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-   -   Momentum distribution in a first class lever (http://www.physicsforums.com/showthread.php?t=289739)

dE_logics Feb4-09 08:00 AM

Momentum distribution in a first class lever
 
http://img16.imageshack.us/img16/7028/tempqk8.png


This is the second time I'm posting this; last time I was told that it was not explicit.

The image consists of 3 masses, m1, m2 and m3, out of which m1 and m3 lies at rest while m2 is in motion in the direction of the arrow.

When m2 collides with the bar, its momentum will be transferred to m1 and m3 or its redistributed between the 3 masses; now how will be this momentum distribution considering the collision is elastic?

dE_logics Feb4-09 09:50 PM

Re: Momentum distribution in a first class lever
 
Now the impulse distribution will be equal to the force distribution cause the time for which the force applies on the 3 masses will be the same.
Cause of the same reason, the force distribution will be = the momentum distribution. By this I conclude that the force on m3 cannot exceeds m2's, cause that way the momentum on m3 will be more than m2.
The force by m2 first applies action to m1, cause if m1 would have not been there, the seesaw would have swinged without effort, once this force has been initiated, the normal reaction given my m2, is traveled along the length of the seasaw where it falls on m3, and since m3 too gives a normal reaction, this normal reaction is distribute between m2 and m1 in accordance to m3's position (I guess). If m3 is in the center, then the normal reaction distributed will be between m2 and m1 will be 50-50%...i.e only half the velocity of m1 will be stopped (but its to be remembered that all the normal reaction is first taken by m3...so the force on it will be equal to the normal reaction of m3...and then it'll pass that normal reaction to m1)
However its to be noted that the force on m1 will continue to fall on m2 (via normal reaction from m3) unless the velocity of m1 is less than m2, which is only possible if all the momentum of m1 is transferred to m2 or the momentum is distributed between m3 and m2; in which case m2 will never gain completely the velocity of m1 cause some of the velocity component of m2 will be transferred to m3; the 50-50 % theory agrees with this part too, in which case, since some of the momentum has been taken by m3; m1 will loose velocity without giving momentum to m2. As we continue with this theory, it might be possible that at a point the velocity of m1 will exceed m2's simply cause its velocity has been taken by m3, but the momentum of m2 will be distributed between m3 and m1 at the same time, and so m2 is bound to stop cause...if (assuming) m1's velocity has 50%...and all has been transferred to m3 (assuming, actually 25% will be transferred to m2 and and rest to m3); only m2 is left to gain momentum from the rest 50%...and cause the masses of m2 and m1 are equal, m1 will have to stop in order to give m2 the complete momentum.

However whatever the force is, it to be noted that as m3 grows heavier, its reluctance to movement will increase, and so the normal reaction will be given with less change in velocity; now the velocity of m3 contributes to the degree of rotation of the stick...that's it, but if the stick rotates by a huge amount, there'll be a considerable resolution of force which will be against the direction of motion of the body; however this can be solved by using the gear arrangement.

According to another theory, the impulse on m3 will be 2 times to that of experienced by m1 or given by m2...the initial part of this one is common with the above, that is first the force travels to m1, then it give a normal reaction to m3, which gives another normal reaction distributed between m1 and m2, but its also to be noted that when force is made to apply on m1, apart from the normal reaction given by m1, m3 also experiences the force given by m2 to m3, so it doubles up.
But if force gets doubled so will be the momentum on m3 or it will be more than that of m2...but its an elastic collision...so that's not possible.

On increasing m3, the velocity increment will decrease, as a result by the formula of K.E, the energy transferred to m3 will too be less and rest of the energy will be transfered to m1 however this doesn't mean that the momentum distribution of m3 will too be less, I think it will be the same cause normal reaction given by it will be a constant irrespective of the mass of m3.

OK...by this I get that momentum distribution will not be equal when m3 is increased; this will mean that more velocities has been given to m1 and so more momentum. You can see this practically!
But this will be against the law of conservation of momentum/the predicted theory...you see the impulse applied on m3 will be the same no matter what the mass; that is same momentum will be transferred to m3. But if the velocity of m1 increases, on increasing m3, then momentum of m1 has been increase cause of the weight increment in m3 and and momentum gained by m3 is static!...that means as the mass increase, the momentum gained by the system.

If m1 and m2 increase by a huge amount, then this will be the same scenario as decrease in mass of m3.


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