Conservation of momnetum or Newtons second law?

In summary, an acrobat of mass 'm' climbs a rope ladder hanging below a stationary balloon of mass 'M'. If the acrobat starts to climb at a speed 'v' (with respect to the ladder), in which direction and with what speed (with respect to earth) will the balloon move? If the acrobat stops climbing, the velocity of the balloon will be 'v + (Mv/2)'.
  • #1
vkash
318
1
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

  • my answer .png
    my answer .png
    22.3 KB · Views: 572
Physics news on Phys.org
  • #2


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.
 
  • #3


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.
 
  • #4


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.
 
  • #5


I would like to clarify that both the conservation of momentum and Newton's second law are relevant in this scenario. The conservation of momentum states that in a closed system, the total momentum remains constant. In this case, the acrobat, rope ladder, and balloon make up a closed system.

According to Newton's second law, the net force acting on an object is equal to its mass multiplied by its acceleration. In this scenario, the acrobat's upward motion on the ladder creates a force on the balloon in the opposite direction, causing it to move.

(a) The direction of the balloon's motion will be in the opposite direction of the acrobat's climb, and its speed can be calculated using the equation: mV = (m+M)V', where V is the acrobat's climbing speed, m is the acrobat's mass, M is the balloon's mass, and V' is the balloon's velocity with respect to the earth.

(b) If the acrobat stops climbing, the balloon will continue to move in the same direction with a constant velocity. This is due to the conservation of momentum, as the total momentum of the system remains constant even when the acrobat stops climbing.

In conclusion, both the conservation of momentum and Newton's second law are important concepts to consider in this scenario.
 

Related to Conservation of momnetum or Newtons second law?

1. What is the conservation of momentum?

The conservation of momentum is a fundamental law in physics that states that the total momentum of a closed system remains constant, unless acted upon by an external force. In other words, the total momentum before a collision or interaction is equal to the total momentum after the collision or interaction.

2. How is momentum calculated?

Momentum is calculated by multiplying an object's mass by its velocity. The formula for momentum is p = mv, where p is momentum, m is mass, and v is velocity. Momentum is measured in units of kilogram-meters per second (kg⋅m/s).

3. What is Newton's second law of motion?

Newton's second law of motion states that the acceleration of an object is directly proportional to the net force acting on the object and inversely proportional to its mass. This can be expressed mathematically as F=ma, where F is force, m is mass, and a is acceleration. This law helps explain how objects move in response to the forces acting on them.

4. How does the conservation of momentum apply in real-life situations?

The conservation of momentum applies in a variety of real-life situations, such as collisions between objects, rocket propulsion, and sports. For example, in a car crash, the total momentum of the car and any other objects involved remains constant, and the force of the impact is distributed among the objects involved according to their masses and velocities.

5. Can momentum be lost or gained?

No, momentum cannot be lost or gained. It can only be transferred from one object to another. This is because of the law of conservation of momentum, which states that the total momentum of a closed system must remain constant. In other words, momentum cannot be created or destroyed, only transferred.

Similar threads

  • Introductory Physics Homework Help
Replies
17
Views
608
  • Introductory Physics Homework Help
Replies
6
Views
3K
Replies
44
Views
1K
  • Introductory Physics Homework Help
2
Replies
42
Views
3K
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
13
Views
1K
  • Introductory Physics Homework Help
Replies
20
Views
2K
  • Introductory Physics Homework Help
Replies
6
Views
1K
Replies
0
Views
1K
  • Introductory Physics Homework Help
Replies
18
Views
6K
Back
Top