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Homework Help: Repeated collision.

  1. Dec 27, 2013 #1
    1. The problem statement, all variables and given/known data
    I'd like to find the minimum mass M which will render possible a second collision between M and the two masses m and the spring in the attachment, and also the time between the two collisions. It is stated that the initial velocity of M is v0 as indicated and that the surface is frictionless. It is also stated that this is an elastic collision, whose time is very short. M doesn't get attached to the masses.

    2. Relevant equations

    3. The attempt at a solution
    I am not quite sure how to formulate a condition on M, which will enable a second collision. I am not even sure how a second collision would be possible unless there were let's say a wall on the other side hitting M back towards the masses and the spring. The kinetic energy of M is turned into an elastic energy of the spring and kinetic energy of both M and the reduced mass = m/2, right (due to conservation of energy)?
    Furthermore, there is conservation of linear momentum, isn't there?
    Must I work in the CM reference frame?

    Attached Files:

  2. jcsd
  3. Dec 27, 2013 #2
    Physically, this is what happens:

    (1) Mass M strikes the left mass m elastically. Speed of mass M is reduced, while mass m acquires a finite speed. At this point in time, the right mass m is unaffected and the spring is still uncompressed.

    (2) As the left mass m moves leftward, the spring compresses, and the right mass m begins to move as well. The system then reaches equilibrium when both masses are moving at the same speed. This speed is obviously less than the original speed of the left mass just after the collision. This is why the mass M can catch up and collide with the system again if conditions are met.
  4. Dec 27, 2013 #3
    Alright, so a fraction of a second after the collision:
    (1/2)Mv0^2 = (1/2)mv^2 (conservation of energy)
    Mv0 = -mv (conservation of momentum)
    Are these correct?
    Now, as the left mass m begins to move, could I write:
    (1/2)mv^2=(1/2)kx^2 + (1/2)*mred*u^2
    where mred = m/2?
  5. Dec 27, 2013 #4
    No. Mass M does not stop after the collision.
    No either. Why do you want to use reduced mass?
    For the second part, conservation of momentum would suffice.
  6. Dec 27, 2013 #5
    Let me then try this:
    (1) (1/2)M(v0)^2 = (1/2)mv^2 + (1/2)M(v1)^2 (conservation of energy)
    (2) (1/2)mv^2 + (1/2)M(v1)^2 = (1/2)kx^2 + (1/2)*2m*u^2
    (3) (M/2m)v0 = [2m*u + M(v1)]/[2m+M]
    Are these somewhat better?
  7. Dec 27, 2013 #6
    No, you should treat the two collisions separately.

    Collision 1
    $$\frac{1}{2}Mv_{0}^{2} = \frac{1}{2}Mv_{1}^{2} + \frac{1}{2}mu_{0}^{2}\\
    Mv_{0} = Mv_{1} + mu_{0}$$

    Collision 2
    $$\frac{1}{2}mu_{0}^{2} = \frac{1}{2}k x^{2} + mu_{1}^{2}\\
    mu_{0} = 2 mu_{1}$$

