Hollie (37 kg) and Donnie (62 kg) ride a toboggan (13 kg) down a hill. When the

Question

Hollie (37 kg) and Donnie (62 kg) ride a toboggan (13 kg) down a hill. When the toboggan reaches the bottom of the slope at B, Hollie gets mad and pushes Donnie 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.

C. What is the velocity of Donnie after he is pushed off?
(include units with answer) Hollie (37 kg) and Donnie (62 kg) ride a toboggan (13 kg) down a hill. When the toboggan reaches the bottom of the slope at B, Hollie gets mad and pushes Donnie 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. C. What is the velocity of Donnie after he is pushed off? (include units with answer)

Explanation / Answer

1st calculate velocity of the system [Hollie (37 kg),Donnie (62 kg) and toboggan (13 kg)]

use conservation of energy

mgh (at top of the hill ) = 1/2m*v^2 (at bottom)

or v = sqrt(2gh) = 6.264 m/sec

Now given in problem that , velocity of Donnie after she is pushed back is 2.1 m/sec relative to the toboggan

Let velocity of Toboggan be V after Donnie is pushed , so velocity of Donnie relative to ground will be (2.1 - V)

Now use conservation of momentum

(Mh + Md +Mt)*6.264 = (Mh + Mt)*V +Md*(-(2.1-V)) ............. -(2.1-V) because this velocity is towards right                                                                                                   and opposite to combined velocity of Hollie                                                     and Toboggan

solving we get V = 7.4265 m/sec   towards left .....answer

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