Recently an experiment was performed in which myself and a partner filled a water balloon and threw it back and forth at each other without breaking it. We gradually increased the distance at which it was being thrown and at one point I ended up on the second floor of one of my high school buildings and when I threw it, my partner caught it. For the last part of my experiment I went up to the third floor of the building and threw it, my partner caught the water balloon (it burst in her hands) I am assuming that the balloon had so much force behind it that the impact caused it to burst. I know that momentum is measured using mass and velocity but I do not know how to calculate the momentum when it reaches my partner. I would say that there is no momentum because once the balloon is caught, it will not be moving. But apparently that is incorrect so I am assuming the momentum will need to be that of the split second right before it is caught. My final question: How do I calculate the momentum of the balloon when it reaches the hands of my partner?
[Physics] How to figure out the momentum of a water balloon when it reaches the person I am throwing it at
classical-mechanicseveryday-lifemomentum
Related Solutions
The asymmetric problem gets into more complicated aspects of the kinematics of rotating bodies than are usually the point when presenting this example.
In the initial, arms out symmetric configuration, the skater's center of mass is directly over the pivot point. If she brings one arm in, and still has the axis of her body strictly vertical, then her center of mass is no longer over the pivot point, and, were she not spinning, she would fall. Now, because she is spinning, you are dealing with a problem similar to that of a gyroscope in a gravitational field whose axis of rotation is non-vertical: the gyroscope precesses.
More step wise
1. (initial condition) the skater is spinning about a vertical axis, both arms outstretched.
2. skater starts pulling her left arm inward, this changes the location of her center of mass.
3. skater starts "falling" towards her outstretched right arm.
4. this is a torque (due to gravity and the friction on the ground that keeps her skate tip at a fixed point on the ice) on a spinning body.
5. this ends up causing her main axis of rotation to precess.
I believe that I can see these kinds of effects in some of the examples here. First thing to note is that the skate is always tracing a circle on the ice. The size of this circle is related to the degree of asymmetry in the skaters body position: more asymmetric - larger circle. This is consistent with the skater needing to manage the location of her center of mass by adjusting her body, in particular her legs, and/or needing to manange rotating about a non-vertical axis in such a way that her overall angular momentum is (very close to) exactly vertical.
This page has a nice summary of rigid body mechanics, which if worked through, could be applied to this situation.
Fluid mechanics is fun!
Nambiar's first paragraph is key, so I'll quote it (with proper attribution)
This is a consequence of Newton's first law. Newton's first law states an object will continue to move in a straight line at a constant velocity unless acted on by an external force.
In fluid mechanics, while this law is still absolutely true, it's a bit trickier to see what will actually happen because fluids can change shape. If instead of a rubber water balloon holding back water, we had a spring holding back something solid like a bowling ball, the outcome would not be to surprising. In that case, we would expect that, once the spring stopped moving, it would take some time for the bowling ball to be stopped by that spring, and the spring would be compressed (if the spring was in front of the ball) or stretched (if the spring was attached behind the ball). The exact same effect is happening with the water and the balloon, it's just being obscured by the more complicated motion a fluid can undertake.
The other key to this is clarifying an assumption, "...the force that was applied to each (the balloon skin and the water) was approximately equal..." If you really look at it, you never applied a force to the water. You apply a force to the outside of the balloon (in an attempt to stop the balloon+water), but you never directly apply a force to the water. The balloon applies the force to the water. It's an intermediary. In many cases, this is a minor detail because it doesn't change the outcome. However, in this case, it does change the outcome because the balloon is elastic, it can stretch under force. If the balloon were not elastic (perhaps it was made out of carbon fiber?) then the behavior of the system would be exactly what you expect: apply force to stop the balloon, it immediately stops the balloon and water.
Incidentally, that immediate stopping of water has a name: water hammer. You can hear it in many houses (especially old ones) when you turn the water on in a bathroom and then suddenly turn it off. While the water was flowing, it had momentum. When you suddenly turn off the faucet, the water has nowhere to go, but it still has momentum. It crashes into the pipes it was flowing through (often moving them visibly), making a loud "bang" noise, and potentially doing damage to the pipes! Thank goodness our balloon is more friendly than that!
So with our elastic balloon, we have to use the laws governing how elastic materials work. The amount of force they can exert is proportional to how stretched they are. So at the moment when you stop the balloon it can't exert much force at all; it's barely stretched enough to hold the water inside it. Accordingly, while you may be able to put a large amount of force into stopping the outside of the balloon, the balloon cannot put any more force into the water than it's current level of stretchedness permits. If this is not enough to stop the water (as in your example), then the water continues moving forward, according to Newton's first law. This motion stretches the balloon, permitting it to exert more force on the water. This process continues until the balloon is stretched enough to fully slow down the water to a stop. (Then, most likely, the balloon will release all of that energy it stored in stretching, moving the water in the other direction!)
What makes the balloon interesting is that the water can take on any shape that is convenient. It could just push right into the front side of the water balloon like a bowling ball pushing against a spring. However, it turns out that is not the most efficient way for the water to move. Because the water can change shape, some of the water near the back of the balloon can actually move outwards. This is where Newton's first law gets interesting with respect to fluids. By Newton's first law, the center of mass of the water must move forward unless resisted by an external force. One way to do that is to have the entire slug of water move forward, but another way is to have the water start spreading out. If the back of the water moves forward and outwards, the center of mass is still moving forward, so Newton's laws are still being obeyed.
So what shape does the water take? Well, it depends on how you're applying your force. If you tug on the string backwards to slow the water balloon, the most efficient shape (minimizing stress in the balloon) will be to stretch the balloon lengthwise until the stretch is sufficient to stop the forward motion of the water. On the other hand, if you were to drop the water balloon on a flat surface, we'd see a different stretch. The water right at the bottom of the balloon would get stopped quickly (because the force is easily transmitted through the balloon into the water), but the water behind it will start to "pile up," pushing sideways on the balloon. This will cause the balloon to assume more of a pancake shape until the balloon is stretched enough to restrain this outward motion. (or, if the balloon is too weak, it will pop!)
Best Answer
Thus, kinetic equations of motion under gravity can be obtained by replacing 'a'(acceleration) by 'g'(acceleration due to gravity) in the kinematic equations of motion known. Accordingly, one of the kinematic equation of motion under gravity can be given as:
$$V_f=V_i+g(t_f-t_i)$$
where,
$V_f$ is velocity at time $t_f$.
$V_i$ is velocity at time $t_i$.
$g$ is acceleration due to gravity.
Initially at time $t_i$=0, $V_i$ will also be equal to 0. Thus, above equation reduces to the below form:
$$V_f=gt_f$$
By knowing the time($t_f$) at which the water balloon is about to reach the hands of partner (stop watch can be used to know the time practically), velocity at that time ($t_f$) can be known.
Thus, momentum ($P$) of the water balloon of mass ($M$) at the time $t_f$ can be given as $P(t_f)=MV_f$. By substituting all the values, the momentum of the water balloon at time $t_f$ can be obtained.