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I was checking a question about projectile and found something quiet interesting.

enter image description here

The question is:

Calculate $R$ if the ball of mass 1kg was split in mid-air evenly.

Looking at the picture, I can imagine this happening in reality, however, physically I cannot understand why the upper part goes further than the lower one even though they had the same speed when splitting and have the same mass.

Can someone please explain this.

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    $\begingroup$ The range depends on other parameters too, besides the speed. $\endgroup$ – nasu Oct 6 '16 at 20:04
  • $\begingroup$ What to explain? Do you know what other parameters determine the range? $\endgroup$ – nasu Oct 6 '16 at 20:19
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    $\begingroup$ The center of mass is right in the middle. You take it from there. $\endgroup$ – Mike Dunlavey Oct 6 '16 at 20:34
  • $\begingroup$ @MikeDunlavey- Sorry but I am new to projectiles and trajectories. Can you explain the effect this might have. $\endgroup$ – Basem Fouda Oct 6 '16 at 20:53
  • $\begingroup$ @nasu what other parameters may affect the range? Can you please name some? $\endgroup$ – Basem Fouda Oct 6 '16 at 20:54
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Whatever force acts on the 2 fragments to make them separate, it is an internal force. It acts equally on both fragments for the same amount of time. So the impulse (=change in linear momentum) is the same for each fragment (ie equal in magnitude, opposite in direction). This is true whether or not the fragments have the same mass.

Because the fragments have the same mass, they also have the same speed (but in opposite directions) relative to the their centre of mass when they split up. But they do not have the same velocity relative to the ground. If they did have the same velocity relative to the ground, they would continue travelling together without moving apart : the projectile would not split up.

While they are still in flight the centre of mass (CM) of the 2 fragments follows the same trajectory as the complete projectile if it had not spit up. The additional momentum given by the internal force will carry one fragment further than the CM and the other fragment not as far as the CM.

However : note that the midpoint of the 2 fragments on the ground is not necessarily the point at which the complete projectile would have hit the ground. If the fragments hit the ground - and stop falling - at different times, then their CM on the ground is likely to be different from the point at which the complete projectile would have landed.

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The assumption is that momentum is conserved during the explosion.
Relative to the original projectile one fragment moves faster whilst the other moves slower whilst conserving momentum.

I would consider motion in the horizontal plane as that will determine the range of fragments.

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  • $\begingroup$ But there is no explosion, and this was emphasised in the question, however I have summarised it. Is there any other possibilities? $\endgroup$ – Basem Fouda Oct 6 '16 at 20:12
  • $\begingroup$ There could have been a spring mechanism inside the projectile. Whatever there was on the projectile / two fragment system there are no forces in the horizontal plane and after the splitting the fragment have different horizontal speeds. $\endgroup$ – Farcher Oct 6 '16 at 20:16
  • $\begingroup$ Again there is no external forces acting on the ball, please check out the answer below $\endgroup$ – Basem Fouda Oct 6 '16 at 20:35
  • $\begingroup$ @BasemFouda The other answer, which is above my answer for me, says the impulses on each fragment are the same which implies conservation of momentum. $\endgroup$ – Farcher Oct 6 '16 at 20:48
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According to Wikipedia, mass is irrelevant so the only discrepancy in where they land will come from differences in the angle of trajectory and initial height caused by how the projectile is split.

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  • $\begingroup$ the initial velocity is only horizontal and there is no vertical component, and the ball was at the same angle. $\endgroup$ – Basem Fouda Oct 6 '16 at 20:37
  • $\begingroup$ When the projectile is split each half will invariably have different trajectory conditions caused by the split. $\endgroup$ – Yogi DMT Oct 7 '16 at 16:26
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When the ball split apart, the two halves had, in the rest frame of the ball, equal nd opposite velocities (imparted by whatever force/impulse drove the half-balls apart).

Since you don't know the direction of the splitting impulse (although if the problem remains planar, you do know the plane it lies in) and you don't know the time of splitting, it is by no means a gimme that the distance $R$ is determined by the problem's conditions. You can, however, assume the direction of the impulse is purely horizontal (because the split in the ball looks to be purely vertical). But the dependency on the time of splitting and the velocity of splitting is still troublesome., because the landing spot of one ball only fixes one parameter. However, if you use the principle of "this is a posed problem, the time of splitting and the velocity must conspire to give one unique answer" then this is an easy solve.

Without the splitting, the ball would have landed at $10\frac{\sqrt{40}}{g} \approx 20.2$ meters. If the splitting happens at the last split-second, the halves will end up equi-distant from the splitting point, which is at $10$ meters -- given -- and $30.4$ meters -- the answer.

In point of fact, as long as the splitting impulse is horizontal, the answer will remain $30.4$ meters. The algebra is not too convoluted.

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  • $\begingroup$ You are indeed right and the answer is as follows: $\endgroup$ – Basem Fouda Oct 6 '16 at 20:28
  • $\begingroup$ If object is projected horizontally, the object moves at the constant velocity horizontally and does at the constant acceleration vertically. From the distance for vertical movement (H=20 = $\frac{1}{2} 10^2 t^2$ ), one can find the time of falling is 2 sec. So, the distance for horizontal movement is 20 m. From the equation of center of mass, one can find the position of the other piece (X), $20=\frac{0.5 * 10 + 0.5*x}{1}$ R=30m $\endgroup$ – Basem Fouda Oct 6 '16 at 20:31
  • $\begingroup$ Can you please explain the last bit because it seems to be the part which has the answer to my question $\endgroup$ – Basem Fouda Oct 6 '16 at 20:32
  • $\begingroup$ It also indicates that there is no external forces $\endgroup$ – Basem Fouda Oct 6 '16 at 20:33
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    $\begingroup$ Shouldn't the result be 30.4? $\endgroup$ – aventurin Oct 7 '16 at 18:29

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