentum and correct separated KE.

But notice if thats the case, then they could launch straight up, and split apart with relative velocity 25 m/s in any horizontal direction (because then every horizontal direction would be effectively identical in that frame), and that would mean that whatever horizontal direction they split, the relative velocity and time to landing would be unchanged. So it would still be 36.5 m at landing.

Notice the change of reference frame to prove this: from the observers frame they moved 10 m/s at peak in a specific direction, switching to a frame of that speed they are moving vertically only and have no horizontal velocity initially, which means we can pick any positive horizontal axis direction and the relative velocity after separation is unchanged by it.

All these frame changes are to other inertial frames, and in any constant horizontal velocity frame, horizontal momentum and relative horizontal velocity are conserved at separation, so we can see that the outcome is unchanged.

  • $\begingroup$ Good answer but (and i did the same) it breaks at the top and speed is 10 there so $KE_i=0.5*5*10^2 = 250$ $\endgroup$
    – Al Brown
    Commented Aug 1, 2021 at 22:53
  • $\begingroup$ Yoire right. How did i write 5 x 10^2 / 2 = 500, not 250......? Fixed! Thank you $\endgroup$
    – Stilez
    Commented Aug 1, 2021 at 23:36
  • $\begingroup$ For some reason i made the same mistake. $\endgroup$
    – Al Brown
    Commented Aug 2, 2021 at 1:30
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    $\begingroup$ There are infinite answers, as i put $\endgroup$
    – Al Brown
    Commented Aug 2, 2021 at 3:34
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    $\begingroup$ I would go even a step further. After deriving the time of fall and the energy of the explosion I'd switch to the reference frame of the exploding body and solve for 0 momentum and 250J kinetic energy… at which point there is only one rotationally symmetric solution. $\endgroup$
    – Jan Hudec
    Commented Aug 2, 2021 at 12:01

Mathematical perspective

The system you are trying to solve is

$$ \begin{align} m_1v_1+m_2 v_2&=(m_1+m_2)u\tag{1}\\ m_1v_1^2/2+m_2v_2^2/2&=(m_1+m_2)u^2/2+\Delta E\tag{2}\\ \end{align} $$

where $v_i$ is the final velocity of mass $m_i$ and $u$ is the initial velocity of the combined masses. $\Delta E$ accounts for the extra energy from collision (here equal to $(m_1+m_2)u^2/2$).

Consider the same system in the primed frame of the COM:

$$ \begin{align} m_1v'_1+m_2 v'_2&=0\tag{3}\\ m_1{v_1'}^2/2+m_2{v_2'}^2/2&=\Delta E\tag{4}\\ \end{align} $$

For this system, clearly if $(v'_1,v'_2)$ is a solution so is $-(v'_1,v'_2)$. This is the reason you obtain two solutions. Back in our original frame this means that $u\pm(v'_1,v'_2)$ are acceptable solutions.

Is the negation of velocities the only kind of transform that leaves the eqns. in COM unchanged? Note that the COM eqns. can be rewritten in vector form as

$$ \begin{align} m_1\mathbf{v'_1}+m_2 \mathbf{v'_2}&=0\tag{5}\\ m_1\mathbf{{v'_1}^T}\mathbf{v'_1}/2+m_2\mathbf{{v'_2}^T}\mathbf{v'_2}/2&=\Delta E\tag{6}\\ \end{align} $$

where $\mathbf{v^T}$ denotes the transposed vector $\mathbf{v}$. In this form any transform

$$\mathbf{v}\to \mathbf{w}=O\mathbf{v}\tag{7}$$ where $O$ is an orthogonal matrix also works. In other words if $\mathbf{u}+\mathbf{v'_i}$ is a solution, so is $\mathbf{u}+O\mathbf{v'_i}$.

Physical perspective

What does this mean physically? It means that as long as the direction of the velocities of the two ejected masses stay opposite to each other, that direction can be arbitrary. The transformation matrix $O$ then represents rotating or inverting this direction.

In $1D$, this degeneracy reflects in the form of the (only possibility) inversion of direction hence the two solutions.

In higher dimensions, rotations are possible in addition, so there would be an infinite no. of solutions - all rotations of one another.

What would happen in a real world experiment?

