Is this question dealing with law of conservation of momentum?

[raek]

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!

 

[stellaman]

You have 2 different parts. At rest and moving, aka before and after, aka primed and unprimed.

 

[nlightner]

Yes, this question does deal with the law of conservation of momentum. This law states that the total momentum of a closed system remains constant, meaning that the total momentum before an event must equal the total momentum after the event. In this case, the man and the canoe can be considered as a closed system, and the law of conservation of momentum can be used to determine what happens to the canoe when the man starts moving.

Using the given equation, we can set up the initial and final momenta for the man and the canoe. The initial momentum is 0 since both the man and the canoe are at rest. The final momentum for the man is (75 kg)(0.50 m/s) = 37.5 kg*m/s. The final momentum for the canoe can be calculated as (120 kg)(0 m/s) = 0 kg*m/s. Since the total momentum must remain constant, the final momentum for the man and the canoe combined must also equal 0 kg*m/s. Therefore, the final momentum for the canoe must be -37.5 kg*m/s in the opposite direction of the man's movement.

In other words, the canoe will move in the opposite direction of the man's movement with a velocity of -0.3125 m/s. This is an example of the law of conservation of momentum in action, as the total momentum of the system remains constant even though the man has started moving.