Hollie (44 kg) and Rick (52 kg) ride a toboggan (11 kg) down a hill. When the to

Question

Hollie (44 kg) and Rick (52 kg) ride a toboggan (11 kg) down a hill. When the toboggan reaches the bottom of the slope at B, Hollie gets mad and pushes Rick off the back with a horizontal velocity of 2.1 m/s to the right relative to the toboggan. Use left as the positive direction and neglect friction in all calculations.

A. What is the velocity of the toboggan immediately before Rick is pushed off?
(include units with answer) B. What is the velocity of the toboggan after Rick is pushed off?
(include units with answer) C. What is the velocity of Rick after he is pushed off?
(include units with answer)

Explanation / Answer

we have to calculate the momentum at bottom when they are moving
but before that we have to calculate their velocity by energy conservation.
Since there is no friction therefore
potential energy get converted into the kinetic energy
mgH = (1/2)mV2
(a)
velocity of toboggan before rick is pushed off
V = 8.2867 m/s
(b) So initial momentum = mass*velocity = (44+52+11)*8.2867
Final momentum = (44+11)*V + 52*2.1
equating the momentum
(44+52+11)*8.2867 = (44+11)*V - 52*2.1
So velocity of toboggan after rick is pushed off
V = 18.106 m/s
(C) velocity of rick after he is pushed off
he is also try to maintain its momentum
(44+52+11)*8.2867 = 52*V
V = 17.05 m/s

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