A 150 kg loaded toboggan is gliding along a flat surface at 2.0 m/s [forward] when one of the 50 kg passengers jumps off towards the back with a velocity of 3.0 m/s [backwards] with respect to the ground. What is the resulting speed of the toboggan and remaining passengers.

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- ganeshie8

do we use conservation of momentum ?

- anonymous

The initial velocity of the toboggan is calculated as
150kg*2.0m/s=300N/s
Since one of the passengers weighing 50kg jumps off back of the boat at 3.0m/s with the frame of reference being the ground,
3.0m/s+2.0m/s=5m/s
Since the passenger acquires a momentum of 250N/s to the opposite direction
300N/s-250N/s=50N/s [forward]
Now the only momentum acting on the boat gliding along is 50N/s
Given the mass of the boat 150kg, and momentum acting on the boat being 50N/s
150kg*v=50N/s
v=0.333m/s
Therefore final velocity of the boat is 0.33m/s

- anonymous

Yeah we do

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- anonymous

Anyway that's my quick shot but I could be wrong.

- anonymous

@ganeshie8 I could be wrong.

- ganeshie8

that doesn't look correct because the initial speed of boat is 2 m/s
how can it reduce if the boat gets a push by unloading a passenger ?

- anonymous

Oh that's true

- anonymous

Perhaps add the 300N/s+250N/s=550N/s
Then increased momentum results in supposedly greater magnitude of velocity so
550N/s=v(150kg)
v=3.66m/s
The final velocity 3.7/s

- anonymous

That seems more or less right

- anonymous

The key is when the passenger jumps towards the back with velocity of 2.0m/s to the ground I get it as 2.0m/s+3.0m/s because of the frame of reference being the ground and also the fact that the boat is gliding along the water to the opposite direction means increased velocity with respect to the boat.

- ganeshie8

looks good to me!

- anonymous

I got it wrong I realized

- anonymous

The momentum that originates from the boat was initially 300kgm/s and more so according to the conservation of momentum originating from one body, which in this case is the boat.

- anonymous

So I bet I should do
300N/s=100kg(v)+50kg(-2m/s)

- anonymous

300N/s=100kg(v)+50kg(-3m/s)

- anonymous

100kg(v)=450kgm/s
v=4.5m/s?

- anonymous

I guess the wording of this question failed.

- anonymous

Boat=momentum of this boat is conserved at all times.... regardless of whatever comes out of it the initial momentum with respect to what comes out of it in one dimensional setting will be equal. Hence addition of -150kgm/s+300km/s=300kgm/s= that's initial momentum of this boat .

- anonymous

I interpreted newton's third law as canceling each other out... which was my fallacy.

- ganeshie8

initial momentum = 150*2 = 300
a mass of 50kg disappears at a speed of 3--2 = 5 m/s with respect to boat, pushing the boat forward
so final momentum is given by 100v and it must satisfy :
100v = 300 + 50*5
solving gives me final speed = 5.5

- anonymous

Right.. I forgot with respect to the ground part.

- ganeshie8

don't trust me on these though
im going by my gut.. its been a very long time me touching physics..

- anonymous

Yeah physics is a full of trickeries.

- anonymous

Frame of reference is such an arbitrary concept.

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