A45.8-kg girl is standing on a 151-kg plank. both originally at rest on a frozen lake that constitutes a frictionless, flat surface. the girl begins to walk along the plank at a constant velocity of 1.46î m/s relative to the plank.
(a) what is her velocity relative to the surface of ice? 1 m/s(b) what is the velocity of the plank relative to the surface of ice? 2 m/s

a. 1.12m/s

b. 0.34m/s

Explanation:

Let Vgi = the velocity of the girl relative to the ice

Let Vgp = the velocity of the girl relative to the plank

Let Vpi = the velocity of the plank relative to the ice

Vgi = Vgp + Vpi

From the question, Vgp = 1.46/s

So, Vgi = 1.46 + Vpi

Conservation of Momentum (Relative to the ice), we have

Initial Momentum = Final Momentum

Because the girl and the plank are at rest initially, the initial Momentum = 0

So, 0 = Mg * Vgi + Mp * Vpi

Where Mg = Mass of the girl = 45.8kg

Mo = Mass of the Plank = 151kg

Make Vpi the subject of the formula, we then have

-Mg*Vgi = Mp*Vpi Divide through by Mp

Vpi = -Mg * Vgi/Mp

Vpi =( -45.8 * Vgi)/151

Vpi = -45.8Vgi/151

Remember that, we have (Vgi = 1.46 + Vpi)

We then substitute the expression of Vpi in the above equation

That is;

Vgi = 1.46 + (-45.8Vgi/151) Open the bracket

Vgi = 1.46 - 45.8Vgi/151 Multiply through by 151

151 * Vgi = 151 * 1.46 - 45.8Vgj

151Vgi = 220.46 - 45.8Vgi Collect like terms

151Vgi + 45.8Vgi = 220.46

196.8Vgi = 220.46 Divide through by 196.8

Vgi = 220.46/196.6

Vgi = 1.1213631739572736

Vgi = 1.12 m/s (Approximated)

So, the velocity of the girl relative to the ice is 1.12m/s

b. Velocity of the plank, relative to the ice

We can solve this using (Vgi = 1.46 + Vpi)

All we need to do is substitute 1.12 for Vgi in the equation

So, we have

Vgi = 1.46 + Vpi becomes

1.12 = 1.46 + Vpi Collect like terms

1.12 - 1.46 = Vpi

-0.34 = Vpi

So, the Velocity of the plank is 0.34m/s to the left

a displacement distance is size of displ

d) physical means can be used to separate the components of air.