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Impluse and momentum

  1. Dec 16, 2004 #1
    A mass M is dropped from height on a floor, so that at the moment of impact its velocity is [tex]\vec{v} = 3\hat{x} - 20\hat{y}[/tex]. The coefficient of friction between the mass and the floor is 0.2. What horizontal distance will the mass pass before stopping. (The mass doesn't lose contact with the floor thruoghout the impact, i.e it doesn't bounce.)

    Here's what I tried to do:

    [tex]\Delta P_x = m(u_x - v_x) = \int{f_kdt} = \int{-\mu N dt}[/tex]
    [tex]\Delta P_y = m(0 - v_y) = \int{N dt}[/tex]

    Dividing:

    [tex]\frac{u_x - v_x}{-v_y} = -\mu[/tex]

    And I get that ux, the mass horizontal speed after the impact, is -1. Obviously that makes no sense...

    I have a feeling that if I knew the total impact time I cuold solve the problem, but it's not given.

    Thanks,
    Chen
     
  2. jcsd
  3. Dec 16, 2004 #2
    I could be wrong on this (someone please correct me if I am), but this is what I would do:

    I looked at only the horizontal component of the change in momentum, which should be 3*M (change in momentum = m*v). I know that the force of kinetic friction doesn't vary, and I know that with a non-varying force impulse just equals force*time. So I did 3M = (0.2)Mg*time and solved for time, to get time = 3M/(0.2Mg) = 15/g seconds.

    I'm not sure if you need to take into account the verticle component of the velocity. It seems that if it doesn't bounce, there is only horizontal motion and that's all that matters, but I don't know.
     
  4. Dec 16, 2004 #3

    dextercioby

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    Think that the velocity vector has a component which is perpendicular to the direction of motion.That means that there is a transfer of momentum (hence a force) which acts on the floor,increasing the pressure which gravity exerts on the floor.In this case,the normal force is greater,and hence the friction force.

    Redo calculations taking into account the hints given.

    Daniel.
     
  5. Dec 16, 2004 #4
    Even if we assume the friction force doesn't vary, it most certainly doesn't equal [tex]\mu mg[/tex] because N != mg during the impact. N is much much greater than mg at that time, because it needs to cancel the vertical momentum the ball has upon impact. Thanks though. :)

    How? I already knew that the normal force is greater during the impact, but I have no way of knowing its magnitude. How exactly would you modify my calculations to include this information? Thanks.
     
    Last edited: Dec 16, 2004
  6. Dec 16, 2004 #5

    Doc Al

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    Interesting problem! I think the best you can do is assume that the vertical component of the collision with the floor resolves in a negligible time, then just the horizontal energy is dissipated by sliding friction. (This is what Silimay assumed.)
     
  7. Dec 16, 2004 #6

    dextercioby

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    Yes,Doc,i'd call it weird,not interesting.
    i)If u assume that the transfer of momentum of the vertical component is produced instantaneously (we're in CM,so we're allowed to do such simplifying assumptions),then you wouldn't fit it in the problem,as it would be almost infinite and the time of this transfer would be another (a priori unknown) variable.BTW,why would the problem supply us with a nonuseful info.regarding the "y" component of the velocity vector.
    ii)Another idea is to assume something rather absurd (what i suggested in my prior post,as rejected 'i)':that this transfer of momentum is produced continuously on the time the box/object moves along the table/floor.In that case,the time interval assumed by the momentum transfer would not be another parameter/unknown,but it would be the quantity the problem was asking about:the time would be linked with the initial speed along "x" and certainly with the acceleration.

    WIERD id the key word.

    Daniel.
     
  8. Dec 16, 2004 #7
    Well, since it doesn't bounce, I'm assuming that we'll treat it essentially as an inelastic collision. Since you're not given any further information about the force experienced during the collision or the time interval [tex]dt[/tex] during which the collision occurs, I think you'd have to make the reasonable assumption that it's vertical momentum changes to 0 effectively instantaneously (although not really). After that occurs and for the remainder of the mass's slide on the floor, the normal force exactly equals the weight or else it would have an acceleration component in the y-direction. So, without further information, I would agree with Doc Al and Silmay and say it's safe to only deal with the x-component of momentum change. I proceeded as follows using the impulse-momentum theorem:

    [tex]
    {\Delta}\vec{p} = \vec{J} = \int_{t_i}^{t_f}{\vec{F(t)}dt}
    [/tex]

    Now, the net force acting on the mass is the kinetic frictional force [tex]-{\mu}N\hat{i} = -{\mu}Mg\hat{i}.[/tex] By the impulse-momentum theorem, the total impulse exerted is the change in momentum (which is, in the x-direction, -3M kg*m/s). So, letting [tex]t_i = 0[/tex] we have,

    [tex]-3M\hat{i} = \int_0^t{-{\mu}Mg\hat{i} = -{\mu}Mgt\hat{i}}[/tex]

    where the units on both the left and right side have dimensions [tex]\frac{ML}{T}[/tex]. Now, everything simplifies down to give the magnitude of the time [tex]t = \frac{3}{{\mu}g} \approx 1.5s[/tex]. Note that this quantity doesn't depend on the mass, which makes sense because an increased mass causes both an increased linear momentum while at the same time an increased normal (and hence frictional) force to retard its motion.

    Cheers.
    ---
    Mike Fairchild
    http://www.mikef.org/
    "Euclid alone has looked on beauty bare."
    --Edna St. Vincent Mallay
     
    Last edited: Dec 16, 2004
  9. Dec 16, 2004 #8
    I'm pretty sure this isn't the way to go, because one of the hints for this problem is "Refer to the normal force that the floor exerts on the ball during the collision time". :smile:

    mjfairch - Thanks for going through the trouble, but I'm afraid we can't do that.
     
