Let thing of a bird standing still in a box on top of a weighing machine that shows a mass $m_0$. Now, imagine that the bird is flying, still in the same box and the same weighing machine shows a mass $m_1$. As the bird when flying, is applying a force towards the weighing machine, could we deduce that $m_0 = m_1$? I'm asking this question because saying that these masses are equal makes me as uncomfortable as saying that they are not equal and can't figure the right answer. So does $m_0 = m_1$ and why?
[Physics] Weighing a flying bird
airmassnewtonian-mechanicsweight
Related Solutions
The easy way to do problems like this is to work in the centre of mass frame. The trick is to add a velocity that makes the total momentum zero, then calculate what happens in the collision, and finally add the centre of mass velocity back on.
Working in the centre of mass frame makes things easy because if the total momentum before the collision is zero the total momentum after the collision must be zero as well. That gives you the useful relation:
$$m_0v_{final 0} + m_1v_{final 1} = 0$$
You can calculate the velocity of the centre of mass frame by requiring that the initial momentum be zero:
$$m_0(v_0 + v_{com}) + m_1(v_1 + v_{com}) = 0$$
which gives you:
$$ v_{com} = - \frac {m_0v_0 + m_1v_1}{m_0 + m_1} $$
More info: in the centre of mass frame the collision looks like:
because we are in the centre of mass frame we know the momentum after the collision is zero, so:
$$m_1 v_1 + m_2 v_2 = 0 $$
the kinetic energy is:
$$ \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 = E $$
where E is the initial kinetic energy ($1/2m_1u_1^2 + 1/2m_2u_2^2$) so you can use the first equation to substitute into the second e.g. you can rearrange the first equation to be:
$$ v_1 = -\frac{m_2}{m_1} v_2 $$
Use this to substitute for $v_1$ in the second equation and after some juggling it gives you:
$$ \frac{1}{2} m_2 \left( \frac{m_2}{m_1} + 1 \right) v_2^2 = E $$
So now you have $v_2$ you can substitute back in the first equation to get $v_1$. Now just subtract off the centre of mass velocity to get back to your initial frame.
You could have done the calculation without using the centre of mass frame, but it makes the algebra easier. NB be careful about the signs of the velocities; remember that velocities in opposite directions have opposite signs. It's easy to lose a minus sign and get the wrong answer!
As this is an exercise, just a couple of guidelines:
You already know the bar is in equilibrium. So you do not need to assume that bar edges are accelerating. For example, set $a_1=0$ in equation 1.
Indeed, the force on the right edge is $T_2+T_3$, but you need to redo your equations (2) and (3) and also figure out the special a-priori relationship between $T_2$ and $T_3$.
As for your 2nd question, the force is tension of string at point of contact. Strictly speaking, this would be $T_4$ on the right side (string to pulley). But you can easily see that $T_4$=$T_3$+$T_2$ when pulley is not accelerating vertically.
Best Answer
The weight will be the same on average over time.
The bird is supported by the air which in turn is supported by the box. To every action there is an equal and opposite reaction so the force supporting the bird must be transferred to the bottom of the box if no air can escape. However as the bird and air move around the overall centre of mass can move up and down so the force will only equal the total weight on average over time.