Problems

A small ball of mass \(m\) is suspended in equilibrium by a light elastic string of natural length \(l\) and modulus of elasticity \(\lambda.\) Show that the total length of the string in equilibrium is \(l(1+mg/\lambda).\) If the ball is now projected downwards from the equilibrium position with speed \(u_{0},\) show that the speed \(v\) of the ball at distance \(x\) below the equilibrium position is given by \[ v^{2}+\frac{\lambda}{lm}x^{2}=u_{0}^{2}. \] At distance \(h\), where \(\lambda h^{2} < lmu_{0}^{2},\) below the equilibrium position is a horizontal surface on which the ball bounces with a coefficient of restitution \(e\). Show that after one bounce the velocity \(u_{1}\) at \(x=0\) is given by \[ u_{1}^{2}=e^{2}u_{0}^{2}+\frac{\lambda}{lm}h^{2}(1-e^{2}), \] and that after the second bounce the velocity \(u_{2}\) at \(x=0\) is given by \[ u_{2}^{2}=e^{4}u_{0}^{2}+\frac{\lambda}{lm}h^{2}(1-e^{4}). \]

Three small spheres of masses \(m_{1},m_{2}\) and \(m_{3},\) move in a straight line on a smooth horizontal table. (Their order on the straight line is the order given.) The coefficient of restitution between any two spheres is \(e\). The first moves with velocity \(u\) towards the second whilst the second and third are at rest. After the first collision the second sphere hits the third after which the velocity of the second sphere is \(u.\) Find \(m_{1}\) in terms of \(m_{2},m_{3}\) and \(e\). deduce that \[ m_{2}e>m_{3}(1+e+e^{2}). \] Suppose that the relation between \(m_{1},m_{2}\) and \(m_{3}\) is that in the formula you found above, but that now the first sphere initially moves with velocity \(u\) and the other two spheres with velocity \(v\), all in the same direction along the line. If \(u>v>0\) use the first part to find the velocity of the second sphere after two collisions have taken place. (You should not need to make any substantial computations but you should state your argument clearly.)

A smooth, axially symmetric bowl has its vertical cross-sections determined by \(s=2\sqrt{ky},\) where \(s\) is the arc-length measured from its lowest point \(V\), and \(y\) is the height above \(V\). A particle is released from rest at a point on the surface at a height \(h\) above \(V\). Explain why \[ \left(\frac{\mathrm{d}s}{\mathrm{d}t}\right)^{2}+2gy \] is constant. Show that the time for the particle to reach \(V\) is \[ \pi\sqrt{\frac{k}{2g}}. \] Two elastic particles of mass \(m\) and \(\alpha m,\) where \(\alpha<1,\) are released simultaneously from opposite sides of the bowl at heights \(\alpha^{2}h\) and \(h\) respectively. If the coefficient of restitution between the particles is \(\alpha,\) describe the subsequent motion.

A ball of mass \(m\) is thrown vertically upwards from the floor of a room of height \(h\) with speed \(\sqrt{2kgh},\) where \(k>1.\) The coefficient of restitution between the ball and the ceiling or floor is \(a\). Both the ceiling and floor are level. Show that the kinetic energy of the ball immediately before hitting the ceiling for the \(n\)th time is \[ mgh\left(a^{4n-4}(k-1)+\frac{a^{4n-4}-1}{a^{2}+1}\right). \] Hence show that the number of times the ball hits the ceiling is at most \[ 1-\frac{\ln[a^{2}(k-1)+k]}{4\ln a}. \]

A particle rests at a point \(A\) on a horizontal table and is joined to a point \(O\) on the table by a taut inextensible string of length \(c\). The particle is projected vertically upwards at a speed \(64\surd(6gc)\). It next strikes the table at a point \(B\) and rebounds. The coefficient of restitution for any impact between the particle and the table is \({1\over 2}\). After rebounding at \(B\), the particle will rebound alternately at \(A\) and \(B\) until the string becomes slack. Show that when the string becomes slack the particle is at height \(c/2\) above the table. Determine whether the first rebound between \(A\) and \(B\) is nearer to \(A\) or to \(B\).

