Slipping and Sliding (Snowy Shenanigans III)

Fraser and Georgina are brother and sister and are playing outside in the snow. They come across a sheet of ice and want to slide on it to collide with each other, at which point they will cling together and let themselves go in whichever direction they go. Fraser's mass is 60 kg 60\text{ kg} and Georgina's mass is 40 kg . 40\text{ kg}. Fraser travels at 2 m/s , 2\text{ m/s}, whereas Georgina travels at v m/s . v\text{ m/s}.

Fraser: "For us to not move after the collision, it must be true that v = 2 3 . v = \frac{2}{3}. "
Georgina: "If v > 3 v > 3 , I will move backwards after the collision."

Just then, their neighbors Owen and Paddy appear to weigh in on the debate.

Owen: "It doesn't matter what v v is. Fraser will move forward as he is heavier."
Paddy: "Fraser will move forwards if v < 3. v < 3. "

Who is correct?

Details and Assumptions:

  • The ice can be assumed to be smooth (i.e. frictionless).
  • The collisions can be assumed to conserve momentum.


This problem is part of the Snowy Shenanigans set.

Fraser Georgina Owen Paddy

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1 solution

Pulak Malhotra
Mar 3, 2018

Considering Fraser's direction as +ve and Georgina's direction as -ve.

Net Momentum before collision = Net Momentum after collision (no non conservative forces, law of conservation of momentum, velocity of center of mass constant)

Net Momentum = Pfraser +Pgeorgia = 60 * 2 + 40 * (-x) = Pnet

Pnet > 0 => x < 3

=> if x < 3 they both will toward Fraser's direction i.e +ve

Oh OK I will do that next time

Sofia Hernandez - 2 years, 8 months ago

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