Even disregarding the chaos and non-linearity of a process as violent as an explosion, no real world body is perfectly homogenous and isotropic e.g. microscopic cracks, chipping, cavities etc. As a result not all directions of ejected masses are equivalent and the system will, in my opinion, fissure along the path of least resistance.

Would the direction be perfectly reproduced on successive repeats of the expt.? Highly unlikely. Even in theory, its basically impossible to predict every microscopic detail of such a chaotic phenomenon. One would generally model this using statistics - instead of predicting the exact direction the daughter projectiles would move along in each trial, one would predict the likelihood of the directions.

What would happen in a thought experiment?

It is impossible to predict which direction the ejected masses will go without further modelling of the fission dynamics (even that may not be enough). At our current level of modelling, all we can say is that the system would spontaneously break the symmetry and choose one of the directions. As remarked earlier, in the real world, the symmetry is only approximate.

On the theme of simple physical systems giving rise to multiple outcomes

This can be surprising at first. Even though Newtonian mechanics is deterministic, it is only as good as the model of nature we use it in. If our model is approximate in the sense that it may not be capturing finer details of a system, sometimes the predicted solutions exhibit multitude.

For example, in your mentioned numerical, if in addition it had been stated that the bulkier daughter projectile fell fore of the original mass, we would have accordingly rejected one of the values of the velocity pair.

Additional knowledge about a system helps reduce the possible space of solutions.

There are many such examples in physics.

  • Consider a projectile freely falling in a parabola. Given a height, at what times does the projectile reach that height? Again there are two time solutions. However, in this case our model does have the extra knowledge needed to reduce (or not) this space - the initial height and velocity.

  • Another similar and simple example is the buckling of a vertical rod under axial stress. The plane in which the rod buckles can't be predicted in advance unless imperfections that would break its axial symmetry are included in our model.

  • An important case where a system initially had multiple solutions yet with further and finer modelling, only one eventually emerged as the 'true' solution was the case of the quantum ground state of the hydrogen atom.

  • Conversely, with better models, sometimes we can discover more solutions to the original problem than previously thought. This can sometimes lead to spectacular predictions that precede experimental discovery as well as lead to new insights. Some examples of such additional solutions include the discovery of positron, quantum tunneling, and slow compression waves in porous media.

  • Consider the case of $2D$ elastic collision of two point masses. Unless the relative orientation of the masses at the moment of collision is also given, the system isn't even solvable. Hence modelling such a collision using point masses fails while using objects of finite extent doesn't. This shows the affect of model assumptions on solubility and nature of solutions.


To expand on my comment. Since the question is quite ambiguous there also exist the interesting scenario where the fragments fly off in the z-axis (see picture). I believe that the only reason for the information that the fragments fly "horizontally" is that we do not need to take into account the loss or gain of GPE. But the direction of the horizontal-trajectory is not specified meaning it could be in the x-axis giving two equally valid solutions has you noticed and as Ankit nicely explained. But it could also be in the z-axis note that it would also here be two valid solutions $v_{1}\approx -7\rm ms^{-1}$ or $v_{1}\approx 7\rm ms^{-1}$. And this is the same reason as Ankit explained, it depends on which way the explosion is oriented.

Note also that it could be any arbitrary direction in zx-plane and it would always give two solutions.

enter image description here


The only additions not covered completely in any other answers are:

1. As was mentioned by Stilez, and as we should expect, it does work in any horizontal direction at any angle from the $\mathrm{5kg}$ object’s horizontal velocity, not just parallel and anti-parallel to it’s velocity, as also suggested by discussion in @ludz’s answer. The math is below.

And, to re-emphasize this:

2. The two solution values for $v_1$, and the two for $v_2$, are centered around 10, ie around $v_i$. The 4kg mass goes backward relatively at 5 (at speed 10-5=5) or forward relatively at 5 (at 10+5=15). And the 1kg relatively at +/- 20 (at 10 +/- 20).

