Question

A pair of figure skaters glide holding hands on a frictionless ice rink, at a speed of 3.0 m/s and heading due East in a straight line. They then push away from each other. Skater A, of mass 48 kg, glides off at a speed of 3.8 m/s and at angle of 29 degrees counterclockwise with respect to the original direction. Skater B has a mass of 72 kg.

a. After they push off from each other, what is the velocity of skater B?

b. What is the velocity of their combined center of mass after they separate?

Answer #1

Here we have given that,

Their initial speed of Vi = 3.0 m/s

Skater A mass 48 kg

final speed of A Vf= 3.8 m/s

angle = 29° degrees counterclockwise

Skater B mass = 72kg.

a. Now using conservation of momentum here we will get,

Conserving the momentum along east direction we will get

Pi = pf

Vi(mA+mB) = mAVAfCos29° + mBVbf

Here only

On plugging the value we will get,

Vbf = 2.78429 m/s

b. Now the velocity of centre of mass after they seperated is given as,

Vcm = mAvAf +mBvBf/mA+mA

on plugging the values we will get,

Vcm = 3.190 m/s

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