Two particles \(P_{1}\) and \(P_{2}\), each of mass \(m\), are joined by a light smooth inextensible string of length \(\ell.\) \(P_{1}\) lies on a table top a distance \(d\) from the edge, and \(P_{2}\) hangs over the edge of the table and is suspended a distance \(b\) above the ground. The coefficient of friction between \(P_{1}\) and the table top is \(\mu,\) and \(\mu<1\). The system is released from rest. Show that \(P_{1}\) will fall off the edge of the table if and only if \[ \mu<\frac{b}{2d-b}. \] Suppose that \(\mu>b/(2d-b)\) , so that \(P_{1}\) comes to rest on the table, and that the coefficient of restitution between \(P_{2}\) and the floor is \(e\). Show that, if \(e>1/(2\mu),\) then \(P_{1}\) comes to rest before \(P_{2}\) bounces a second time.

\(\,\)

+ - ↺

A straight staircase consists of \(N\) smooth horizontal stairs each of height \(h\). A particle slides over the top stair at speed \(U\), with velocity perpendicular to the edge of the stair, and then falls down the staircase, bouncing once on every stair. The coefficient of restitution between the particle and each stair is \(e\), where \(e<1\). Show that the horizontal distance \(d_{n}\) travelled between the \(n\)th and \((n+1)\)th bounces is given by \[ d_{n}=U\left(\frac{2h}{g}\right)^{\frac{1}{2}}\left(e\alpha_{n}+\alpha_{n+1}\right), \] where \({\displaystyle \alpha_{n}=\left(\frac{1-e^{2n}}{1-e^{2}}\right)^{\frac{1}{2}}}\). If \(N\) is very large, show that \(U\) must satisfy \[ U=\left(\frac{L^{2}g}{2h}\right)^{\frac{1}{2}}\left(\frac{1-e}{1+e}\right)^{\frac{1}{2}}, \] where \(L\) is the horizontal distance between the edges of successive stairs.

A uniform smooth wedge of mass \(m\) has congruent triangular end faces \(A_{1}B_{1}C_{1}\) and \(A_{2}B_{2}C_{2},\) and \(A_{1}A_{2},B_{1}B_{2}\) and \(C_{1}C_{2}\) are perpendicular to these faces. The points \(A,B\) and \(C\) are the midpoints of \(A_{1}A_{2},B_{1}B_{2}\) and \(C_{1}C_{2}\) respectively. The sides of the triangle \(ABC\) have lengths \(AB=AC=5a\) and \(BC=6a.\) The wedge is placed with \(BC\) on a smooth horizontal table, a particle of mass \(2m\) is placed at \(A\) on \(AC,\) and the system is released from rest. The particle slides down \(AC,\) strikes the table, bounces perfectly elastically and lands again on the table at \(D\). At this time the point \(C\) of the wedge has reached the point \(E\). Show that \(DE=\frac{192}{19}a.\)

Show Solution

---

Solution: Conservation of energy, tells us that \(2m \cdot g \cdot 4a = 8amg\) is equal to \(\frac12 m v_{wedge}^2 + \frac12(2m)v_{particle}^2\). Conservation of momentum (horizontally) tells us that \(m v_{wedge}+2mv_{particle, \rightarrow} = 0 \Rightarrow v_{particle, \rightarrow} = -\frac12 v_{wedge}\).

