A 65kg ice skater is gliding with his 50kg partner at a velocity of +3 m/s. He t

Question

A 65kg ice skater is gliding with his 50kg partner at a velocity of +3 m/s. He then pushes her away from him such that she is now gliding with a velocity of +5 m/s. Assume that the friction due to the ice is negligable. (a)[4 pts] How fast is the 65kg skater moving after the push? (b)[7 pts] What is the momentum and kinetic energy of the system before and after the push? (c)[1 pt] What was the energy added from the push? (d)[1 pt] Where did the energy come from? (e)[1 pt] What is the coeficient of restitution? (f)[1 pt] What kind of collision is this?

Explanation / Answer

Initial momentum of the system, p = m1*v1+m2*v2 = 65*3+50*3 = 345 kgm/s

a)Momentum of the system will be conserved

=> p1 = p

but p1 = 65*v+50*5 = 345 =>

v = 95/65 = +1.46 m/s

b)Momentum after the collision = 345 kg-m/s

Kinetic energy before the collision = 0.5*50*3^2+0.5*65*3^2 = 115*4.5 = 517.5 J

Kinetic energy after the collision = 0.5*50*5^2+0.5*65*1.46^2 = 694.27 J

C) Energy added by the push = 694.27-517.5 = 176.77 J

d) In this collision some work will be done by 65kg skater in pushing, that will translate into the additional energy.

e) coefficient of restitution = sqrt(K.E.final/K.E.initial) = sqrt(694.27/517.5) = 1.158

f) COR is >1 => energy is added in collision, it can be termed as exergonic collision. example of such collision is an explosion.

Please comment if you have any doubt in the solution.

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