Question

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.

Answer 1

do we use conservation of momentum ?

Answer 2

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

Answer 3

Yeah we do

Answer 4

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

Answer 5

@ganeshie8 I could be wrong.

Answer 6

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 ?

Answer 7

Oh that's true

Answer 8

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

Answer 9

That seems more or less right

Answer 10

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.

Answer 11

looks good to me!

Answer 12

I got it wrong I realized

Answer 13

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.

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

I guess the wording of this question failed.

Answer 18

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 .

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

Yeah physics is a full of trickeries.

Answer 24

Frame of reference is such an arbitrary concept.