yrelhan4
  • yrelhan4
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
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
yrelhan4
  • yrelhan4
@Vincent-Lyon.Fr
shubhamsrg
  • shubhamsrg
*
yrelhan4
  • yrelhan4
@shubhamsrg ??

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shubhamsrg
  • shubhamsrg
i just sort of book marked it since now i'll get notifications whenever this question is discussed..
yrelhan4
  • yrelhan4
right!
shubhamsrg
  • shubhamsrg
i just have a doubt : since A is at a higher position on the rope , will the system be in eqbm ?
shubhamsrg
  • shubhamsrg
as there will be a difference in pot energy of both A and B
yrelhan4
  • yrelhan4
^ yes.. but i got no idea about this question.. even the the rope is flexible.. so....
anonymous
  • anonymous
(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****
yrelhan4
  • yrelhan4
@Diwakar answer to part a is 2h/19..
anonymous
  • anonymous
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.

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