Physics Puzzle: Balls on a Stick

Posted on April 1, 2008
Filed Under Physics, Physics Puzzles |

There’s an interesting physics puzzle hiding behind that painful title.

Consider an infinitely long stick, with masses m initially placed along it (at rest) at distances d:

Balls on a stick

 

A constant force F is applied to the first, leftmost mass. Assume completely inelastic collisions. Can you compute, after a long enough time, the velocity of the initial mass? Can you compute the velocity of the rightmost mass (that has been struck)?

Now - can you repeat the computation for the case of completely elastic collisions? Can you compute the velocity of the initial, leftmost mass? Can you compute the velocity of the “shock wave” that will be created?

Notes: Neglect gravity & gravitational forces, assume the problem is one dimensional and that the masses are point-objects, and that they are free to move without friction on the infinite stick.

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One Response to “Physics Puzzle: Balls on a Stick”

  1. admin on April 4th, 2008 6:36 am

    SOLUTION, ADDED 4/4/2008:

    Let’s look at the asymptotic quantities.

    INELASTIC COLLISIONS
    ——————–

    This is the “easy” case. Both the leftmost mass and the rightmost mass struck will move with the same speed. This “super-mass”, made out of lots of balls stuck together, will continually accelerate, and then lose a bit of speed because of hitting the next mass in line. If we want some form of “average” view of the situation, let’s replace our masses by a mass density, Q=m/d, so

    F = dp/dt = dm/dt * v + m*dv/dt

    Asymptotically, dv/dt = 0, so

    F = dm/dt * v

    However, dm = Q*dx = Q*v*dt, and thus

    F = Q*v^2 —> v = sqrt(F/Q) = sqrt(Fd/m)

    ELASTIC CASE
    ————

    This is a bit trickier. There are two insights here to be considered:

    1.) When a mass m1 with speed v collides with a stationary mass m2 with no velocity, then (assuming m1=m2), m1 will come to a halt and m2 will continue with velocity v.

    2.) In the case of an infinite number of balls/masses, the initial mass will move forwards, accelerate up to a velocity v, hit next mass and stop. This will take it a time DT, given by

    F = ma
    –> a = F/m
    –> x(t) = F*t^2/(2m) (integrate)
    –> x(DT) = d = F*DT^2/(2m)
    –> DT = sqrt(2md/F)

    The velocity of our mass, vf, when hitting this next mass is

    F = ma
    –> v = F/m*t
    –> vf = F/m*DT = sqrt(2Fd/m)

    After hitting the next mass, our initial mass will stop and starting accelerating from 0 velocity once more.
    Meanwhile, the next mass will move forwards and hit the mass just after it and stop as well, ad infinitum, making a shock wave move forwards with velocity vf.
    The first mass, meanwhile, will accelerate once again (thanks to the force F), and once again will hit the stationary mass in front of it after a time DT.
    This process will continue indefinitely, creating a shock wave moving with velocity vf. Our initial leftmost mass will continue accelerating and stopping in the same manner all the time, and so, will have an average velocity of

    vf/2

    which, by the way, is sqrt(2) slower than the inelastic case (but the shock wave’s speed, vf, will be sqrt(2) faster!)

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