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Balls Collision |
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| Jun16-12, 12:37 AM | #1 |
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Balls Collision
1. The problem statement, all variables and given/known data
A ball of mass m = 1.9 kg moving with a speed of v0 = 20.1 m/s strikes an identical ball which was initially at rest. After the collision, the incoming ball (ball 1) goes off at q1 = 26° relative to its original direction and the struck ball (ball 2) moves off at q2 = 36° as shown in the diagram. What magnitude of kinetic energy was lost (to sound, heat, etc.) in the collision? 2. Relevant equations ΔKf-ΔKi 3. The attempt at a solution The final velocity of the billiard balls came up to be ball 1= 13.29 m/s and ball 2= 9.92 m/s so...ΔK= 1/2 m(13.292+9.922)-1/2m(20.12) m=1.9 am i doing this right? |
| Jun16-12, 01:07 AM | #2 |
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Your method is correct to get the change of energy from the speeds. The velocity is not a number, but a vector. So you calculated the speeds of the balls as 13.29 m/s and 9.92 m/s. How did you get these values?
ehild |
| Jun16-12, 01:40 AM | #3 |
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20.1= v1f cos26°+v2f cos 36° plug in v1f= 1.34v2f and solve for v2f then plug in the result for v2f in the first equation to get v1f |
| Jun16-12, 02:02 AM | #4 |
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Balls CollisionRegarding v1f= 1.34v2f, do not round off the intermediate results too much. At the end, you have to subtract the two energies. Your speed values are accurate only for the first decimal. The energies itself will be even less accurate. If they do not differ much, your end result would be quite inaccurate. The data are given with two significant digits, you need to give the result also with two significant digits, and should not introduce errors with the calculation. Do not round off too early!!! ehild |
| Jun16-12, 02:04 AM | #5 |
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The acclaimed, research-based four-step problem-solving
framework ISEE(Identify, Set Up, Execute, and Evaluate). 1. Identify. a. Conservation of energy b. Conservation of momentum. |
| Jun16-12, 10:32 AM | #6 |
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Pi=Pf m1v1i+m2v2i=m1v1f+m2v2f So are you saying that i should re-do the problem (look for the velocities) except without rounding off to early, and plug in the new velocities into ΔK= 1/2m(v1f2+v2f[SUP]2[SUP] - ΔKi |
| Jun16-12, 10:46 AM | #7 |
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hm, okay im getting new velocities and they also seem to be right when I type them in.
v1f=13.46 v2f=10.05 should i add them up to get the final kinetic energy? 1/2(1.9)(13.46^2+10.05^2) ? |
| Jun16-12, 10:49 AM | #8 |
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pfffffffffffffff!. alright I finally got the answer. Thanks!
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| Jun16-12, 12:00 PM | #9 |
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You helped to yourself I see...
![]() ehild |
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