Is this question dealing with law of conservation of momentum?

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SUMMARY

The discussion centers on the application of the law of conservation of momentum in a scenario involving a 75-kg man moving in a 120-kg canoe. The relevant equation is m1v1 + m2v2 = m1v1' + m2v2', which represents the conservation of momentum before and after the man begins to move. The initial momentum of the system is zero, as both the man and the canoe are at rest. When the man moves forward at 0.50 m/s, the canoe will move in the opposite direction to conserve momentum.

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Homework Statement


A 75-kg man sits in the back of a 120-kg canoe that is at rest in a still pond. If the man beings to move forward in the canoe at .50m|s relative to the shore, what happens to the canoe?


Homework Equations


The equation they gave me is m1v1+m2v2 = m1v1'+m2v2'
I don't get it.


The Attempt at a Solution


I didn't really make an attempt. All I have on the paper is
(120 kg)(.50m|s) + ... I don't know what to put there. If it's even right!
 
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You have 2 different parts. At rest and moving, aka before and after, aka primed and unprimed.
 

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