How Does Relative Velocity Affect Motion on a Frictionless Surface?

[gaimon]

I have done a bit of work and am not seeing where I am going wrong in this. So here's a try:
A 46.8 kg girl is standing on a 130 kg plank. The plank, originally at rest, is free to slide on a frozen lake, which is a flat, frictionless surface. The girl begins to walk along the plank at a constant velocity of 1.75 m/s relative to the pank.
a)what is her velocity relative to the ice surface?
b)what is the velocity of the plank relative to the ice surface.

What I've done is find the power of the girl (46.8)(1.75) to get 81.9.
Then to find the velocity of the plank relative to the ice, take that power and divide by the mass of the plank so 81.9/130=.63 m/s.
Then the velocity of the girl on the plank minus the plank on the ice: 1.75-.63=1.12
Problem is..that's not right.

Any help would be appreciated!

 

[OlderDan]

You are sort of on the right track, but the quantity you are dealing with is momentum, not power. You need to work in terms of the girl's velocity relative to the ground, not her velocity relative to the plank. You don't know what that is. What you do know is that if you call her velocity v, and the velocity of the plank u, both relative to the ground, then their difference will be 1.75 m/s (either u or v will be negative, depending on which direction you choose to be positive) If you want to make both u and v positive as you did in your attempt you can, but then it is their sum that is 1.75 m/s. If both are positive, then you are correct to equate the mass*velocity products as you have done. This will give you two equations involving v and u, which you can solve to get the answer.

 

[gaimon]

I see. So what I have done is then say 46.8(1.75+y)+130y=0; y=.463 which I then subtract from the velocity against the plank, 1.75-.463=1.29 m/s.
Which is right!
Then, for part B, I tried to divide that by the girls mass times that of the plank (46.8*130)/1.29=.4644, it seemed to make sese to me as then m^2/(m/s) would be m/s. Is that even reasonable?

 

[OlderDan]

I see. So what I have done is then say 46.8(1.75+y)+130y=0; y=.463 which I then subtract from the velocity against the plank, 1.75-.463=1.29 m/s.
Which is right!
Then, for part B, I tried to divide that by the girls mass times that of the plank (46.8*130)/1.29=.4644, it seemed to make sese to me as then m^2/(m/s) would be m/s. Is that even reasonable?

From your equation, your result for y must be a negative number. You then correctly added the negative number to 1.75 to find the answer to a).

For part b) you would want the product of the girls mass times her speed divided by the mass of the plank. What you did is not what you said you did, and as you observed it is dimensioanlly incorrect. That should be a clue that you did something wrong. Now if you think a bit more about what you are being asked in part b), and what you did to find the answer to part a), you should realize that there is an alternative to this multiplying and dividing stuff. What is the y in part a)?