Momentum transfer in a first class lever

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dE_logics
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We have a seesaw arrangement, a mass m3 supports the fulcrum, we have mass m2 on one end, and a mass m1 fall on the other end with velocity v.

Considering the distance between the fulcrum and m1 is d and that of m2 is r, what will be the momentum distribution?
 
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I think -

How much the force is applied to m3 depends on m2's mass, cause the reason why force is applying on m3 is cause of m2's reluctance to change in motion (or cause of inertia).
 
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dE_logics said:
We have a seesaw arrangement, a mass m3 supports the fulcrum, we have mass m2 on one end, and a mass m1 fall on the other end with velocity v.

Considering the distance between the fulcrum and m1 is d and that of m2 is r, what will be the momentum distribution?

Hi dE_logics! :smile:

i think nobody's answering because we don't understand the question :redface: …

what do you mean by the "momentum distribution"?

what is the original question? :smile:
 
Now how can I define momentum distribution?

When 2 bodies collide momentum gets distributed...i.e they undergo momentum distribution.