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Is momentum conserved? 
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#1
Mar2312, 04:28 PM

P: 963

In a ballistic pendulum test, the mv is added to the system and that total remains constant over time. This what conservation of Momentum states. But at the end of the swing velocity is zero. Thus no momentum. What is conserved here?
As in energy total energy always remains the same. In momentum its nor true even in ordinary motion with resistance, the velocity decreases thus momentum. To me it is just a Law of Union/Separation. After Union/Separation its not use/applicable. 


#2
Mar2312, 04:35 PM

P: 886

The momentum is transferred into the stand and then into the earth.



#3
Mar2312, 05:37 PM

P: 2

Thank you aziz and khashishi ,I agree with both of your answers with what everyone may know that momentum is conserved.With a rule that states that "momentum before is equal to momentum after".
If anyone is inquiring about energy then the rules that YES energy is also conserved but in a different form. 


#4
Mar2312, 05:39 PM

P: 2

Is momentum conserved?
YES it is.The rules states that momentum before is equal to momentum after.√



#5
Mar2312, 07:29 PM

P: 963

Thank you. I mean here i see the Conservation of Momentum only applies to union and separation of items. In other instances other laws apply. When a body moves, it follows the Newton's first Law. When a body moves and changes direction or/and magnitude, Newton 2nd law is used and sometimes we use KE and PE, conservation of energy.
Thus COM only applies only JUST before the union/separation and JUST after that. Conservation law should be at any instances/places. 


#6
Mar2312, 10:39 PM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,338

Conservation of momentum holds only as long as there is no "exernal" force. The force of gravity is an external force.



#7
Mar2412, 02:42 AM

P: 789




#8
Mar2412, 07:08 AM

P: 963

I agree that with no external force, means there's no change in momentum. Total sum of momentum remains constant. I still cannot figure out why we need conservation of momentum in solving eg. ballistic pendulum. Initial mv=final mv then we get velocity. Then KE to PE. Why not just KE incoming bullet to PE without resorting to COM if we assume no energy expended in the process(I've been making this error frequently). Conservation of energy is intuitively easy for accept. 


#9
Mar2412, 07:21 AM

P: 789




#10
Mar2412, 05:54 PM

P: 963

ok now i understand why we have to resort to COM.
In a collision, [itex]mv_{in}=mv_{out}[/itex] where no external forces involve. [itex]KE_{in}\geq KE_{out}[/itex] So applying COM is the most ideal transformation for a collision. 


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