Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Hitting a Block Off-Center

  1. Sep 22, 2014 #1
    Hello everyone!

    I was watching a Professor Lewin lecture

    from about 15 minutes onwards
    Where he talks about striking a block off-center.
    What is confusing me is that it seems that by striking it in a different place you are giving it more energy.
    This feels wrong to me because you are giving it the same impulse.
    Is it perhaps because we assume that during the impulse the block remains in the same place?

    Thanks in advance,

    A bit more description: When you hit the block through the center the velocity = Impulse/Mass. This is the same for when you hit it off-center but it also has rotational energy.
  2. jcsd
  3. Sep 22, 2014 #2


    User Avatar
    Science Advisor

    The equation ##mv = F \Delta t = I## tells you that the impulse determines the linear momentum transfer to the block.

    However, the equation for energy imparted to the block (work) is ##F \Delta x##, that is, the work depends on the distance, not the time, over which the force acts.

    To produce both the same linear velocity as the on-center impulse, plus the angular velocity, the off-center impulse has to do more work on the block while applying the same impulse.

    Intuitively, you can image that the block tends to "give" more easily (by spinning away) when you hit it off center; therefore when you apply the force for the same (very short) amount of time, you actually apply it over a longer distance.
  4. Sep 22, 2014 #3
    Brilliant! Thank you, this bit at the end was the sort of vague feeling I had about it but this makes it clear. The idea that it can require more effort to give the same impulse is nice.
    Although there is something still bugging me. If WD = Fx what is the x that is different for the off-center?
    Is this where the assumption that the block stays still whilst applying the impulse comes in?
    Thanks again
  5. Sep 22, 2014 #4


    User Avatar
    Science Advisor

    I don't think you really have to assume that the block stays still. If you apply the impulse at the center, then the block moves a tiny distance ##\Delta x## (whatever that is) during the application of the impulse over ##\Delta t##, which is fine since the block ends up with some translational kinetic energy. If you instead apply the impulse at d, then there is ##\Delta x## due to both the shift of the center of mass and the rotational shift around the center of mass during that ##\Delta t##.

    Anyway this gets hand-wavy quickly, so I'd just solve the conservation equations and realize that the same linear momentum is imparted whether you strike the block on-center or off-center, but the total energy imparted is not the same.
  6. Sep 22, 2014 #5
    Yeah okay dokay, that is nice. You're right, I think the assumption was to make the particular question easier and not to make the physics work.
    I'm pretty sure I understand it now, I've ran through the equations as well.

    Thanks again though, the blocking part for me was not realising that one impulse can impart different amounts of energy than another. Although I still feel a bit odd about this. I like intuitively the idea of a really long rod resisting my push more at the end than in the middle, but do you have any 'non-rotational' examples of this idea?
  7. Sep 22, 2014 #6


    User Avatar
    Science Advisor

    No problem.

    Maybe it will help to imagine a simple inelastic collision between two objects ##m## and ##M##.

    Suppose ##m##, which had initial velocity ##v_i##, smashes into ##M##, which was at rest. Since the collision is inelastic, ##m## and ##M## stick together after the collision and share a velocity, call it ##v_f##.

    By conservation of momentum, ##mv_i = (m+M)v_f##, i. e., the entire momentum ##mv_i## has been transferred to the new "blob" of combined mass.

    If ##m = M##, then ##v_f = v_i/2##. However, if ##m \ll M##, then ##v_f \ll v_i##.

    Since the kinetic energy of the resulting blob is ##\frac{1}{2}(m+M)v_f^2##, you can see that the kinetic energy of the resulting system can be a lot larger if the masses are more nearly equal than if you have a smaller mass crashing into a much larger one. What happens in the second case is that a lot of energy gets dissipated (usually as heat) in absorbing the impact of ##m## into the much larger and more "resistant" ##M##.
  8. Sep 22, 2014 #7
    Thanks for your example, I've been studying this sort of collision longer and am more familiar with it.

    This is also weird to me! I've never thought about it like that before. I've always led with the idea that momentum is conserved and therefore the relevant speeds and energies fall out. But it seems to me that it is 'more likely' that two balls the same size would stick together and move on. I guess the idea of throwing a lump of blue tac at a wall and it kind of splatting into it and being deformed into a pancake helps the idea that when a object hits a much more massive object there is more energy lost.

    I guess what feels weird to me about this is that I feel you wouldn't need an as inelastic object to move on with one the same size but for one much larger you would need a more inelastic object. Regardless, I am happy with this example because the energy loss is easily accounted for in my head.

    However, it does not rid my confusion about the original question. I am fine looking at the problem from the side you walked me through and that has really helped me think about it. But, i'm still having the thought that sometimes it can require more energy to deliver the same impulse. And I can't really think of any other instances where this is the case.

    Thanks again
  9. Sep 23, 2014 #8


    User Avatar
    Science Advisor

    Glad this is helping. It's an interesting problem (especially as stated) and it takes some time to sort through the implications.

    Yes—it's possible that different amounts of energy need to be transferred to the block to be consistent with the impulse specified in the problem.

    The usual "realistic" version of your problem involves shooting a bullet into a wooden block. The bullet has some momentum and kinetic energy which are independent of where you shoot the block. The momentum of the bullet is transferred to the block (the bullet ends up embedded in the wood) and hence it delivers a certain impulse to the block. Part of the energy of the bullet is carried off in the kinetic energy of the block (translating and spinning), but part of the energy is dissipated as heat as the bullet penetrates the block. If you strike the block on-center, the bullet penetrates deepest and dissipates the most energy. Anywhere off-center and some of the energy goes into spinning the block; hence there is less penetration and energy dissipation.

    Another real-world example is spacecraft engines—different types of rockets consume different amounts of propellant and energy to produce a certain amount of thrust (force). There is a Wikipedia article on Specific impulse which has an interesting discussion about the trade-offs.
  10. Sep 23, 2014 #9


    User Avatar
    Science Advisor
    Gold Member

    That is demonstrated here:

    Which was a response to this:

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook