I don't understand what the following problem is trying to say.
A 45 kg girl is standing on a 150 kg plank. The plank, originally at rest is free to slide on a frozen lake, which a flat, frictionless surface. The girl begins to walk along the plank at a constant velocity of 1.5 m/s to the right relative to the plank. What is her velocity relative to the ice, what is the velocity of the plank relative to the ice?
Are they trying to say that when she walks 1.5 m/s to the right that the plank will advance at .45 m/s since the weight of the girl divided by the weight of the plank is .3 which times 150 is .45? Couldn't she walk in a way that would transfer the energy directly downwards through the plank into the ground...?
Update:Alright, the book got a different answer to me so they must have been mistaken.
Answers & Comments
Verified answer
To answer it, you mutiply her velocity relative to the plank by her mass, then divide that by the plank mass. That gives the velocity of the plank relative to the ice.
By subtracting that from the girl's velocity, you find her velocity relative to the ice.
(1.5 x 45)/150 = -0.45m/sec. plank velocity relative to the ice;
(1.5 - 0.45) = +1.05m/sec. girl's velocity relative to the ice.
This is a law of conservation of linear momentum problem. Since the system is initially at rest, and there is no friction, the momentum of the moving girl must be equal and opposite to the momentum of the plank.
E.g. sum(mass*velocity).initial = sum(mass*velocity).moving
0 = m.girl*v.girl + m.plank*v.plank
The plank velocity you found is correct, but it's important to note that it is negative (i.e. the plank is moving towards the left.)
The lack of friction between the plank and the lake means there is no force parallel to the plank to resist the plank's movement. As long as the girl is only touching the plank, it doesn't matter how she walks.