A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 one year ago
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.
anonymous
 one year ago
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.

This Question is Closed

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1do we use conservation of momentum ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The 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/s250N/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
 one year ago
Best ResponseYou've already chosen the best response.0Anyway that's my quick shot but I could be wrong.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@ganeshie8 I could be wrong.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1that 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
 one year ago
Best ResponseYou've already chosen the best response.0Perhaps 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
 one year ago
Best ResponseYou've already chosen the best response.0That seems more or less right

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I got it wrong I realized

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The 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
 one year ago
Best ResponseYou've already chosen the best response.0So I bet I should do 300N/s=100kg(v)+50kg(2m/s)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0300N/s=100kg(v)+50kg(3m/s)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0100kg(v)=450kgm/s v=4.5m/s?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I guess the wording of this question failed.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Boat=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
 one year ago
Best ResponseYou've already chosen the best response.0I interpreted newton's third law as canceling each other out... which was my fallacy.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1initial momentum = 150*2 = 300 a mass of 50kg disappears at a speed of 32 = 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
 one year ago
Best ResponseYou've already chosen the best response.0Right.. I forgot with respect to the ground part.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1don't trust me on these though im going by my gut.. its been a very long time me touching physics..

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah physics is a full of trickeries.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Frame of reference is such an arbitrary concept.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.