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Homework Help: Cart on an inclined ramp, with rebound

  1. Nov 17, 2007 #1
    [SOLVED] Cart on an inclined ramp, with rebound

    A 500 g cart is released from rest 1.0 m from the bottom of a frictionless, 30° ramp. The cart rolls down the ramp and bounces off a rubber block at the bottom. The figure (below) shows the force during the collision.

    After the cart bounces, how far does it roll back up the ramp?


    It'd be cool if we could get this solved relatively soon. It's due at midnight EST. And no, I'm not going to be a jerk if it's not solved by then, it's my fault for leaving it late :)
    In theory, I have this solved, but I've made a mistake somewhere as it says my answer is wrong. This is what I did:

    m = 500g = 0.5kg
    x = 1.0m
    [itex]\theta[/itex] = 30°

    Starting from rest, the cart travels 1.0m at the acceleration of gravity (Altered by the incline of the ramp).
    [itex]v_f^2 = v_i^2 + 2ad[/itex]

    Assuming that downwards is the negative direction, therefore [itex]v_f[/itex] and g will be negative.
    If the incline was 90°, the cart would freefall.
    If the incline was 0°, the cart wouldn't be affected by gravity at all.
    Therefore, we need a coefficient on g. Since sin(90) = 1 (Full gravity) and sin(0) = 0 (No effect), we use sin(30) = 0.5 to make this alteration.
    [itex]v_f = \sqrt{0 + 2ad}[/itex]
    [itex]v_f = -\sqrt{2(sin(30)g)d}[/itex]

    sin(30) = 0.5, which cancels out the 2. g = 9.8, and d = 1.0 so can be ignored.
    [itex]v_f = -\sqrt{g} = -3.1305[/itex]
    This is the speed of the cart immediately before it hits the rubber block.

    Next, [itex]J = F\Delta t = \Delta P[/itex]
    [itex]F\Delta t = 200 * \frac{26.7}{1000} = 5.34 Ns[/itex]
    [itex]5.34 Ns = \Delta P = | mv_0 - mv_f |[/itex]
    Where [itex]v_0, v_f[/itex] are the velocities immediately before and after rebounding, respectively.

    The velocity before rebounding, [itex]v_0[/tex], we just calculated to be -3.1305 (The [itex]v_f[/itex] of rolling down the ramp).

    [itex]5.34 Ns = | 0.5 * -3.1305 - 0.5 * v_f |[/itex]
    [itex]5.34 Ns = | -1.56525 - 0.5*v_f |[/itex]
    For the absolute value to equal 5.34, what is inside of the absolute bars must be 5.34, or -5.34. The only way for it to become 5.34 is if [itex]v_f[/itex] is large enough to bring -1.56525 above positive, and into 5.34 (Meaning [itex]v_f[/itex] would need to be somewhere around 12m/s, which doesn't make sense, as that would imply it actually gained speed in rebounding).

    So, we assume the absolute bars to be equal to -5.34, and solve [itex]v_f[/itex] accordingly.
    [itex]-5.34 Ns = -1.56525 - 0.5*v_f[/itex]
    [tex]v_f = \frac{-5.34 + 1.56525}{0.5}[/tex]
    [itex]v_f = -7.55[/itex]
    Since [itex]v_f[/itex] is a rebound up the incline, it is taken to be a positive value, 7.55m/s.
    This is also above our initial collision value of 3.13, which also isn't possible, so my mistake is probably before this point.

    However, just to solve the problem (And to show how to do so):
    [itex]v_0, v_f[/itex] now respectively represent the velocity just as you begin your rebound, and the velocity just as you finish your rebound (Once gravity has slowed the cart back down to 0m/s, before it begins falling down the ramp again)
    [itex]v_f = v_i + at[/itex]
    [itex]0 = 7.55 + (-sin(30)gt)[/itex]
    [itex]7.55 = 0.5*9.8t[/itex]
    [itex]t = 1.54 sec[/itex]

    And lastly, finding out how far the cart can recoil in 1.54 seconds, with an initial speed of 7.55m/s and a gravitational force of -sin(30)g on it.
    [tex]d = \frac{v_i + v_f}{2} * t[/tex]
    [tex]d = \frac{7.55 + 0}{2} * 1.54[/tex]
    [itex]d = 5.82m[/itex]

    By the grading system, I'm told this answer is wrong. This makes sense too, as it shouldn't rebound farther than it initially started. My error is most likely in calculating the 7.55m/s.
    Last edited: Nov 17, 2007
  2. jcsd
  3. Nov 17, 2007 #2


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    Homework Helper

    This is fine.

    Your basic concept of the impulse appears correct. Now, look at the shape of the force profile during the collision. One way of dealing with this is to ask what is the average magnitude of the force during the 26.7 msec collision and then multiply that average by the collision duration. Another way is to find the total area of the force profile during the collision, which gives the magnitude of the impulse (if you've had calculus, we'd say that the integral of F(t) with respect to time equal J).

    I'm not entirely clear what the question(s) is/are that you are to answer, but take things from there.
  4. Nov 17, 2007 #3
    Posted all that, and forgot to post the question, sorry.
    Edited original post to show the question.

    Tried the thing with a different value for J (Dividing old value by 2, since the area is a triangle), still got a wrong answer.
  5. Nov 17, 2007 #4


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    OK, I guess we're also agreed now that the impulse is given by

    If we divide the impulse by the mass of the cart, we get a velocity change of

    (2.67 kg-m/sec) / 0.5 kg = 5.34 m/sec .

    Since the cart was headed downhill at 3.13 m/sec just before impact, the velocity of the cart after the collision is

    -3.13 m/sec + 5.34 m/sec = +2.21 m/sec (so plainly the collision with the bumper is inelastic).

    I'd avoid using the absolute value since that leads to ambiguous solutions, one of which is unphysical. The important point is that if we call the cart's velocity negative before contact, the force from the bumper being positive means that the acceleration on the cart is positive and thus directed uphill. So a positive velocity change would be added to a negative velocity, giving here a positive (uphill) velocity after contact.

    I'm taking it from the method you used to finish the problem that you've had kinematics, but not energy yet. Have you had the "velocity-squared" equation? (You are using a version of it here: [itex]v_f = \sqrt{0 + 2ad}[/itex]). The acceleration back up the slope is -g sin 30º , as you remarked earlier, the initial speed is now 2.21 m/sec and the final speed will be zero. So we have

    (v_f)^2 = (v_i)^2 + 2ad giving

    0 = (2.21 m/sec)^2 + 2 · (-4.91 m/(sec^2)) · d ,

    d being the distance the cart runs back up the slope.

    As I look at your step involving absolute values, I believe the value you'd want to use is

    [itex]2.67 Ns = -1.56525 - 0.5*v_f[/itex] , rather than -2.67 N·sec . (As I say, my preference is to avoid absolute value expressions and to look instead at directions of the velocities and directions of the changes.)
    Last edited: Nov 17, 2007
  6. Nov 17, 2007 #5
    Went through your advice, computed some things, tried it, and it was correct. Thank-you :)
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