Kvant Physics Problem 67

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Verdicts: PASS + PASS
Solve time: 23m07s
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Problem

F67. A buffer device (Fig. 2) consists of a rod $A$, a spring $B$ mounted on the rod, and a guide sleeve $C$. The sleeve can move inside a channel made in a massive wall $D$. As the sleeve $C$ moves, a friction force of constant magnitude $F_{\text{тр}}$ acts between its outer surface and the wall. The rod inside the sleeve and the spring along the rod move without friction.

A ball of mass $M$ strikes the end surface of the rod $A$, having velocity $v_0$ before the collision. With what velocity will the ball rebound?

The masses of the rod, spring, and sleeve are negligibly small compared with the mass of the ball. The spring constant is $k$.

V. K. Peterson, 31st Moscow State University School Physics Olympiad, 1970.

Description of Fig. 2 (according to the problem text): the buffer device consists of a rod $A$, a spring $B$ mounted on the rod, and a guide sleeve $C$ moving inside a channel in a massive wall $D$.

Setup and Assumptions

A ball of mass $M$ moves along the axis of a buffer device and strikes the end of a massless rod attached to a massless spring of stiffness $k$. The spring is connected to a guide sleeve that can slide inside a channel in a massive wall. The sleeve experiences a constant friction force of magnitude $F_{\text{тр}}$ opposing its motion, while the rod and spring slide without friction along the axis. The analysis is performed in the frame of the wall, which is inertial.

Because the rod, spring, and sleeve are massless, they cannot store kinetic energy and cannot sustain force imbalances. This implies that at every instant the internal forces adjust so that any massless element has zero net force.

Physical Principles

While the sleeve is at rest, the spring force is $F = kx$, where $x$ is the compression. The sleeve begins to slide when the spring force reaches the friction threshold $kx = F_{\text{тр}}$.

Once sliding starts, the sleeve has zero mass, so its equation of motion reduces to instantaneous force balance along the axis. The forces acting on the sleeve are the spring force $kx$ in one direction and the friction force $F_{\text{тр}}$ in the opposite direction. For zero net force at every instant during sliding, these must satisfy $kx = F_{\text{тр}}$, which fixes the spring compression at a constant value

$x_* = \frac{F_{\text{тр}}}{k}.$

Since $x$ remains constant during sliding, the spring energy remains fixed at

$U_* = \frac{1}{2}k x_*^2 = \frac{F_{\text{тр}}^2}{2k}.$

The crucial point is that during sliding the spring does not change its energy; all additional mechanical energy supplied by the ball is transferred into work against friction of the sleeve.

Derivation

The ball initially has kinetic energy

$E_0 = \frac{1}{2} M v_0^2.$

Compression begins with the sleeve at rest. While $kx < F_{\text{тр}}$, the sleeve is fixed and the motion is purely elastic, so energy is stored in the spring. When $kx$ reaches $F_{\text{тр}}$, the compression reaches $x_*$ and further compression becomes impossible because any increase would violate force balance on the massless sleeve.

From this moment onward, the kinematic constraint $x = x__$ freezes the spring energy at $U__$. The ball may still be moving, and its motion is now coupled to the sleeve through the rod. Any further forward displacement of the ball necessarily produces displacement of the sleeve because the rod is rigid and massless.

Let the sleeve move a total distance $s$ during the entire interaction after the onset of sliding. During this motion the spring force on the sleeve is constant and equal to $F_{\text{тр}}$, so the work done against friction is

$A_{\text{fr}} = F_{\text{тр}} s.$

No other irreversible mechanism is present, so this work must equal the total mechanical energy lost by the ball–spring system beyond the energy stored in the spring at the onset of sliding. The only energy available to be dissipated after that instant is the excess

$E_0 - U_*.$

This equality is not an assumption but follows from global energy accounting: the spring energy cannot change while $x = x_*$, and no kinetic energy remains in the massless elements, so the only sink for additional mechanical energy is friction in the sleeve. Therefore,

$F_{\text{тр}} s = E_0 - U_*.$

The motion continues until the ball reverses direction. At the turning point of the ball, its kinetic energy is zero, so all remaining recoverable energy of the system is stored in the spring. At that instant the sleeve is still at the same compression constraint $x = x__$, so the spring still stores exactly $U__$.

When the ball begins moving backward, the spring releases energy. During this stage the sleeve is already at rest because friction has absorbed all energy associated with its displacement, and no further constraint enforces motion of the sleeve. As a result, the spring energy $U_*$ is transferred entirely into the kinetic energy of the ball without additional losses.

Thus the rebound kinetic energy of the ball equals $U_*$:

$\frac{1}{2} M v_{\text{reb}}^2 = U_*.$

Solving for the speed gives

$v_{\text{reb}} = \sqrt{\frac{2U_*}{M}} = \sqrt{\frac{F_{\text{тр}}^2}{kM}} = \frac{F_{\text{тр}}}{\sqrt{kM}}.$

This result is independent of the initial energy once the threshold for sleeve motion has been exceeded, because all excess energy is irreversibly converted into frictional work during the constrained sliding stage, while the spring is locked at a fixed energy level.

For lower initial energies where the condition $E_0 \le U_*$ holds, the sleeve never moves, no friction work is performed, and the interaction reduces to a purely elastic compression and release, giving $v_{\text{reb}} = v_0$.

Result

The rebound speed of the ball is

$$v_{\text{reb}} = \begin{cases} v_0, & \frac{1}{2} M v_0^2 \le \frac{F_{\text{тр}}^2}{2k}, \[1em] \frac{F_{\text{тр}}}{\sqrt{kM}}, & \frac{1}{2} M v_0^2 > \frac{F_{\text{тр}}^2}{2k}. \end{cases}$$

The rebound velocity vector is directed opposite to the initial motion,

$\vec v_{\text{after}} = - v_{\text{reb}} , \hat{\mathbf{x}}.$

Sanity Checks

The expression $F_{\text{тр}}^2/k$ has dimensions of energy since it scales as force squared divided by spring stiffness. The combination $F_{\text{тр}}/\sqrt{kM}$ has dimensions of velocity, consistent with the final result.

In the limit $F_{\text{тр}} \to 0$, the threshold energy $U_* \to 0$, so any nonzero impact immediately enters the sliding regime and all energy is dissipated, giving vanishing rebound speed. In the opposite limit $F_{\text{тр}} \to \infty$, the sleeve never moves, so no energy is lost to friction and the interaction reduces to an ideal elastic reflection from a rigid spring system, yielding $v_{\text{reb}} = v_0$.

The threshold condition $\frac{1}{2} M v_0^2 = U__$ marks the transition between regimes where the sleeve either remains locked or enters constrained sliding. At this boundary the two expressions coincide in energy accounting, since in both cases the spring energy at release is exactly $U__$.