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MV or 1/2MV^2by mtworkowski@o
Tags: 1 or 2mv2 
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#1
Jun2708, 09:58 PM

P: 207

My experience is that because a lighter bowling ball can be thrown faster, the force that the ball has is altered. Should I use MV or 1/2MV^2?



#2
Jun2708, 10:07 PM

HW Helper
P: 2,616

Neither. Force is not described by 1/2mv^2 or mv.



#3
Jun2708, 10:11 PM

Sci Advisor
HW Helper
P: 1,276

Can you maybe clarify your question? Neither MV nor 1/2MV^2 represents a force in any way.



#4
Jun2708, 10:17 PM

P: 172

MV or 1/2MV^2
MV = P which is Momentum.
Ke = 1/2 mv^2 which is Kinetic Energy which can be exchanged for Potential energy. Force is typically defined as , F=ma. 


#5
Jun2708, 11:22 PM

P: 207




#6
Jun2708, 11:55 PM

P: 2,954

Pete 


#7
Jun2808, 12:04 AM

P: 35

Just though I would throw in the more accurate formula for momentum (or so it seems)...
p = ɣmv ɣ = (1v^2/c^2)^(1/2) 


#8
Jun2808, 12:19 AM

P: 207




#9
Jun2808, 02:33 AM

P: 1,133

xArcherx's perspective is essentially the same as what you've been learning, but in general form. In classical physics, we mostly deal with what we can experience, which is one of the things that makes modern physics different (it tells us we've only been seeing part of the picture, and that phenomenon start acting differently over different scales).
If you've noticed, there is a c in the equations, which is the speed of light. Compared to v (the velocity of the bowling ball), its probably much bigger, so the formula for momentum can more or less be regarded as p = mv for this situation with the bowling ball. 


#10
Jun2808, 06:16 AM

Mentor
P: 16,101

A bowling ball might travel at 20 mph. That means we are talking about a correction in the 15th decimal place. Not only is this totally unmeasurable, it's tiny compared to the correction to the mass of the bowling ball as it travels and picks up and drops off various bits of this and that. So not only is it an unmeasurable correction, it's not even the biggest unmeasurable correction. 


#11
Jun2808, 07:44 AM

P: 207

But when looking at a trade off between speed and weight, we use MV and not 1/2MV^2?



#12
Jun2808, 07:47 AM

Mentor
P: 8,297

Note that the two things you have formulae for are momentum and kinetic energy, respectively. 


#13
Jun2808, 02:36 PM

P: 207




#14
Jun2808, 03:10 PM

Mentor
P: 15,055

Firstly, it is both conservation of momentum and conservation of energy that dictate what happens. I good starting point is to assume perfectly elastic collisions. Together, conservation of momentum and conservation of energy tell what happens.
Secondly, your physiology rather than physics is the determining factor here. You are missing a key factor, which is throwing a ball accurately. Suppose that you can throw a 16 lb ball fast enough to get more pin action than you can get with a 15 lb ball. That does not necessarily mean you should prefer the heavier ball. If the heavier ball makes you less accurate you might want to use the lighter ball. 


#15
Jun2808, 03:20 PM

P: 207




#16
Jun2808, 04:53 PM

P: 35




#17
Jun2908, 12:41 AM

P: 207

to all. you know my question is perfectly clear. If you can't answer it than i'm out. you can just debate it for the next thousand years, which i'm sure you will. By The idea of using relativistic physics on a bowling bowl. Ha Ha Hal. What a joke.



#18
Jun2908, 08:29 PM

P: 85

[tex]F\propto m[/tex]. Here's why. [tex]\frac{1}{2}mv^2=W=\int_0^d ma dx=ma\int_0^d dx=mad[/tex] So... [tex]\frac{1}{2}mv^2=Fd[/tex] and finally, [tex]F=\frac{1}{2d}mv^2[/tex] with d and v constants, [tex]F\propto m[/tex]. 


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