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Homework Help: Impulse and Momentum: Ball and cart connected by a cord

  1. Dec 7, 2018 #1
    1. The problem statement, all variables and given/known data

    A 1kg cart, a 0.1kg ball, and an elastic cord connecting the ball and the cart, are placed in a weightless environment; see Figure. The cart has the shape of a symmetric trapezoid with its left and right edges at a 60-deg angle from the vertical. It is also attached with four rollers (two on top and two on bottom) so that it can move freely horizontally. At t = 0, the system is in static equilibrium with the elastic cord fully stretched to exert a force Fs = -1 N (to the left) on the cart. The ball is then released from rest in its t = 0 position and hits the cart at t = 2 seconds. Note that the attachment of the elastic cord is at exactly the same level as the centre of mass of the cart. Answer the following questions. (a) Will the cart move right after the release of the ball (t = 0+)? Provide your reason(s). (b) What are the impulsive forces at the instant the ball hits the cart? (c) Assume the ball sticks to the cart surface, what is the speed of the cart right after the ball hits the cart? Neglect the mass of the string.

    M = 1kg
    m = 0.1kg
    θ = 60

    Screen Shot 2018-12-07 at 3.07.07 PM.png

    2. Relevant equations

    F = ma
    p = mv


    3. The attempt at a solution

    a) The cart will want to move after the ball is released because the ball and cart are connected by the cord. The force pulling the ball to the cart is equal and opposite to the force pulling the cart to the ball. However, there is a ledge blocking the cart's motion, so it will stay put.

    b) Right of the bat I'm not really sure what's meant by impulsive forces. I thought to find an impulsive force a time frame over which the force acts must be known? Anyways, I figure the ball is pulled in by a force of 1N over two seconds, so the acceleration and velocity after 2s can be given by:

    am = 1/m
    vm = 2/m

    The impulse is given by the change in momentum. The ball goes from vm = 2/m stationary at the moment of contact, so:

    Impulse = 2/m

    and impulsive force would be this divided by some time? Not sure.

    Well, whatever the impulsive force is, it acts normal to the cart surface and can be broken down into x and y components. the x component is proportional to cos(60) and the y component to sin(60).

    Another thing - I feel that the vm value is incorrect, because wouldn't the force change over time, if the cord is elastic?

    c) The ball has momentum mvm before it hits the cart. A proportion cos(60) of this momentum is carried forward, so:

    mvm = (m + M)*v*cos(60)

    mvm / (( m + M)*cos(60)) = v

    where v is the velocity of the ball and cart stuck together.

    Any ideas?






     
  2. jcsd
  3. Dec 7, 2018 #2

    haruspex

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    The question asks whether the cart will tend to move to the right just after the ball is released. Intuitively, would it? Can you think of a reason why it might?
    It is elastic. The force will not be constant.
    The cart is free to move. Why would the ball become stationary?
    This is awkward. At this point we have not been told that the ball sticks to the cart, so I suspect your answer is the one expected. But when we factor in that it sticks, the picture gets rather complicated. I do not like this question.
     
  4. Dec 7, 2018 #3

    CWatters

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    +1

    Impulse is the change in momentum. Look it up on Wikipedia.

    I don't think you can answer b) without knowing if the collision is elastic or inelastic. I think I would assume its an elastic collision for b) and inelastic for c).

    However I don't think you can answer c) without knowing more about the roller since they are rotating? Perhaps you are meant to assume the rollers are massless?
     
  5. Dec 8, 2018 #4
    There is a normal force acting on the cart from the wall, equal to the pulling force from the cord. Just after the ball is released, perhaps the pulling force decreases, and the normal force pushes the cart to the right. Is that reasoning valid?
     
  6. Dec 8, 2018 #5

    haruspex

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    That fits with my intuition, and there is a good reason for it.
    In the real world, everything has a bit of elasticity, including the barrier that is preventing the cart moving left. So when the ball is released, the recoil from the barrier will push the cart right.
    That said, I am not at all sure what the questioner expects. Maybe they want you to take the barrier as completely rigid.
     
