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2-D Elastic Collision: finding final velocities

  1. Nov 27, 2014 #1
    This is a modified problem from the one posted a few days ago. I’ve been unsuccessfully trying to solve it since then, so maybe it’s time to ask for some help.

    I’ve changed the quantities :)


    1. The problem statement, all variables and given/known data


    The mass of puck 1 is ##0.2\ kg## and that of puck 2 is ##0.25\ kg##. The initial velocities are ##v_1 = 1.5\ m/s## and ##v_2 = 0.8\ m/s##.

    Assuming no friction between the pucks, only normal forces during collision, find their final velocities.

    collision2.jpg

    Data:

    ##m_1 = 0.2\ kg##

    ##v_1 = 1.5\ m/s##

    ##m_2 = 0.25\ kg##

    ##v_2 = 0.8\ m/s##

    ##\theta: -35º##

    2. Relevant equations

    ##p_{1x} + p_{2x} = p_{1x}’ + p_{2x}’##

    ##p_{1y} + p_{2y} = p_{1y}’ + p_{2y}’##

    ##KE_{initial} = KE_{final}##

    3. The attempt at a solution

    For a while I thought I was done, but I wasn’t. I’m doing something wrong. Whether is the wrong coordinate system, the signs and/or the math, I don’t know. I appreciate any guidance.

    First, I chose a coordinate system so that the line of impact lies along the x-axis. After the collision, one ball will move away from the origin along the line ##y = 0##, while the other will bounce in the sort of wall formed by the y-axis:

    collision_a.jpg

    Having established that, ##\theta = 215º##, ##v_2 = -0.8\ m/s## and ##v_2’## will be in the positive direction.

    x-axis:

    ##m_1v_1\cos(\theta) + m_2v_2 = m_1v_1’\cos(\beta) + m_2v_2’##

    y-axis:

    ##m_1v_1\sin(\theta) = m_1v_1’\sin(\beta)##

    ##v_1\sin(\theta) = v_1’\sin(\beta)##

    Conservation of internal KE:

    ##\frac{1}{2} m_1v_1^2 + \frac{1}{2} m_2v_2^2 = \frac{1}{2} m_1v_1’^2 + \frac{1}{2} m_2v_2’^2##

    ##m_1v_1^2 + m_2v_2^2 = m_1v_1’^2 + m_2v_2’^2##

    Knowing that the overall momentum is conserved ##(p_1 + p_2 = p_1’ + p_2’)##, I used this equation to solve for ##v_2’## and substitute it in the KE equation. Finally, I applied the quadratic formula which gives me the following results:

    ##v_{1(a)}' = \frac{80+28}{72} = 1.5\ m/s##

    ##v_{1(b)}' = \frac{80-28}{72} = 0.7\ m/s##

    Assuming that ##v_{1(a)}'## is not the answer, being the velocity before the collision, I took ##v_{1(b)}'## as the final velocity. Then, I tried to find ##v_2’##:

    ##\Delta p_1 = \Delta p_2##

    ##m_1(v_1’-v_1) = m_2(v_2’-v_2)##

    ##v_2’ = \frac{m_1(v_1’-v_1)}{m_2}+v_2 = \frac{0.2\ kg (v_1’-1.5\ m/s)}{0.25\ kg}-0.8\ m/s##

    If one replaces ##v_1’## by the value obtained in ##v_{1(b)}'##, then ##v_2’## is negative, meaning that ##m_2## is moving in the same direction before the collision (it doesn’t make sense).

    If one replaces it by the value obtained in ##v_{1(a)}'##, then ##v_2’## is the same as before the collision ##(-0.8\ m/s)##, meaning that ##m_2## didn’t hit ##m_1## (it doesn’t make sense either).

    I haven’t used the equations of momentum along the x or y axis, but I get in trouble when trying to do so. By taking into account that ##v_1\sin(\theta) = v_1’\sin(\beta)##, using the Pythagorean formula/identity and by combining equations I still can’t find a way to reduce the number of unknowns.

    Thanks !!
     
    Last edited: Nov 27, 2014
  2. jcsd
  3. Nov 27, 2014 #2

    haruspex

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    Your new theta is still 35 degrees.
    I don't get that, so please post your working.
    As remarked on the previous thread, this problem will be simpler if you consider v1 as two separate quantities, the x component and the y component. Since the y velocity is conserved, the contribution to the KE of the y velocity is conserved, so the total KE from x axis motion is conserved. This reduces it to a 1D problem.
     
  4. Nov 28, 2014 #3
    True :)

    image.jpg

    I think I got the velocities. This last time (and a couple of times before) I solved the total CoM equation for ##v_1'## and then subtituted it in the KE equation to find ##v_2'##. I carried out the process renaming the variables with letters, to make sure I wasn't making a mistake (It seems I was making mistakes with the math).

