IPhO 1994 · Theory problem 3

Collision of Discs with Surface Friction

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A homogeneous disc A of mass \(m\) and radius \(R_A\) moves translationally on a smooth horizontal \(x\)-\(y\) plane in the \(x\) direction with a velocity \(V\) (see the figure on the next page). The center of the disk is at a distance \(b\) from the x-axis. It collides with a stationary homogeneous disc B whose center is initially located at the origin of the coordinate system. The disc B has the same mass and the same thickness as A, but its radius is \(R_B\). It is assumed that the velocities of the discs at their point of contact, in the direction perpendicular to the line joining their centers, are equal after the collision. It is also assumed that the magnitudes of the relative velocities of the discs along the line joining their centers are the same before and after the collision.

1) For such a collision determine the X and Y components of the velocities of the two discs after the collision, i. e. \(V'_{AX}\), \(V'_{AY}\), \(V'_{BX}\) and \(V'_{BY}\) in terms of \(m\), \(R_A\), \(R_B\), \(V\) and \(b\).

2) Determine the kinetic energies \(E'_A\) for disc A and \(E'_B\) for disc B after the collision in terms of \(m\), \(R_A\), \(R_B\), \(V\) and \(b\).

Figure 1

Official English text of the problem.

Answers

Let \(D=R_A+R_B\). For an impact, \(0\le b<D\), with A above the x-axis as shown.

  1. The velocity components are \[ \boxed{V'_{AX}=\frac{5Vb^2}{6D^2}},\qquad \boxed{V'_{AY}=\frac{5Vb\sqrt{D^2-b^2}}{6D^2}}, \] \[ \boxed{V'_{BX}=V\left(1-\frac{5b^2}{6D^2}\right)},\qquad \boxed{V'_{BY}=-\frac{5Vb\sqrt{D^2-b^2}}{6D^2}}. \]

  2. Including both translation and rotation, \[ \boxed{E'_A=\frac{3mV^2b^2}{8D^2}}, \qquad \boxed{E'_B=\frac{mV^2}{2} \left(1-\frac{11b^2}{12D^2}\right)}. \]

Solution

1) Velocities after the collision

Treat the collision as instantaneous, so the line joining the centers remains fixed during the impact. The smooth horizontal plane supplies no horizontal impulse or torque about the vertical axis. Initially neither disc rotates.

Define \[ s=\frac{b}{D},\qquad c=\sqrt{1-s^2},\qquad D=R_A+R_B. \] At first contact, the center of A is at \((-Dc,b)\), while the center of B is at the origin. Thus the unit vectors along the line from A to B and perpendicular to it are \[ \mathbf n=(c,-s),\qquad \mathbf t=(s,c). \] The initial velocity of A is \[ \mathbf v_A=Vc\,\mathbf n+Vs\,\mathbf t. \]

Let the impulse exerted on B be \[ \mathbf J=J_n\mathbf n+J_t\mathbf t. \] The impulse on A is \(-\mathbf J\). Linear impulse–momentum therefore gives \[ v'_{An}=Vc-\frac{J_n}{m},\qquad v'_{Bn}=\frac{J_n}{m}, \] \[ v'_{At}=Vs-\frac{J_t}{m},\qquad v'_{Bt}=\frac{J_t}{m}. \]

The normal relative velocity reverses direction while retaining its magnitude: the discs approach before impact and separate afterward. Consequently, \[ v'_{Bn}-v'_{An}=Vc, \] so \[ \frac{2J_n}{m}-Vc=Vc \quad\Longrightarrow\quad J_n=mVc. \] Hence \[ \boxed{v'_{An}=0,\qquad v'_{Bn}=Vc}. \]

