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Billiard ball collision 
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#1
Jan2014, 12:05 PM

P: 3

In an inelastic collision of a moving billiard ball with a second ball at rest, I understand that the two balls have the same velocity after contact, correct?
But at the instant of contact how do the velocites of the two balls change instantaneously? Doesn’t that imply an infinite force? What am I missing here? Thanks. 


#2
Jan2014, 12:08 PM

P: 3

Sorry. I mistitled my previous post. It should have read Inelastic collision.



#3
Jan2014, 01:06 PM

P: 1,000

Rather than try to measure the exact details of a collision, one can often summarize the relevant effects of a collision in terms of quantities such as "impulse" or "coefficient of restitution". 


#4
Jan2314, 08:08 PM

P: 3

Billiard ball collision
briggs444: Thanks for your post. It clarifies my problem but not completely. May I explain my problem and the approach I have been taking?
I am trying to exam the effect of a large, fast moving mass (an airplane weighing 25,000 kg moving at 350 m/sec) hitting a stationary object (say a a rod 1 square cm in cross section and 1 meter long that weighs 1 kg). I want to compute the force of the collision on the windshield.. Assuming that I know the strength of the windshield material (i.e. the value of the force (in pounds /sq. inch) that the winshield can withstand), I could then estimate how heavy the rod must be to break the windshield. First, is this an elastic or an inelastic collision? My approach By the law of conservation of momentum, I compute the velocity of the airplane (and the rod) after impact: Momentum before impact = Momentum after impact. Thus VaMa + 0 = V’Ma + V’(Mass of rod) = V’ (Ma +1) ...... where Va and V’ are the velocity of the airplane before and after the collision . V’ = VaMa/Ma+1 = Va [1/(1+1/Ma] = Va [ 11/Ma] = 350 [10.04 x 10*3] = 350 (1 4x10*5) So the airplane speed falls by 1400x10*5 m/sec = 1.4 x 102 m/sec = 1.4 cm/sec. Now, I compute the force that will slow the airplane speed down by this amount – using F=M x acceleration where acceleration = 1.4 cm/sec / delta t. But, in light of your post, what delta t do I use here... in a practical problem? I am stumped. Needless to say I welcome your (and anyone elses) comments. ...wdc 


#5
Jan2314, 08:35 PM

Engineering
Sci Advisor
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Thanks
P: 7,298

Another way to get the time is by experiment: hit the windscreen with a projectile that contains a force transducer, or an acclerometer, or make a high speed video (maybe 10,000 frames per second) and measure what happens. Then use that data to model a different situation with a different projectile. For "low tech" impact testing (like the shock resistance of laptop computers, cellphones, etc if they are accidentally dropped), usually the calculation is done on the basis of past experience, using the maximum acceleration the object is meant to survive on impact. The accelerations can be very large  e.g. thousands of times the acceleration due to gravity. For impacts that have "serious" consequences (e.g. car crashes, as well as impacts on aircraft), computer models that track the deformations and stresses in the structures over time are used. One of the industry standard programs used for this is LSDYNA3D. You might be interested in some of the animations on their website, e.g. click the "animated result" tab at http://www.dynaexamples.com/examplesmanual/misc/airbag  and explore around the rest of the site to find more. Note, those models are only small demos to show how the program works. "Real world" models may be literally thousands of times bigger, and take several days to run to simulate an impact lasting a few milliseconds. 


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