contestada

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

Respuesta :

Kazeemsodikisola

Answer:

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.

This options are not written well

Check attachment for correct option

Ver imagen Kazeemsodikisola
ACCESS MORE
ACCESS MORE
ACCESS MORE
ACCESS MORE

Otras preguntas