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The problem is as follows:

The diagram from below shows a sphere is hanging from a ceiling by a cable whose length is R as is shown in the diagram from below. A bullet whose mass is $m$ is fired and has a speed of $v$ and travels in a straight line and gets embedded in the sphere whose mass is also $m$. As a result of the collision the sphere swings and makes a circular orbit. What should be the minimum speed of the bullet such the sphere does a circular orbit?.

Sketch of the problem

The alternatives given are as follows:

$\begin{array}{ll} 1.&2\sqrt{2Rg}\\ 2.&2\sqrt{Rg}\\ 3.&4\sqrt{Rg}\\ 4.&3\sqrt{Rg}\\ 5.&\sqrt{0.5Rg}\\ \end{array}$

What I thought to solve this problem was that I can relate the momentum before and after the collision and with that speed I can relate it with the conservation of the mechanical energy when the bullet is embedded in the sphere.

This would become into:

$mv= \left(m+m\right) v_f$

$v_f=\frac{1}{2}v$

This speed can be used to relate it with the conservation of mechanical energy as follows:

$E_{k1}=E_u+E_{k2}$

$\frac{2m}{2}\left(\frac{1}{2}v\right)^2=(2m)gR+\frac{2m}{2}v_{t}^2$

$v_{t}=\textrm{Tangential speed when the ball swings}$

Then:

$\frac{v^2}{4} = 2gR+v_{t}^2$

However I end up with not knowing how to get the tangential speed or relating it to the speed which should had the bullet. How can I find it?. I'm guessing that it could be the second option.

But for that to happen would imply that:

$\frac{mu^2}{R}=2mg$

Then

$u=\sqrt{2Rg}$

So by the conservation of momentum would imply:

$p_i=p_f$

Therefore $mv=2mu$

$v=2\sqrt{2Rg}$

So that would mean that is the first option. But would that be the answer?. That would mean that the sphere rises to the top making half a circle.

Can an answer include some drawing to explain for a better view what it was intended in this problem? please. I'm stuck!.

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  • $\begingroup$ The speed must be such that at the top of its orbit its downward acceleration is $g$ and must equal $v^2/R$. Can you see why? $\endgroup$ – David G. Stork Dec 12 '19 at 5:58
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    $\begingroup$ I find it amusing that the author decided to hang the sphere directly from a ceiling rather than hanging it in any of the ways that would actually work. (How is the sphere supposed to reach the top of its circular orbit when there's a ceiling in the way?) $\endgroup$ – David K Dec 12 '19 at 6:15
  • $\begingroup$ @DavidK I was actually trying to understand that part as well because it didn't make sense to me. $\endgroup$ – Chris Steinbeck Bell Dec 13 '19 at 21:20
  • $\begingroup$ @DavidG.Stork I think you are referring because at that point the acceleration due gravity is the same as the centripetal acceleration. Do you mean this?. I'd appreciate that you may post an answer or something to help me clearing out the doubt as I mentioned in my question. $\endgroup$ – Chris Steinbeck Bell Dec 13 '19 at 21:23
  • $\begingroup$ Yes... that is exactly what I mean. Find the speed such that there is no tension in the rope at the apex. $\endgroup$ – David G. Stork Dec 13 '19 at 21:28
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The energy the instant after the bullet strikes is purely kinetic energy:

$$E_1 = \frac{1}{2} (2 m) \left( \frac{v}{2} \right)^2$$

The energy when the mass is at the top of the loop is the sum of its kinetic energy and potential energy:

$$E_2 = \underbrace{\frac{1}{2} (2m) V^2}_{kinetic} + \underbrace{(2m)g(2 R)}_{potential}$$

where $V$ is the velocity at the top.

By conservation of energy $E_1 = E_2$.

For the pendulum to make the full circle loop, the centripetal acceleration at the top must equal the acceleration due to gravity. (Otherwise, there would be tension in the rope and you could find a lower velocity that would enable the looping.)

So we must have $V^2/R = g$.

Put these together and solve for $v$.

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