Let $y$ be vertical direction, $_i$ denotes just before explosion and $_f$ denotes after:

$m_1=4kg, m_2=1kg, m_{i}=5kg,$

$v_{xi}=5m/s, v_{zi}=0$

For any explosion angle $\theta$:

$v_{1x}= 10 + 5\sin({\theta})$

$v_{1z}= 5\cos(\theta)$

$v_{2x}= 10 - 20\sin(\theta)$

$v_{2z}= -20\cos(\theta)$

$$mv_{tot x}= m_1v_{1x} + m_2v_{2x}=(40+10)+(20\sin\theta-20\sin\theta)=50=m_iv_{xi}$$

$mv_{tot z}= m_1v_{1z} + m_2v_{2z} = 20\cos\theta - 20\cos\theta = 0$

$KE_f=\frac{ [m_1|v_1|^2 + m_2|v_2|^2]} {2}$

$= \frac{m_1 [(10 + 5\sin\theta)^2+ (5\cos\theta)^2] }{2} +\frac{ m_2 [(10 - 20\sin\theta)^2+ (-20\cos\theta)^2] }{2}$

$= \frac{[400+ 100(\sin^2\theta+\cos^2\theta)+ 400\sin\theta]}{2}+ \frac{[100+ 400(1)-400\sin\theta]}{2}$

$=500 = KE_i$

Since this is for all theta $0$ to $2\pi$, at each angle from $0$ to $180^O$ it covers $m_1$ and $m_2$ splitting off both ways.

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    $\begingroup$ I am unable to understand how you got the value of $KE_i$=1000 , isn't it 250? $\endgroup$ Commented Aug 1, 2021 at 8:44
  • $\begingroup$ @tanishadaharwal yes youre right. I thought it’s speed was 20. It’s 250👍🏻 $\endgroup$
    – Al Brown
    Commented Aug 1, 2021 at 22:38
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    $\begingroup$ @ACB thanks look forward to learning whatever u did to edit $\endgroup$
    – Al Brown
    Commented Aug 2, 2021 at 3:39
  • $\begingroup$ Both these points are covered in my answer,as of yesterday. But nice to see them focused on, and a mathematical (vs reference frame) answer added on those points. $\endgroup$
    – Stilez
    Commented Aug 2, 2021 at 10:17
  • $\begingroup$ @Stilez youre right. My mistake, didnt read the bottom of yours well. Editing $\endgroup$
    – Al Brown
    Commented Aug 2, 2021 at 10:45

The solution will always be unique if you provide the initial position and momentum (or equivalently, velocity) of the particles. In your example, the energy of the particles is what is given, which is quadratic on momentum. Therefore, there are two possible initial conditions that satisfy the energy requirement that the problem provided.

I would say it is not a real physical issue, but a sort of riddle developed by whoever wrote the problem and gave you the energy of the particles rather than the initial condition that determines the physical system. You can think that this ambiguity comes from the "ignorance of the experimentalist", who does not have access to full information about the system.


Assume we are working in right hand 3d coordinate system with xy plane along the screen and z axis out of the screen.

In an explosion , the two parts can fly in any directions according to the mass distribution of the object, point of origin of explosion within the object and so on.

Now let us deduce which direction the question is about.

  • Since it mentions the horizontal direction, it means there is no y-component of velocity. That is, only xz plain remains.
  • Still in xz plane, there are uncountable directions. Judging by the nature of problem, it is a very basic problem taught in elementary mechanics course, to teach conservation laws and not complicated 3d coordinates. Also, the problem does not mention any z component of initial velocity. If these assumptions are true, then only possible directions remain are along x axis.

Now since the solution involves quadratic equations, it will obviously give two possible roots. (These cases of more than two roots(solution) are very common in physics.) The maths does not take into account what is actually possible physically. For instance, there are many problems where we get two values of time, distance etc. We interpret these numbers according to physical reality. We won't take negative time and distance as it is impossible. Now in your case, there are two possible solutions. Notice that in first case, one mass has positive and other mass has negative velocity. This is very probable event as commonly explosion leads to masses flying in opposite directions. But in the second case, both masses are flying in same direction. This seems improbable. But on thinking further, this solution is only possible if lighter mass is facing toward positive direction of x-axis at the time of explosion as masses cannot pass through one another.


Ankit has given a great explanation, and his images reminded me something. Here in your question, you have pre-assumed that the velocities after the collision should be along the horizontal direction. But the question doesn't verify a direction (But I personally think that they wanted you to do what you did). Hence there are many possiblities. That parts can go up or down as well, according to the nature of collision. Or they can be inclined directions too!

EDIT: Noted that in your question they state, "The fragments separate horizontally after the explosion" : |

Math is beautiful, isn't it? You even didn't expect to get two answers (me as well) but math knew that! If she was able to tell you, she would surely explain which one is happening at what time.


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