IPhO 1983 · Theory problem 1

Jumping particle

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A particle moves along the positive axis \(Ox\) (one-dimensional situation) under a force that’s projection on \(Ox\) is \(F_x=F_0\) as represented in the figure below (as function of \(x\)). At the origin of \(Ox\) axis is placed a perfectly reflecting wall.

A friction force of constant modulus \(F_f=1{,}00\,N\) acts anywhere the particle is situated.

The particle starts from the point \(x=x_0=1{,}00\,m\) having the kinetic energy \(E_c=10{,}0\,J\).

a. Find the length of the path of the particle before it comes to a final stop

b. Sketch the potential energy \(U(x)\) of the particle in the force field \(F_x\).

c. Draw qualitatively the dependence of the particle speed as function of his coordinate \(x\).

Figure 1

Official English text of the problem.

Answers

  • a. The total path length is \(\boxed{L=20.0\,\mathrm m}\). The final resting position is the wall, \(x=0\).
  • b. Choosing \(U(0)=0\), \[ \boxed{U(x)=(10.0\,\mathrm N)x,\qquad x\ge0.} \] Its graph is a straight line through the origin with slope \(10.0\,\mathrm{J\,m^{-1}}\).
  • c. The speed–position graph consists of square-root branches representing successive outward and inward passages. Each successive outer turning distance is \(9/11\) times the preceding one; successive wall speeds decrease by \(\sqrt{9/11}\). The branches converge to \((x,v)=(0,0)\). For initial motion toward \(+x\), the outer turning points are \[ \boxed{a_n=\frac{21}{11}\left(\frac9{11}\right)^n\mathrm m,\qquad n=0,1,2,\ldots.} \] If the initial motion is toward the wall, replace \(21/11\) by \(19/11\), after the initial inward passage.

Solution

Write the field force as \(F_x=-F\), where \[ F=10.0\,\mathrm N,\qquad f=F_f=1.00\,\mathrm N. \] Friction opposes the motion. Consequently, the net force is \(-(F+f)\) during outward motion and \(-(F-f)\) during inward motion. A perfectly reflecting wall reverses the velocity without changing the kinetic energy.

a. Total path length

Since \(F>f\), the particle cannot remain at rest at any \(x>0\): whenever it reaches an outer turning point, it starts moving back toward the wall.

To establish the eventual resting position, consider a turning point at distance \(A\) from the wall. During the inward trip, the field does work \(FA\) and friction does work \(-fA\). Thus the kinetic energy on reaching the wall is \[ K_{\mathrm w}=(F-f)A. \] After reflection, this energy is spent moving outward against both the field and friction. The next turning distance \(A'\) therefore satisfies \[ (F+f)A'=(F-f)A, \] so \[ \frac{A'}A=\frac{F-f}{F+f}=\frac9{11}<1. \] The turning distances and wall speeds tend to zero. The final resting position is therefore \(x_f=0\). Moreover, each half-trip takes a time proportional to the square root of its turning distance; the remaining times form convergent geometric sums. Thus, in this idealized model, infinitely many reflections accumulate in a finite time.

Now apply the work–energy theorem to the entire motion. The total work of the constant field depends only on the endpoints: \[ W_x=F_x(x_f-x_0)=(-10.0\,\mathrm N)(-1.00\,\mathrm m) =10.0\,\mathrm J. \] Friction does work \(-fL\), while the elastic reflections cause no kinetic-energy loss. Hence \[ 0-E_c=W_x-fL. \] Therefore \[ \boxed{ L=\frac{E_c+Fx_0}{f} =\frac{10.0\,\mathrm J+10.0\,\mathrm J}{1.00\,\mathrm N} =20.0\,\mathrm m. } \] This result is independent of the initial direction of motion.

b. Potential energy

The potential energy associated with the field obeys \[ F_x=-\frac{dU}{dx}. \] Since \(F_x=-F\), \[ \frac{dU}{dx}=F, \qquad U(x)=Fx+C. \] With \(U(0)=0\), \[ \boxed{U(x)=(10.0\,\mathrm N)x,\qquad x\ge0.} \]