  10. Dec 16, 2004 #9

    Andrew Mason

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    Ok. I think I agree with you. But I would first of all observe that this is really an energy problem.

    You can determine the horizontal component of its speed the instant it hits the floor. That gives you its kinetic energy at that point. Then, for a brief period of time, the normal force (which determines the force of friction) is greater than mg. What really matters is the distance that force operates over.

    What, then, is the distance over which this extra force operates? Obviously, it is [itex]\Delta s = v_x \Delta t[/itex] where [itex]\Delta t[/itex] is the time it takes for the floor to reduce the vertical motion to 0. So we are looking for:
    (1)[tex]\mu_kF_{normal}v_x\Delta t_[/tex]

    Now here is the tricky part. We know, from conservation of momentum, that [itex]F_{normal}\Delta t = mv_y[/itex]. So just use that value [itex]F_{normal}\Delta t[/itex] and plug it into (1). That gives you the energy loss while the vertical momentum is being absorbed. The rest of the kinetic energy is consumed by the regular friction force times stopping distance ([itex]\Delta KE = \mu_k mgd[/itex]).

    AM
     
    Last edited: Dec 16, 2004
  11. Dec 16, 2004 #10
    Andrew: I like your energy approach, but I'd like to discuss a little further.
    But during the time interval [itex]{\Delta}t[/itex] that the vertical impulse is acting, the mass is sliding across the floor with a forward horizontal speed, and it is also simultaneously experiencing the kinetic fricional force opposing its motion. Thus its velocity is not constant as you imply in your previous equation, but is instead changing due to the acceleration from the friction. What we need is a representation of the normal force as a function of time (e.g. [itex]N(t)[/itex]) that decays rapidly down to the "standard" normal force [itex]N=mg[/itex] after the small time interval [itex]{\Delta}t[/itex] during which the vertical impulse is still acting.

    Unless further information is given regarding the time interval [itex]{\Delta}t[/itex] during which the vertical impulse acts, I don't think this problem can be solved explicitly without the assumption we made neglecting the vertical component of the impulse; rather it will have to be left in implicit form involving that [itex]{\Delta}t[/itex] we keep mentioning.

    Assuming we did know that time interval, then we can break the problem into two parts. (1) the energy loss while the vertical impulse is still acting (using our [itex]N(t)[/itex] for the normal force obtained above), and (2) the remainder of its journey where the frictional force is computed with the "standard" normal magnitude of [itex]N=mg[/itex].

    Cheers.
    ---
    Mike Fairchild
    http://www.mikef.org/
    "Euclid alone has looked on beauty bare."
    --Edna St. Vincent Mallay
     
  12. Dec 17, 2004 #11

    Andrew Mason

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    Yes. I assumed [itex]\Delta v_x << v_{x0}[/itex] (ie. that the total time of stopping was much larger than [itex]\Delta t[/itex]). We don't know [itex]\Delta t[/itex] and we can't find out what it is. It depends upon the hardness of the block and floor. We can only determine [itex]F\Delta t[/itex] (or more exactly [itex]\int_0^t F\delta t[/itex]) where F is the vertical force in excess of Mg. So I should not have assumed anything about the time of impact.

    Just looking at it again, I see that one can determine the loss of horizontal momentum due to the extra horizontal friction from the normal force of the vertical impact: [itex]M\Delta v_x = \mu_kF_{normal}\Delta t[/itex] where [itex]F_{normal}[/itex] is the normal force in excess of Mg. Since [itex]Mv_{y0} = F_{normal}\Delta t[/itex],
    [tex]\Delta v_x = \mu_kv_{y0}[/tex]

    Since [itex]\mu_k = .2[/itex] and [itex]v_{x0}/v_{y0} < .2[/itex] the loss of horizontal speed is completed during the impact.

    The equation for d would be:

    [tex]1/2(v_{x0} - \mu_kv_{y0})^2)/\mu_kg = d [/tex]


    So the answer in this case is effectively 0. The horizontal motion stops in roughly the same distance as the vertical motion, which is negligible. Looking back, this is similar to what Chen concluded with his d = -1 result.

    AM
     
  13. Dec 17, 2004 #12
    Thank you very much for your in-depth explanations. :smile:

    Andrew - it was also my conclusion that the distance is 0, because vx0/vy0 is smaller than the coefficient of friction and the mass comes to a halt in the X axis before it does so in the Y axis. I'll attempt to submit this answer today and let you know how it went.

    By the way, what if the ratio vx0/vy0 was bigger than the coefficient? We would definitely need the dt of the impact, right? Another reason why this might be just a trick question...

    Thanks again,
    Chen
     
  14. Dec 17, 2004 #13

    Andrew Mason

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    Not necessarily. The formula in my last post should apply if the time it takes to stop is greater than dt.

    The total stopping distance, d, will be non-zero (non-negligible) if [itex]v_{x0}>> \mu_kv_y[/itex]. In fact, the formula should give the exact distance d, because it includes the uncertain distance travelled during impact which is determined by the average speed during that time:
    [tex]s=(v_{x0} - \Delta v_{ximpact}/2)\Delta t[/tex] or
    [tex]s=(v_{x0} - (\mu_kv_y + \mu_kg\Delta t)/2)\Delta t[/tex]

    Note in the formula [itex]\Delta v_x[/itex] was defined as the loss of speed due only to the extra horizontal friction from impact. It did not include the friction due to [itex]\mu_kMg[/itex]. So technically, in order for the formula to give the exact distance, the inequality [itex]v_{x0} > (\mu_kv_y + \mu_kg\Delta t)/2[/itex] has to apply.

    AM
     
    Last edited: Dec 17, 2004
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