Relative motion and conservation of momentum

In summary: If you do that, then the balloon would not be at rest in this frame. If you take the initial and final momentum of the system relative to this frame, then the initial and final momentum would not be equal.In summary, the conversation discusses the problem of a 52kg man climbing on a ladder hanging from a 450kg balloon. The man begins to climb at a speed of 1.2m/s relative to the ladder and the direction and speed of the balloon is questioned. The conversation also includes a discussion of different methods for solving the problem, including the use of conservation of momentum and Galilean transformations. However, there is a misconception in the reasoning when transforming to a frame tied to the balloon, as
  • #1
stfz
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Homework Statement


A 52kg man is on a ladder hanging from a balloon that has a total mass of 450kg (including the basket passenger). The balloon is initially stationary relative to the ground. If the man on the ladder begins to climb at 1.2m/s relative to the ladder, (a) in what direction does the balloon move? (b) At what speed (with respect to the ground) does the balloon move? (c) If the man stops climbing, what is the speed of the balloon?

Homework Equations


Centre of mass velocity: ##v_{CM}=\frac{1}{M}\sum m_i v_i##
Galilean transformation: ##v'=v-v_{frame}##

The Attempt at a Solution


My question is with part (b). I know that, relative to the balloon, the man is moving up at 1.2m/s. However, the balloon is also moving down so I must find the man's velocity relative to the ground (and hence the desired absolute velocity of balloon). I've been able to compute the correct answer using a different method (##v_{balloon}=-0.124m/s##).

Nevertheless, I'm not sure why the following method is not working - I have a misconception somewhere.
I'm going to assume that momentum is conserved (i.e. the balloon and man are in equilibrium, with the air supporting them against gravitational forces). The system is also initially stationary, hence total momentum cannot change.

Thus, ##mv_m+Mv_b=0##, where ##v_m, v_b## are absolute velocities of man and balloon respectively, with masses ##m, M## respectively.

I have 2 unknowns, but I know that, in balloon frame, ##v_m'=1.2##. So I transform to balloon frame by subtracting ##v_b##:
##v_b'=0## and ##v_m'=v_m-v_b##.
In this new frame of reference, I know that momentum is still conserved. Also, initially, when both balloon and man are stationary, the momentum in this frame is zero.

Hence, ##mv_m'+Mv_b'=mv_m'=m(v_m-v_b)=0##.
This seems to imply ##v_m=v_b## which is unrealistic and incorrect.

I've made a mistake somewhere in my reasoning above, in particular when I began to transform. Could someone enlighten me as to what I've misunderstood?

The method I used (which worked) was to recognize that ##v_m'=v_m-v_b=1.2## and hence I have a linear relationship between v_m and v_b. Solving along with the momentum conservation condition yields ##v_b=-0.124##.
Thanks!
Stephen
 
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  • #2
stfz said:
Hence, ##mv_m'+Mv_b'=mv_m'=m(v_m-v_b)=0##.
Is it correct to assume that the total momentum in the primed frame is zero?
 
  • #3
According to my (flawed) reasoning, initially, before the man moves, there is zero momentum in the primed frame. Also, I know that momentum cannot change, so during/after motion, momentum in primed frame is zero.
I know I'm wrong somehow, but I still can't see it :(
 
  • #4
stfz said:
According to my (flawed) reasoning, initially, before the man moves, there is zero momentum in the primed frame. Also, I know that momentum cannot change, so during/after motion, momentum in primed frame is zero.
I know I'm wrong somehow, but I still can't see it :(
Doesn't the velocity of primed frame change (with respect to the ground)?
 
  • #5
stfz said:
According to my (flawed) reasoning, initially, before the man moves, there is zero momentum in the primed frame.
Before the man starts climbing, there is zero momentum in the earth frame. What is the total momentum in the primed frame before the man begins to climb? Hint: What are the velocities of the man and balloon in the primed frame before the man starts climbing?
 
  • #6
Well total momentum in primed frame would be ##p'=\sum m_i v_i'##.
However, initially the balloon is stationary with respect to the ground, and so is the man. Hence the man is stationary with respect to the balloon?

Yes, the velocity of the balloon (initially)in the primed frame is initially zero. However, as the man begins to move, the velocity of the balloon in the primed frame still zero because we're in balloon reference frame... is that because the frame I'm using is non-inertial and I can't apply conservation of momentum here?
 
  • #7
It is not generally true that initial momentum equals final momentum if you calculate initial momentum with respect to one inertial frame and calculate the final momentum in a different inertial frame. You have the Earth frame (where the balloon is at rest before the man starts climbing) and you have the primed frame (where the balloon is at rest after the man starts climbing). These are two different inertial frames. If you want to use the primed frame where the balloon is at rest after the man starts climbing, then you must calculate both the initial and final momentum of the system relative to this particular inertial frame.

If you pick a reference frame that is always tied to the balloon, then that would be a non-inertial frame.
 

1. How does relative motion affect the conservation of momentum?

Relative motion refers to the movement of an object in relation to another object. The conservation of momentum states that in a closed system, the total momentum remains constant. This means that the relative motion of objects does not affect the overall momentum of the system.

2. What is the role of inertia in relative motion and conservation of momentum?

Inertia is a property of matter that describes an object's tendency to resist changes in its motion. In the context of relative motion and conservation of momentum, inertia plays a crucial role in maintaining the momentum of a system. Objects with greater mass have more inertia and thus require more force to change their motion, which affects the relative motion and momentum of the system.

3. How does the mass of an object affect its relative motion and momentum?

The mass of an object directly affects its momentum, as stated by the equation p=mv, where p is momentum, m is mass, and v is velocity. In relative motion, the mass of an object also determines its inertia and thus its resistance to changes in motion. Objects with greater mass have more inertia and require more force to change their motion, which affects the relative motion and momentum of the system.

4. Can relative motion be used to explain the law of conservation of momentum?

Yes, relative motion can be used to explain the law of conservation of momentum. When two objects interact, their momenta are affected, but the total momentum of the system remains constant. This is because the relative motion of the objects does not affect the overall momentum of the system, as long as there are no external forces acting on the system.

5. How do external forces affect relative motion and conservation of momentum?

External forces, such as friction or air resistance, can affect the relative motion and conservation of momentum in a system. These forces can change an object's velocity and thus its momentum, which can affect the overall momentum of the system. In order for the law of conservation of momentum to hold true, the system must be closed and free from external forces.

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