    ##a = m_ 1 = 0.2\ kg##
    ##d = v_1 = 1.5\ m/s##
    ##x = v_1' =\ ?##
    ---
    ##b = m_2 = 0.25\ kg##
    ##f = v_2 = -0.8\ m/s##
    ##y = v_2' =\ ?##

    CoM:

    ##m_1v_1 + m_2v_2 = m_1v_1' + m_2v_2'##

    ##ad + bf = ax + by##

    ##x = \frac{b(f-y)}{a}+d \hspace{35pt} (1)##

    KE:

    ##m_1v_1^2 + m_2v_2^2 = m_1v_1'^2 + m_2v_2'^2##

    ##ad^2 + bf^2 = ax^2 + by^2##


    ##ad^2 + bf^2 = a\left(\frac{b(f-y)}{a}+d\right)^2 + by^2\hspace{35pt} ## <-- Replace by ##(1)## here.


    ##ad^2 + bf^2 = a\left(\frac{b^2(f^2-2fy+y^2)}{a^2} + \frac{2bd(f-y)}{a}+d^2\right) + {by^2}##


    ## \underline{ad^2} + bf^2 = \frac{b^2f^2 - 2b^2fy + b^2y^2}{a} + 2bdf - 2bdy+ \underline{ad^2} + by^2##


    ##0 = b^2f^2-2b^2fy + b^2y^2 + 2abdf - 2abdy+ aby^2 - abf^2##


    At this point I found the coefficients of the polynomial separately:

    ##(b^2 + ab)y^2 = (0.25^2+0.2\cdot0.25)y^2 = 0.1125y^2##

    ---

    ##-(2b^2f+2abd)y = -(2\cdot0.25^2\cdot(-0.8)+2\cdot0.2\cdot0.25\cdot1.5)y= -0.05y##

    ---

    ##b^2f^2 + 2abdf - abf^2 = 0.25^2\cdot0.8^2+2\cdot0.2\cdot0.25\cdot1.5\cdot(-0.8)-0.2\cdot0.25\cdot0.8^2= 0.04-0.12-0.032=-0.112##


    Then:

    ##0 = 0.1125y^2-0.05y-0.112##


    The quadratic formula:

    ##y= \frac{0.05 \pm \sqrt{0.05^2-4\cdot0.1125\cdot(-0.112)}}{0.225}##

    ##y= \frac{0.05 \pm \sqrt{0.0529}}{0.225}##

    ##y_1 = 1.24\ m/s##

    ##y_2 = -0.8\ m/s##

    ##y_2## turns out to be the velocity before the collision, so ##v_2' = 1.24\ m/s##. Then ##v_1'## is found to be ##-1.06\ m/s##. These values agree with a CoM of ##0.1\ kg\dot\ m/s## and a Co(KE) of ##0.305\ J##.

    When vectors lie right in the x or y axis it's easy to spot their sign, but I have trouble associating the sign with vectors that don't lie along one of the axis (It was a problem when finding ##v_1'## with the quadratic, instead of ##v_2'##). Do I have to take a look at their slopes to figure that out?

    The direction of ##v_1'## is ##-54.6º##. Is my diagram wrong? Or do I have to interpret that as ##180º -54.6º = 125.4º##?

    Do you mean to take ##v_1cos(\theta)## as a total velocity, to say ##u_1##, and then use CoM and KE with ##u_1## and ##u_1'##?

    By writing the KE equation without distributing the masses in their respective velocities, and dividing it by the CoM equation, I got the coefficient of restitution.

    I'm not sure if either the approach you propose or the CoR can be useful to avoid finding a velocity without using the quadratic formula (which was the messiest part).
     
    Last edited: Nov 28, 2014
  5. Nov 28, 2014 #4

    haruspex

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    Yes.
    Coefficient of restitution? Up to this point, you'd given the impression that the collision is purely elastic. If not, you need another piece of information.
    You can certainly avoid the quadratic.
    ##m_1u_1+m_2u_2 = m_1v_1+m_2v_2## (1)
    ##m_1(u_1-v_1) = m_2(v_2-u_2)## (1')
    ##m_1u_1^2+m_2u_2^2 = m_1v_1^2+m_2v_2^2## (2) (KE)
    ##m_1(u_1^2-v_1^2) = m_2(v_2^2-u_2^2)## (2')
    ##m_1(u_1-v_1)(u_1+v_1) = m_2(v_2-u_2)(u_2+v_2)## (2'')
    From 1' and 2'':
    ##u_1 + v_1 = v_2 + u_2## (3) (Newton's Experimental Law for the case where the CoR = 1)
    You can then use (1) and (3) to solve.
     
  6. Dec 1, 2014 #5
    Sorry for the late reply.

    /: I've used the wrong wording. I meant the equation you've written below.

    Well, at the moment when I found Newton's EL on my own, I didn't see its usefulness because I made a mistake there with the math too /:

    What is still bugging me is the direction of the angle of v_1'. I calculated it again:

    ##\beta = \arcsin\left(\frac{v_1\cdot sin(\thetaº)}{v_1'}\right) = \arcsin\left(\frac{1.5\cdot sin(35º)}{1.24}\right) = 47.74º##

    That's because of my chosen coordinate system, right? should it be understood as ##47.74º## from the y-axis?
     
  7. Dec 1, 2014 #6

    haruspex

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    No, it's from the x axis (line of centres at collision), as drawn. Why do you think it should be from the y axis?
     
  8. Dec 2, 2014 #7
    Ah, true. I was just muddling the y-axis up with the x-axis (:.
     
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