Quantcast

A community for students. Sign up today!

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

yrelhan4

  • one year ago

Two men A and B each of mass m sit on loops at the ends of a light flexible rope passing over a smooth pulley, A being h metres higher than B. In B’s hand is placed a ball os mass m/10 which he instantly throws up to A, so that it just reaches him. Calculate (a) The distance through which A moves by the time he catches the ball. (b) The total distance moved by A when he ceases to ascend

  • This Question is Closed
  1. yrelhan4
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @Vincent-Lyon.Fr

  2. shubhamsrg
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    *

  3. yrelhan4
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @shubhamsrg ??

  4. shubhamsrg
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i just sort of book marked it since now i'll get notifications whenever this question is discussed..

  5. yrelhan4
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    right!

  6. shubhamsrg
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i just have a doubt : since A is at a higher position on the rope , will the system be in eqbm ?

  7. shubhamsrg
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    as there will be a difference in pot energy of both A and B

  8. yrelhan4
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ^ yes.. but i got no idea about this question.. even the the rope is flexible.. so....

  9. Diwakar
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    (a) First note that since the masses of the men are same there will be no acceleration as long as A does not catch the ball. Suppose that B throws ball with velocity,u. By conservation of momentum,B moves with velocity u/10. Due to constrained relation, A moves up with velocity=u/10. Now , time taken by the ball to stop=u/g The man catches the ball when it stops Therefore distance travelled by A before he catches it=u^2/10g.............(1) We find the value of u by equating the distance travelled as follows Or u^2/10g +h=u^2/2g u^2=5gh/2 We will get the distance travelled by A=h/4 by putting in eqn (1) (b) After catching the ball we get the value of common velocity=u/10, the same as before. [by conserving momentum] However ,now there will be an acceleration in the system since the two masses on the different sides of the pulley are different. The acc will be g/21. You can find it by using constraint eqns. The mass 11m/10 [A+ball] is moving upwards but the acc is downwards. The distance travelled by it before it stops will be u^2/2a where a is acc i.e. g/21. This will give us the distance as 21h/80. Note that value of u^2 is 5gh/2. ***PLEASE TELL ME IF I AM WRONG WHICH IS HIGHLY PROBABLE****

  10. yrelhan4
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @Diwakar answer to part a is 2h/19..

  11. gleem
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Let vi=velocity of the ball when thrown d= the distance it has to travel for A to catch it vb = the velocity of B and A after the ball is thrown Then the equations you need are for the distance the ball has moved at time t when caught by A who is moving at velocity vb. \[d=-\frac{ 1 }{ 2 }g t ^{2}+v _{i}t\] and the distance A has moved at time t as he catches the ball relative to the position of the ball when thrown \[d=h+v _{b}t\]. and\[v=-g t+v _{i} \] velocity of the ball at time t. And finally we impose the condition that when the ball reaches A a distance d it is traveling at a speed of vb also the speed of A \[v=v _{b}\] We still need one more relation that is the speed of the ball relative to the speed of A or B. Use conserv. of mom. Since A and B are connected they act as one mass so when the ball is tossed its momentum must equal the momentum of A and B. ie\[2mv _{b}= \frac{ 1 }{ 10 }mv _{i}\] Solve for d-h. I think this will work for part a. For part b the increase in PE from the point of the catch until A stops must equal the KE available from the total mass of 2.1m initially traveling at speed vb the speed of all bodies when A catches the ball.

  12. Not the answer you are looking for?
    Search for more explanations.

    Search OpenStudy
    • Attachments:

Ask your own question

Ask a Question
Find 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
  • 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.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.