Weird Physics Problem.... Conservation of momentum.?

Question

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 is 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.

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 the ice?

I'm confused. I'm pretty sure I'm doing it right, yet I'm getting the wrong answer.
Here's what I'm doing:

M(girl)=45kg
M(plank)=150kg
V(girl)[initial]=0
V(plank)[initial]=0
V(girl)[final]=1.5
V(plank)[final]=?

M(girl)V(girl)[initial] + M(plank)V(plank)[initial] = M(girl)V(girl)[final] + M(plank)V(plank)[final]

So: 0 + 0 = (45)(1.50) + (150)(V)
So: 0 = 67.5 + 150(V)
So: 67.5 = -150(V)
So: V = -.45 <- Final Plank velocity.

So if the girl is moving 1.50 m/s, and the plank is moving -.45 m/s. Wouldn't her velocity relative to the ice be 1.50 - .45 = 1.05?

Book answers:
a) 1.15 m/s
b) -0.346 m/s

Oh my heck....
I failed to realize that the weight of the plank includes her weight because she's kinda standing on it. :D

But I got it!!
Thank you guys!!

Answer 1

As the girl proceeds forward at speed V1...the momentum must be conserved.

A) m1V1=m2V2 note (m2 = mass of the girl + mass of the plank since the girl is moving on the plank and with the plank)
V2= (m1/m2)V1
Vr= V1-V2
Vr= V1 - (m1/m2)V1 = V1(1 - m1/m2)
Vr= 1.5( 1 - 45/(150+45)) = 1.15m/s

B) V2= (m1/m2)V1
V2=(45/(150 +45)) 1.5= 0.346 m/s

Answer 2

velocity of plank rel to ice should be
velocit of girl X her weight = (weight of girl + weight of plank)Xvelocity of plant rel to ice

you will get .346 m/s

vel of girl w r t plank will be 1.5-.346 = approx 1.15 m / s