Intergalactic Bowling

Intergalactic Bowling

1) The centers of two bowling balls are positioned 1 meter apart in intergalactic space (ie so that gravitational effects from other sources are negligible). On Earth, one ball weighs 12 lbs (5.443 kg) and the other ball weighs 10 lbs (4.536 kg). Each ball is 8.5 inches (0.216 m) in diameter.

        a) How long will it take for the balls to collide?

        b) How far will each ball have traveled at the point of impact?


2) In a different part of intergalactic space, 10 bowling pins are positioned in a typical triangular arrangement with the centers of the 3 corner pins located at the apices of an equilateral triangle measuring 36 inches (0.914 m) on a side. A bowling ball is then positioned with the center of the bowling ball 60 feet (18.288 m) from the center of the closest pin. On Earth, the bowling ball weighs 16 lbs (7.257 kg) and each pin weighs 3.5 lbs (1.588 kg). (In this problem, assume the position of each pin to be fixed [ie so that neither the gravitational force of its neighbors nor of the approaching ball causes a change in its position]. Also, the diameters of the bowling ball and pins can be disregarded).

        a) How long will it take for the bowling ball to strike the closest pin?

        b) What will be the velocity of the ball at the time of impact?

Note: Just like in physics class, please show your calculations (ie not just the answer).
WeThotUWasAToadAsked:
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d-glitchCommented:
This appears to be an academic assignment.  As such, you can get help but not answers.
Post your work or ask queztions about things you don't underdtand.

Do you know the law of gravitation (with the capital G)?
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WeThotUWasAToadAuthor Commented:
LOL I'm 56 years-old and a physicist wannabe but unfortunately not taking any courses right now. A friend of mine and I came up with these questions out of curiosity. However, I can certainly see how someone in school could attempt to get their homework done by someone at EE.

Most of the gravitation problems I've played around with involve an object on Earth. Thus the attraction and displacement is primarily in one direction. Maybe my problem is just with the algebra but I'm not quite sure how to apply Newton's laws when two objects of roughly the same size are involved.

To answer your question, yes, I am familiar with the laws and formulas involved:

F = G•Bb/d^2

where B & b represent the masses of the 2 objects, d is the distance between them & G is the gravitational constant:  

6.67384x10^-11 m^3/kg-s^2 (where m = meters, kg = kilograms, s = seconds)

F = G (5.45 kg)(4.54 kg)/m^2 = G (24.74) kg^2/m^2

F = 6.67384x10^-11 m^3/kg-s^2  (24.74) kg^2/m^2

F = 1.65x10^-9 kg-m/s^2 (magnitude of the force)

Now, rearranging F = ma (force = mass x acceleration) to a = F/m and applying that to each object, I got:

Acceleration for the larger ball is:
        a(B) = (1.65x10^-9 kg-m/s^2)/(5.45 kg) = 3.03x10^-10 m/s^2

and acceleration for the smaller ball is:
        a(b) = (1.65x10^-9 kg-m/s^2)/(4.54 kg) = 3.64x10^-10 m/s^2

which makes sense since (intuitively) the same force will cause a lighter object to accelerate at a greater rate than it will a heavier object.

But now is where things aren't so clear. I've got that the three kinematic equations for constant acceleration (which is the case here):
Fig.1I know that the sum of the distance traveled by each object is 1m:

        d(B) + d(b) = 1 m

and the traveling time for each object is the same:

        t(B) = t(b) = t

but that's where I get hung up.

It's the algebra, right? I've tried rearranging things and substituting but I feel like I'm chasing my tail but hopefully my physics is accurate.

Thanks
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phoffric\Commented:
The distance between the objects is getting smaller and so the forces are getting larger, and so the acceleration is getting larger also. Equations that assume constant acceleration will not work.

Another consideration for part b is to realize that since there are no external forces acting on this system of two objects, then their center of mass will remain constant. So compute the CM, and then put the objects together in a way so that when they touch, the CM has not changed. In this case, when they collide, the CM will be within one of the two objects. Here are some details:
http://www.experts-exchange.com/Other/Math_Science/Q_25280369.html#a27439063
And you will find other answers that will help you in part a.
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d-glitchCommented:
Please excuse the challenge.  A quick look at your Question History would have indicated that you probably are not a student.

Thank you phoffric for remembering that earlier question.  It was a long time ago.

I have reworked the earlier spread sheet for the current problem.
The 12 lb ball is the Earth (with an e subscript), and 10 lb ball is the Moon (with an m subscript).
It shows a collision at approx 40750 seconds.

Note that the values for force and acceleration in the first line (t=0) match your calculations.

The non-numerical way to solve this involves a double integral:
     The displacement of each ball from its initial position is the integral of its velocity...
     Which is the integral of its acceleration...
     Which is a function of its position with respect to the other ball.

Bowling-in-Space.xlsx
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ozoCommented:
Time to collision calculation might be approached more easily by thinking of the balls as being in highly elliptical orbits.

Final velocity calculation might approached more easily by considering conservation of energy.

Distance traveled might approached more easily from the center of mass frame.
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WeThotUWasAToadAuthor Commented:
Thanks for the insights & comments.
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ozoCommented:
The time for a circular orbit of two balls can be determined from straight forward force and circular motion calculations.
Based on that, Kepler's third law can give you the time for an orbit with eccentricity 1.
But you need to adjust that because the size of the balls means that contact occurs before periapsis.
An easy way to do that is to use Kepler's second law.  If you stretch the infinitesimal minor axis back into a circle, the area ratios are unaffected and formulas for the area of a circular segment give you the time ratios.
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