[nj_bartel]

The problem reads:

A 45.0 kg girl is standing on a 142 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 1.50 m/s relative to the plank.

(a) What is her velocity relative to the surface of the ice?

(b) What is the velocity of the plank relative to the surface of ice?



To solve this, I used conservation of momentum and set m1v1 = m2v2, or
45(1.5) = 142(v2).  I then solved for v2 and got 0.4754, so -0.4754 m/s
would be the velocity of the plank relative to ice, which would be the
answer to part (b) (which was wrong by 10%-100%).

I then took the difference between the speed of the plank relative to
the ice and the speed of the girl relative to the plank (1.5 m/s -
0.4754 m/s), and came up with 1.0246 m/s, which was also wrong by 10% -
100%.

Where am I going wrong?

Thanks

[enahs]

I am sure there is some way to answer this, based on how you are doing problems in class.
But this question is highly flawed.

Does the plank actually move? It depends on how you walk and the direction of the force. If you step properly virtually all the force will be directly down and there will be no significant lateral motion, or you can walk in a manner to generate a lot of lateral motion.

So, I can answer A as 1.5 and B as 0, or I can walk in a manner such that A is .75 and B is .75, or anywhere in between....

[nj_bartel]

I believe you're to take it very simply and assume all motion is lateral.

[Borek]

But this question is highly flawed.

Does the plank actually move? It depends on how you walk and the direction of the force.

It is told that lake surface is frictionless, so yes, the plank moves. 45*1.5=142*v_plank.

[nj_bartel]

Anything else Borek?  That's how I set it up and it came out wrong =/

[Borek]

My bad, looks like we have both mixed frames of reference. I think what I wrote will work if the speed of the girl is 1.5 relative to the ice.

Sorry, no time for that now, will try to remember to check it later.

[nj_bartel]

Grad student figured it out.  It's conservation of momentum, but you have different frames of reference in the given info, so you have to set up 2 equations and solve for one of the velocities then substitute to get the same reference frame.

[Borek]

Just got back to my computer; more or less that's what I expected - just the details were blurry.