  7. Dec 14, 2018 at 1:23 PM #6
    When I Read the question I took "right after" to mean "immediately after".

    This is my attempt at a solution. What do you think? I used the Dirac delta function to find the impulses. Thanks for the help, I appreciate it.

    a) The cart will move to the right after the ball is released. When the tension in the cord is relieved, the normal force from the wall will push the cart.


    b) The ball feels an impulsive force when it is released from a still position, giving it a velocity:


    ##J\quad =\quad \int { Fdt } \quad =\quad 1\int { \delta (t)dt } \quad =\quad 1\\ J\quad =\quad m\Delta v\quad =\quad m({ v }_{ b }-\quad 0)\\ 1\quad =\quad m{ v }_{ b }\\ { v }_{ b }=\frac { 1 }{ m }##



    Similarly, for the cart’s velocity due to the normal force:


    ##{ v }_{ c }=\frac { 1 }{ M }##


    When the ball hits the cart at t = 2, it experiences a change in velocity and thus an impulse:


    ##\Delta v\quad =\quad ({ v }_{ c }\quad -\quad { v }_{ b })\quad =\quad (\frac { 1 }{ M } \quad -\quad \frac { 1 }{ m } )\\ J\quad =\quad F\int { \delta (t-2)dt } \quad =\quad m\Delta v\quad =\quad m(\frac { 1 }{ M } \quad -\quad \frac { 1 }{ m } )##

    The impulsive force has componets:


    ##{ F }_{ x }=\quad m(\frac { 1 }{ M } \quad -\quad \frac { 1 }{ m } )\cos { (60) } \\ \\ { F }_{ y }=\quad m(\frac { 1 }{ M } \quad -\quad \frac { 1 }{ m } )\sin { (60) }##


    c) After the collision, we have:


    ##m{ v }_{ f }\cos { (60) } \quad +\quad M{ v }_{ c }\quad =\quad (M\quad +\quad m)v\\ \\ v\quad =\quad \frac { \cos { (60) } \quad +\quad 1 }{ (M\quad +\quad m) } \quad =\quad 0.15m/s##
     
  8. Dec 14, 2018 at 1:40 PM #7

    gneill

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    Ever watch a slinky drop?
    <iframe width="948" height="543" src="" frameborder="0" allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen></iframe>
     
  9. Dec 14, 2018 at 2:15 PM #8

    haruspex

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    This is inadequate. If the barrier is rigid the normal force will do no work. Reread what I wrote before.

    For b and c, if the impulse is purely normal to the surface of the block, what happens to the component of the ball's velocity that was parallel to that surface?
     
  10. Dec 14, 2018 at 4:05 PM #9
    Ok, so looking at the slinky I can assume that when the ball is released, the tension in the cord will not be relieved near the cart.

    And if the barrier is rigid, then the normal force will do no work.

    So, another attempted solution:

    a) The cart will not move when the ball is released, assuming the barrier is rigid.

    b) The ball has a velocity before it hits the cart given by:

    ##J\quad =\quad \int { Fdt } \quad =\quad 1\int { \delta (t)dt } \quad =\quad 1\\ J\quad =\quad m\Delta v\quad =\quad m({ v }_{ b }-0)\\ 1\quad =\quad m{ v }_{ b }\\ { v }_{ b }=\frac { 1 }{ m } ##

    When the ball hits the stationary cart we have an impulsive force that can be broken down into x and y components:

    ##J\quad =\quad F\int { \delta (t-2)dt } \quad =\quad m{ v }_{ b }\\ F\quad =\quad m{ v }_{ b }\quad =\quad 1N\\ { F }_{ x }=\quad m{ v }_{ b }\cos { (60) } =\quad 0.5N\\ { F }_{ y }=\quad m{ v }_{ b }\sin { (60) } =\quad 0.87N##

    I'm taking x to be horizontal and y to be vertical, and I'm thinking the y component of the force just goes away. It would increase the normal force of the ball on the track and thus the friction, but if the objects are totally rigid this wouldn't happen, right? All I can think of is that the component of the ball's velocity that is parallel to the surface would make it squish in a certain way.

    c) To find the final velocity of the cart we can use the x component of the impulsive force:

    ##\int { { F }_{ x }\delta (t-2)dt } \quad =\quad (M+m)v\\ v\quad =\quad \frac { m{ v }_{ b }\cos { (60) } }{ (M+m) } \quad =\quad 0.05m/s##
     
  11. Dec 14, 2018 at 4:23 PM #10

    gneill

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    Not instantaneously, no.
     
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