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Which mass to use in a momentum change problem?

  1. May 2, 2016 #1
    1. The problem statement, all variables and given/known data

    Just a general physics question that I was thinking about. I posted it here because I assumed it is a general question that is easy to most experts. I apologize if something similar was posted before, but I couldn't find anything asking in a way I needed to think about it. Let's say I'm doing an egg drop type of project. Make a container to put an egg in, drop it from some height, calculate some important info. Let's say I know an egg cracks at around 60 N of force. If I used the Ft = m (change in) v, which mass would I use to calculate the momentum change: the mass of the egg itself or the mass of the egg + the container?

    If the answer is egg only, then the mass of the container doesn't really matter for whether an egg can survive a fall, only the collision time it takes to change its (only the egg's) momentum. If the answer is egg + container, then a higher collision time is needed for the egg to survive in a container compared to the egg alone due to a bigger momentum change.

    This got me thinking about Force in a way that I don't totally understand. An egg strapped to a container colliding with the ground. Does the egg experience the same momentum change and force as the rest of the container? Does every part and every molecule of an object experience the same amount of force when it collides with something as every other part?

    2. Relevant equations

    Ft = m (change in) v

    3. The attempt at a solution

    Thanks for the insight
  2. jcsd
  3. May 2, 2016 #2


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    If you assume the container stops almost instantly and the egg inside keeps moving so that it's energy is absorbed by padding inside the container... then m will be the mass of the egg only. Perhaps model it as..


    No. Momentum = mass * velocity. They might be going the same velocity but the mass of each is different.

    Personally I prefer to think about stopping distances in impacts. If something stops in a short distance then it's deceleration must be high. Newton says that F = ma so high acceleration (or deceleration) implies large forces. In other words short stopping distances imply high forces.

    So what matters is the thickness and elasticity of the padding. The thickness effectively sets the stopping distance. The stopping distance of the egg can't be greater than the thickness of the padding or the egg will hit the inside of the container.

    In theory if you have enough data about the egg it's possible to calculate the minimum stopping distance and hence the minimum thickness of the padding using the SUVAT equations. You could also calculate the spring constant of the padding (how hard or soft it needs to be).

    If I was doing the egg drop experiment I would use a very large box full of soft packaging but not too soft. Sadly the rule usually prevent you using a box the size of a packing crate.
  4. May 2, 2016 #3


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    Correct. The "collision time" and the stopping distance are related.

    You mentioned 60N force and you know the mass of the egg so you can use F=ma to calculate the max deceleration of the egg to keep the force below 60N. You can calculate the initial (impact) and final (zero) velocities and use SUVAT to estimate the stopping time and or distance.
  5. May 3, 2016 #4


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    It's not the elasticity, at least not in the usual sense.
    The ideal is a material that offers a constant resistance, not too much less than the maximum force the egg can bear. That will bring the egg safely to a stop in the shortest distance and time. Elastic materials tend to offer resistance proportional to the extent of compression. Constant resistance is more readily achieved by materials that are crushed by the impact.

    An example would be a tightly wound spool of paper in which the centre has been pushed out to one side to produce a cone. Pushing the centre back in would take a roughly constant force, but you would want to arrange that the centre cannot reach its original position before other parts. That could be done by placing a stiff plate over the thin end of the cone. Hope that's clear without a diagram.
  6. May 3, 2016 #5
    Thanks, everyone. That is more clear now.

    Let me check to make sure I understand this, so in a collision, each part of the object may not necessarily experience the same momentum change (due to different mass) and force. That was not easy for me to really get. So in a car crash, the person and the car feel a different amount of force. Very interesting.
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