3
$\begingroup$

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?

$\endgroup$
  • 3
    $\begingroup$ To simplify the situation, why don't you just take into consideration the force of gravity acting on the balloon and use the equations of motion for linear acceleration to find out the velocity of the balloon (assuming you know its mass, of course). $\endgroup$ – user28355 Jan 22 '14 at 4:10
  • 1
    $\begingroup$ The momentum is the same as for any object of same mass (thrown w/ same force in same direction, of course). Unfortunately (or fortunately :-) ), balloons tend to have a range of tensile strength and asymmetric weak points, so you'll have to throw MANY balloons to get a good estimate of the mean bursting force. (Wet partner; hilarity ensues :-) ) $\endgroup$ – Carl Witthoft Jan 22 '14 at 12:51
  • $\begingroup$ Just a comment as an aside. A friend of mine thought he would drop a water balloon off the top of a building onto one of the parked cars below. He expected it to splash off the car's roof. Which it did, but also put a big dent in it. $\endgroup$ – user56903 Nov 4 '14 at 13:05
1
$\begingroup$

Following assumptions are made to calculate velocity (at the desired time) of the balloon (and thus momentum),

  • Balloon has been considered to be a point object or particle. While studying the motion of an object; sometimes, its dimensions are of no importance. For example, if one travels from one place to another distant place by a bus, the length of the bus may be ignored as compared to the distance traveled. In other words, although the bus has a finite size, yet for the study of the motion of the bus along the road; its motion may be considered as the motion of a point or a particle.
  • In the present calculation, the motion of water balloon is considered to be along a straight line (rectilinear motion).
  • Water balloon has been assumed to be uniformly accelerated towards the center of the earth with an acceleration of $9.8 ms^{-2}$.
  • Air resistance has been ignored.

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.

$\endgroup$
1
$\begingroup$

The momentum is given by $mv$. This can be calculated by knowing the initial speed and acceleration due to gravity (See Godparticle's answer for a more detailed explanation).

However I think the your conclusion that the momentum causes the balloon to burst is not entirely correct. It is more to do with the force, i.e. change in momentum experienced by the balloon. This is influenced by both the momentum the balloon has and how quickly you stop it.

Consider a slightly different experiment where rather than catching the balloon you drop it onto a surface. If you drop it onto the hard floor it bursts at a certain height. But if you put a foam mat or other soft material there you will need a larger height as the balloon is decelerated by a lower force (over a longer distance/time).

Having said this, presuming your partner moves their hand the same amount each time you are correct that momentum there is a correlation with momentum.

How the balloon is caught may also be a factor as it can create local spots of high pressure on the balloon as it is gripped. Think bursting the balloon in your hand. This is probably difficult to control for in your experiment.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.