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Spring 2009 20 Math

Beads of equal mass $m$ are strung at equal original distances $d$ on a long horizontal wire. The beads are initially at rest but can move along the wire without friction. The leftmost bead is continuously accelerated (towards the right) by a constant force $F$. The other beads do not feel $F$, but do undergo collisions with the leftmost bead and each other.

As a result of the collisions, a compression wave propagates to the right down the wire. What are the speeds of the leftmost bead and the front of the ‘shock wave’ after a long time, if the collisions of the beads are:

(a) completely inelastic,

(b) perfectly elastic?

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Answer

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(a) Let $v_0$ denote the asymptotic common speed.

In a given time interval $\Delta t$, the cluster collides with $v_0 \Delta t/ d$ further beads, which increases its mass by $\Delta m = m v_0 \Delta t / d$ and its momentum by $\Delta p = v_0 \Delta m = m v_0^2 \Delta t / d$. According to Newton's law of motion,

(1)which yields $v_0 = \sqrt{Fd / m}$ for the ultimate speed in the case of inelastic collisions.

(b) In an elastic collision between two equal mass bodies with one of them initially at rest, their velocities are exchanged.

The body initially moving with velocity $v$ stops, while the second one moves away with velocity $v$. The leftmost bead accelerates uniformly and reaches a speed of

(2)before the first (elastic) collision takes place. It then transfers its speed to the second bead and stops, after which it starts accelerating again as a result of the external force. The second bead moves at a constant speed $v_1$, collides with the third bead and stops. The third and subsequent beads behave similarly, and a ‘shock wave’ propagates forward at speed $v_1$.

Meanwhile, the leftmost bead is again accelerated to speed $v_1$, collides with the second bead, which is now at rest, and the process is repeated, thus starting a new ‘shock wave’. The speed of the leftmost bead varies uniformly from zero to $v_1$, its average value is $v_1 / 2 = v_0 / \sqrt{2} = \sqrt{Fd / (2m)}$.

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