Conservation of momnetum or Newtons second law?

AI Thread Summary
The discussion centers on the interaction between an acrobat climbing a rope ladder and a stationary balloon. When the acrobat climbs at speed 'v', the balloon moves downward due to the conservation of momentum, as there are no external forces acting on the system. The center of mass of the acrobat-balloon system remains stationary, despite the acrobat's movement. The acrobat's initial acceleration and subsequent constant velocity do exert a force on the balloon, affecting its motion. Ultimately, the key takeaway is that while the balloon moves in response to the acrobat's actions, the overall center of mass remains unchanged.
vkash
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conservation of momnetum or Newtons second law?

Homework Statement



An acrobat of mass 'm' clings to a rope ladder hanging below a balloon of mass 'M'. the balloon stationary with respect to ground
(a) If the acrobat begins to climb the ladder at a speed 'v' (with respect to the ladder), in what direction and with what speed (with respect to earth) will the balloon move?
(b) If the acrobat stops climbing, what will be the velocity of the balloon.

Homework Equations





The Attempt at a Solution



I put my answer and figure in attachment.

you tell me where am i wrong
 

Attachments

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vkash said:

Homework Statement



An acrobat of mass 'm' clings to a rope ladder hanging below a balloon of mass 'M'. the balloon stationary with respect to ground
(a) If the acrobat begins to climb the ladder at a speed 'v' (with respect to the ladder), in what direction and with what speed (with respect to earth) will the balloon move?
(b) If the acrobat stops climbing, what will be the velocity of the balloon.

Homework Equations





The Attempt at a Solution



I put my answer and figure in attachment.

you tell me where am i wrong


In the absense of any unbalanced external force, the centre of mass will remain stationary.
If you alter the position of part of the mass of the system, you will probably alter the position of other parts of the system. If some part of the system began moving at some speed, the rest of the system would also move at a certain speed

Imagine if the man and the balloon had the same mass, so the centre of Mass was exactly half way between them, and the ladder was 100m long.
 


u r wrong in thinking that the man, as moving with const. velocity, exerts no force on balloon, where the actual case is the man starts from rest and acquires a velocity and in doing so acquires some accln. also, until he reaches a const. velocity and hence applies force to the balloon also. this force is to be counted as an external force for the balloon, but there is no external force for the balloon+ man system. hence it follows that although the balloon moves downward as the man moves upward the centre of mass of man+ balloon does not move.
 


PeterO said:
In the absense of any unbalanced external force, the centre of mass will remain stationary.
If you alter the position of part of the mass of the system, you will probably alter the position of other parts of the system. If some part of the system began moving at some speed, the rest of the system would also move at a certain speed

Imagine if the man and the balloon had the same mass, so the centre of Mass was exactly half way between them, and the ladder was 100m long.
bjd40@hotmail.com said:
u r wrong in thinking that the man, as moving with const. velocity, exerts no force on balloon, where the actual case is the man starts from rest and acquires a velocity and in doing so acquires some accln. also, until he reaches a const. velocity and hence applies force to the balloon also. this force is to be counted as an external force for the balloon, but there is no external force for the balloon+ man system. hence it follows that although the balloon moves downward as the man moves upward the centre of mass of man+ balloon does not move.
thanks to both persons.
I got the point.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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