The sketch is a straight, rising line through \((0,0)\), passing through \[ (x,U)=(1.00\,\mathrm m,10.0\,\mathrm J) \] and \[ (x,U)=(2.00\,\mathrm m,20.0\,\mathrm J). \] The wall restricts the accessible region to \(x\ge0\). Friction is dissipative and is not included in this potential.

c. Speed as a function of position

Let the particle mass be \(m\), and define its initial speed by \[ v_0=\sqrt{\frac{2E_c}{m}}. \] Since no mass is supplied, \(v_0\) provides a convenient vertical scale.

First take the initial velocity to point toward \(+x\). On the initial outward passage, work–energy gives \[ \frac12mv^2=E_c-(F+f)(x-x_0). \] The first turning point is consequently \[ a_0=x_0+\frac{E_c}{F+f} =1.00\,\mathrm m+\frac{10.0\,\mathrm J}{11.0\,\mathrm N} =\frac{21}{11}\,\mathrm m \approx1.91\,\mathrm m. \]

Using the ratio found in part a, all subsequent turning points are \[ a_n=a_0r^n,\qquad r=\frac9{11}. \]

For an outward passage ending at \(a_n\), the remaining kinetic energy equals the work needed to reach that turning point: \[ \boxed{ v_{\rightarrow,n}(x) =\sqrt{\frac{2(F+f)}m(a_n-x)}. } \] For \(n=0\), only \(x_0\le x\le a_0\) is traversed. For \(n\ge1\), the full interval \(0\le x\le a_n\) is traversed.

On the inward passage from \(a_n\), the net force toward the wall has magnitude \(F-f\), giving \[ \boxed{ v_{\leftarrow,n}(x) =\sqrt{\frac{2(F-f)}m(a_n-x)}, \qquad 0\le x\le a_n. } \] Both expressions denote speed, so both are nonnegative.

At the wall, the inward branch from \(a_n\) joins the next outward branch continuously in speed: \[ v_{\mathrm w,n} =\sqrt{\frac{2(F-f)a_n}{m}} =\sqrt{\frac{2(F+f)a_{n+1}}{m}}. \] Thus \[ \frac{v_{\mathrm w,n+1}}{v_{\mathrm w,n}} =\sqrt{\frac9{11}}. \]

The qualitative sketch can be constructed as follows:

  • Start at \((x,v)=(1.00\,\mathrm m,v_0)\). Follow a decreasing, concave-down square-root curve to \((1.91\,\mathrm m,0)\).
  • Return along the inward branch, moving left and upward on the graph, to \[ (x,v)=\left(0,\sqrt{\frac{189}{110}}\,v_0\right) \approx(0,1.31v_0). \]
  • After reflection, follow the next outward branch to \[ (x,v)=(a_1,0),\qquad a_1=\frac{189}{121}\,\mathrm m\approx1.56\,\mathrm m. \]
  • Repeat with successively smaller turning distances. Every branch has a vertical tangent at its outer turning point. For the same turning distance, the outward branch lies above the inward branch.
  • The sequence converges to the final point \((0,0)\).

There are generally several speeds at the same position, corresponding to different passages; the complete graph is therefore not a single-valued function of \(x\).

If the initial velocity instead points toward the wall, the initial branch is \[ v(x)=\sqrt{\frac2m\left[E_c+(F-f)(x_0-x)\right]}, \qquad 0\le x\le x_0. \] The first wall encounter has kinetic energy \[ E_c+(F-f)x_0=19.0\,\mathrm J, \] so its speed is \(\sqrt{1.90}\,v_0\). After this reflection, the turning points are \[ b_n=\frac{19}{11}\left(\frac9{11}\right)^n\mathrm m. \] The same outward and inward branch formulas apply with \(a_n\) replaced by \(b_n\).

The solution was checked against the official solution.