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Torque and momentum.

  1. Aug 21, 2012 #1
    Sorry the title should be : Impulse and Momentum

    3000Solved Problems in Physics-Shaum's

    A camper lets fall a heavy mallet of mass M from the height y upon the top of a tent stake of mass m and drives it into the ground a distance d. Find the resistance of the ground, assuming it to be constant and the stake and mallet stay together on impact.

    Solution given:

    The speed of the mallet on just striking the stake is v= √(2gy). Momentum is conserved at the instant of collision so that Mv=(M+m)v', where v' is the speed of the stake plus mallet just after impact.

    My question is why the momentum is conserved since there is external force acting on the system, that is gravity.
    Last edited: Aug 21, 2012
  2. jcsd
  3. Aug 21, 2012 #2
    Mind the wording. They write "at the instant..." or "just after". That means that the time interval is extremely small, infinitesimal. The time just before impact and just after impact are extremely close. In such a short instance of time gravity has done no work on the system, thus momentum is conserved.

    Take for example "the instant of time" you drop something from rest. You've already dropped it, but it's speed is still zero. At least a measurable interval of time has to pass before gravity can change the momentum.
    Last edited: Aug 21, 2012
  4. Aug 21, 2012 #3
    Thank you.
    But I still believe how small the time interval is, the gravity does not stop from exerting force.
  5. Aug 21, 2012 #4
    Okay. It's true there's still a gravitational force, but the change in momentum is equal to the force times the time interval:

    [itex] \frac{\mathrm{d}p}{\mathrm{d}t} = F [/itex]
    [itex]\Delta p = \int F\;\mathrm{d}t = F \Delta t = 0[/itex]

    So, even when there is a force (which I have never denied), momentum is conserved as long as this force doesn't do any work.
  6. Aug 21, 2012 #5


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    Staff: Mentor

    Sure, but if the duration of the collision is very short, the force of gravity is a tiny fraction of the force of the collision. Consider what happens if you just rest the hammer on the stake (nothing).
  7. Aug 21, 2012 #6
    I think FΔt≠0 since Δt≠0
    You can assume it is so small that its product is negligible.
    But if F is large value then the product should have the effect on momentum.
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