To determine the tangential impulse, we must include rotation. The moments of inertia are \[ I_A=\frac12mR_A^2,\qquad I_B=\frac12mR_B^2. \] Take counterclockwise angular velocity as positive. The contact points lie at \(R_A\mathbf n\) relative to A's center and \(-R_B\mathbf n\) relative to B's center. Angular impulse–momentum gives \[ I_A\omega'_A=-R_AJ_t,\qquad I_B\omega'_B=-R_BJ_t. \] Therefore, \[ \omega'_A=-\frac{2J_t}{mR_A},\qquad \omega'_B=-\frac{2J_t}{mR_B}. \]

The tangential velocities of the material points in contact are \[ u'_{At}=v'_{At}+R_A\omega'_A =Vs-\frac{3J_t}{m}, \] \[ u'_{Bt}=v'_{Bt}-R_B\omega'_B =\frac{3J_t}{m}. \] The stated equality of these velocities requires \[ Vs-\frac{3J_t}{m}=\frac{3J_t}{m}, \] which yields \[ \boxed{J_t=\frac{mVs}{6}}. \] Thus \[ \boxed{v'_{At}=\frac56Vs,\qquad v'_{Bt}=\frac16Vs}, \] and the angular velocities, needed for part 2, are \[ \boxed{\omega'_A=-\frac{Vs}{3R_A},\qquad \omega'_B=-\frac{Vs}{3R_B}}. \] Both discs rotate clockwise for the geometry shown.

Returning to Cartesian coordinates, \[ \mathbf v'_A=\frac56Vs\,\mathbf t =\left(\frac56Vs^2,\frac56Vsc\right), \] \[ \mathbf v'_B=Vc\,\mathbf n+\frac16Vs\,\mathbf t =\left(V\left(c^2+\frac{s^2}{6}\right),-\frac56Vsc\right). \] Using \(c^2=1-s^2\), \(s=b/D\), and \(c=\sqrt{D^2-b^2}/D\) gives the four components in the Answers section. Their sums satisfy conservation of linear momentum: \[ V'_{AX}+V'_{BX}=V,\qquad V'_{AY}+V'_{BY}=0. \]

2) Kinetic energies after the collision

Each disc's kinetic energy includes translation of its center and rotation about its center: \[ E'_i=\frac12m|\mathbf v'_i|^2+\frac12I_i(\omega'_i)^2. \]

For A, \[ \begin{aligned} E'_A &=\frac12m\left(\frac56Vs\right)^2 +\frac12\left(\frac12mR_A^2\right) \left(\frac{Vs}{3R_A}\right)^2\\ &=\left(\frac{25}{72}+\frac{1}{36}\right)mV^2s^2\\ &=\frac38mV^2s^2. \end{aligned} \] Therefore, \[ \boxed{E'_A=\frac{3mV^2b^2}{8(R_A+R_B)^2}}. \]

For B, the normal and tangential velocity components are perpendicular, so \[ |\mathbf v'_B|^2=V^2c^2+\frac{V^2s^2}{36}. \] It follows that \[ \begin{aligned} E'_B &=\frac12mV^2\left(c^2+\frac{s^2}{36}\right) +\frac12\left(\frac12mR_B^2\right) \left(\frac{Vs}{3R_B}\right)^2\\ &=\frac12mV^2c^2+\frac1{24}mV^2s^2\\ &=\frac12mV^2\left(1-\frac{11s^2}{12}\right). \end{aligned} \] Thus \[ \boxed{E'_B=\frac{mV^2}{2} \left(1-\frac{11b^2}{12(R_A+R_B)^2}\right)}. \]

As a consistency check, the total kinetic-energy loss is \[ \frac12mV^2-(E'_A+E'_B) =\boxed{\frac{mV^2b^2}{12(R_A+R_B)^2}}\ge0. \] This loss results from tangential friction; the prescribed elastic condition applies only to normal relative motion. For a central collision, \(b=0\), A stops, B acquires velocity \(V\), neither disc rotates, and no kinetic energy is lost.

The solution was checked against the official solution. Where this solution differs from the official solution, the official solution contains an error or a misprint.