    For collision 2, the COE equation is not useful because you don't need the spring compression and you cannot directly compute it without already knowing the initial and final speeds.
  8. Dec 27, 2013 #7
    But mass M is not at rest after collision 1 takes place, as you've pointed out. How come it is not included in any of the equations referring to collision 2?
  9. Dec 27, 2013 #8
    Sorry if I haven't been clear, but collision 2 refers to the 'collision' between the two masses m, and not the repeated collision between mass M and the spring-mass system.
  10. Dec 27, 2013 #9
    Right, but whilst writing your COE equations, aren't you expected to refer to the system as a whole?
  11. Dec 27, 2013 #10
    You could, but nothing happens to the mass M in the collision 2 that I am referring to. So the terms before and after would just cancel each other out.
  12. Dec 27, 2013 #11
    Fair enough. How could I now set a condition on mass M to make a second collision possible? You wrote "if conditions are met". Which are the conditions?
  13. Dec 27, 2013 #12
    Well, to solve for the conditions is the goal of the question isn't it?
    Well, you need the mass M to be able to catch up and hit the spring-mass system. So that means that ##v_{1}## must be greater than ##u_{1}##.
  14. Dec 27, 2013 #13
    Did you not mean that v1 must be less than u1? If v1 were greater, M would "flee" away before the spring-mass system could hit it again.
  15. Dec 27, 2013 #14
    Why don't you just solve for the motion of each of the three masses as a function of time using Newton's second law on each of them (and applying the appropriate initial conditions for conservation of energy and momentum at t = 0)? Then there will be no guessing involved. You already have written the equations for establishing the initial conditions on M and the first mass m.
  16. Dec 27, 2013 #15
    Mass M must be moving leftwards even after the first collision in order for the repeated collision to occur. This is because the spring-mass system will always be moving leftwards.
  17. Dec 27, 2013 #16
    Alright, I have managed to find the condition: Mmin = 2m.
    However, I am not quite sure how to determine the time elapsed between the two collisions.
    Need I to transfer to the center of mass frame of reference, determine xCM(t) before and after the first collision?
  18. Dec 27, 2013 #17
    Rereading the previous replies, should I instead find x(t) for M and x(t) for the reduced mass with the spring k and find out the times at which they are equal? Or, even more simply, determine when x(t) for the reduced mass is zero?
  19. Dec 27, 2013 #18
    Shouldn't the time between the two collisions simply be pi/omega, where omega=sqrt(k/reduced mass)?
  20. Dec 27, 2013 #19
    Let the subscript M refer to the mass M, and let the subscripts 1 and 2 refer to the first mass and the second mass. From the equations you've already derived, the initial velocities of the masses after the initial collision are vM=v0(M-m)/(M+m), v1=2mv0/(M+m), and v2=0. The initial locations of the masses at the initial collision are xM=x1=0, and x2=L, where L is the undeformed length of the spring. The velocity of the mass M does not change after this. Therefore, xM=v0(M-m)t/(M+m).

    The force balance equations for two small masses are:
    If we add these two equations, we get
    The solution to this equation, subject to the initial conditions is :
    If we subtract the two force balance equations, we obtain:
    where [itex]δ=(x_1-x_2+L)[/itex].
    Solve this equation for δ subject to the initial conditions δ=0 and dδ/dt=2mv0/(M+m) at t = 0.
  21. Dec 27, 2013 #20
    Okay, I got:
    δ = (2mv0/[ω(M+m)]) * sin(ωt)
    where ω = √(2k/m)
    Finding x1 and equating x1(t) and xM(t) I obtained:
    which can be solved graphically to give t≈1.9/ω
    Would you agree?
  22. Dec 27, 2013 #21


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    You can do better than resorting to a graphical solution.
    You have reduced the problem to solving an equation pair like
    y = at, y = sin(ωt)
    for some t > 0.
    But you are specifically looking for the lowest (most negative) a for which there is a solution.
    What, geometrically, is the relationship between the straight line and the sine curve in that case?
  23. Dec 27, 2013 #22
    Your solution for δ looks right, but what happened to all the M's and m's in the solution? You're trying to find the value of M that barely allows two hits. Please show us your work. Also, as Haruspex was alluding to, not only must the displacements match, but also something else must match.

  24. Dec 27, 2013 #23
    Mind you, your expressions for v1 and vM would not be accurate unless M were 2m. I have substituted them in all the equations and an equality was only attained upon equating M to 2m. This is what I ended up substituting. In any case, I did make a mistake earlier. Redoing some of the work, I obtained:
    x1 = (mv0t)/(M+m) + δ/2 = xM = (v0(M-m)t)/(M+m)
    which yielded sin(ωt)=0, hence t = π/ω.
  25. Dec 27, 2013 #24


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    Is that in response to Chester's analysis? You certainly should not be making such an approximation. It isn't necessary.
    That is not the right answer.
  26. Dec 27, 2013 #25
    If I do not make such assumption/approximation, I obtain:
    t(M-2m) = (m/w)sin(wt)
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