Question

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

Answer 1

@Vincent-Lyon.Fr

Answer 2

*

Answer 3

@shubhamsrg ??

Answer 4

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

Answer 5

right!

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

(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****

Answer 10

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

Answer 11

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.