Kinetic Energy Loss of Colliding Billiard Balls

AI Thread Summary
The discussion focuses on calculating the kinetic energy lost during a collision between two billiard balls. The initial ball, with a mass of 1.9 kg and a speed of 20.1 m/s, strikes a stationary identical ball, resulting in specific angles and speeds for both balls post-collision. Participants emphasize the importance of using vector components for velocity in energy calculations and maintaining significant figures to ensure accuracy. The conservation of momentum and energy principles are highlighted as essential for solving the problem correctly. Ultimately, the correct approach involves recalculating final velocities without premature rounding to accurately determine the kinetic energy lost.
SalsaOnMyTaco
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Homework Statement



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?

Homework Equations



ΔKf-ΔKi

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?
 
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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
 
ehild said:
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

v1f= 1.34v2f

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
 
SalsaOnMyTaco said:
v1f= 1.34v2f

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

To make your solution nice, you should write down the basic steps and the principle behind. From what law do your equations follow? (they are correct , but you should explain)

Regarding 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
 
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.
 
ehild said:
To make your solution nice, you should write down the basic steps and the principle behind. From what law do your equations follow? (they are correct , but you should explain)

Regarding 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

P=mv therefore since the billiard balls go different directions, this is an elastic collision, so:

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+v2f2 - ΔKi
 
hm, okay I am 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) ?
 
pblackblackfff!. alright I finally got the answer. Thanks!
 
You helped to yourself I see...:smile:

ehild
 
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