Relative motion and conservation of momentum

[stfz]

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

 

[TSny]

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?

 

[stfz]

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 :(

 

[SammyS]

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)?

 

[TSny]

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?

 

[stfz]

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?

 

[TSny]

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.