Two ice skaters, Skater X and Skater Y, are at rest on a horizontal surface made of ice. The friction between the ice and the skaters is negligible. Skater X, who has a smaller mass than Skater Y, pushes Skater Ysuch that Skater Y travels with a speed of 2vo to the right. Which of the following indicates the direction of the velocity of Skater X and the direction of the velocity of the center of mass of the two-skater system after the push?

a. Skater X Center of Mass
To the left To the right

b. Skater X Center of Mass
To the left Zero

c. Skater X Center of Mass
Zero To the right

d. Skater X Center of Mass
Zero Zero

Explanation:

Both skaters are initially at rest

Then,

Ux = Uy = 0m/s

It is assumed that it is a frictionless surface

Mass of skater X is less than Y

Mx < My

My > Mx

This shows that,

My = a•Mx,

Such that a > 1

Skater X pushes Y such that it moves with velocity.

Vy = 2Vo to the right, I.e positive x axis

Vy = 2Vo •i

We want to find direction of the velocity of skater X

Using conservation of momentum

Initial momentum =Final momentum

Mx•Ux + My•Uy = Mx•Vx + My• Vy

Ux = Uy = 0,

also My = aMx. And Vy = 2Vo •i

0 + 0 = Mx•Vx + a•Mx•2Vo •i

0 = Mx•Vx + 2a•Mx•Vo •i

Mx•Vx = —2a•Mx•Vo •i

Divide through by Mx

Vx = —2a Vo •i

Therefore, since a is always positive

Then, Vx is in the negative direction or opposite direction to Vy

Now, center of Mass

The center of mass calculated using

Vcm = 1 / M Σ Mi•Vi

Where

M is sum of all the masses

M = Mx + My = Mx + aMx = (a+1) Mx

Then,

Vcm = (Σ Mi•Vi ) / M

Vcm = (MxVx+MyVy) / (a+1)Mx

Vcm=Mx•(-2aVo) +aMx(2Vo)/(a+1)Mx

Vcm = -2a•Mx•Vo + 2a•MxVo / (a+1)Mx

Vcm = 0 / (a+1) Mx

Vcm = 0

So, conclusion

Skater X direction is to the left and the centre of mass is 0.

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