+ - ↺

We know that the particle must remain on the slope, and so \(v_{particle,\downarrow} = \frac{4}{3} \frac{3}{2} v_{wedge} = 2v_{wedge}\). In conclusion, we have: \begin{align*} && 8amg &= \frac12 m v_{wedge}^2 + \frac12 (2m)\left ((-\tfrac12 v_{wedge})^2 + (2v_{wedge})^2 \right ) \\ &&&= \frac{19}{4}mv_{wedge}^2 \\ \Rightarrow && v_{wedge}^2 &= \frac{32}{19}ag \end{align*}. To calculate the time the ball bounces for, note that: \(s = ut + \frac12 at^2 \Rightarrow 0 = 2v_{wedge} - \frac12 gt \Rightarrow t = \frac{4v_{wedge}}{g}\). During this time, the wedge (and ball) who horizontally are moving apart with speed \(\frac32 v_{wedge}\) we have they move apart by: \begin{align*} && DE &= \underbrace{\frac32 v_{wedge}}_{\text{speed they move apart}} \cdot \underbrace{\frac{4v_{wedge}}{g}}_{\text{time they are moving apart for}} \\ &&&= \frac{6}{g} v_{wedge}^2 \\ &&&= \frac{6}{g}\frac{32}{19}ag \\ &&&= \frac{192}{19}a \end{align*}

A smooth uniform sphere, with centre \(A\), radius \(2a\) and mass \(3m,\) is suspended from a fixed point \(O\) by means of a light inextensible string, of length \(3a,\) attached to its surface at \(C\). A second smooth unifom sphere, with centre \(B,\) radius \(3a\) and mass \(25m,\) is held with its surface touching \(O\) and with \(OB\) horizontal. The second sphere is released from rest, falls and strikes the first sphere. The coefficient of restitution between the spheres is \(3/4.\) Find the speed \(U\) of \(A\) immediately after the impact in terms of the speed \(V\) of \(B\) immediately before impact. The same system is now set up with a light rigid rod replacing the string and rigidly attached to the sphere so that \(OCA\) is a straight line. The rod can turn freely about \(O\). The sphere with centre \(B\) is dropped as before. Show that the speed of \(A\) immediately after impact is \(125U/127.\)

A smooth billiard ball moving on a smooth horizontal table strikes another identical ball which is at rest. The coefficient of restitution between the balls is \(e(<1)\). Show that after the collision the angle between the velocities of the balls is less than \(\frac{1}{2}\pi.\) Show also that the maximum angle of deflection of the first ball is \[ \sin^{-1}\left(\frac{1+e}{3-e}\right). \]

Show Solution

---

Solution:

+ - ↺

Set up the coordinate frame so that the \(x\)-direction is the line of centres of the spheres. Then if the initial velocities are \(\displaystyle \binom{u_x}{u_y}\) and \(\displaystyle \binom{0}{0}\). Then the final velocities must be: \(\displaystyle \binom{v_{x1}}{u_y}\) and \(\displaystyle \binom{v_{x2}}{0}\) where \(mu_x = mv_{x1}+mv_{x2}\) by conservation of energy and \(\frac{v_{x1}-v_{x2}}{u_x} = -e\). \begin{align*} && \begin{cases} v_{x1}+v_{x2} &= u_x \\ v_{x1}-v_{x2} &= -eu_x \\ \end{cases} \\ \Rightarrow && 2v_{x1} &= (1-e)u_x \\ \Rightarrow && v_{x1} &= \frac{(1-e)}{2} u_x \\ && v_{x2} &= \frac{1+e}{2} u_x \end{align*} Notice that since \(0 < e < 1\) we must have \(v_{x1} > 0\) and so the ball on the left is still continuing in the positive direction, therefore the angle will be less than \(\frac12 \pi\). The angle the first ball is deflected through is the angle between: \(\displaystyle \binom{u_x}{u_y}\) and \(\displaystyle \binom{\frac{1-e}{2}u_x}{u_y}\). We can scale the velocities so \(u_y = 1\). So we are interested in the angle between \(\displaystyle \binom{x}{1}\) and \(\displaystyle \binom{\frac{1-e}{2}x}{1}\). To maximise \(\theta\) we can maximise \(\tan \theta\), so: \begin{align*} && \tan \theta &= \frac{\frac{2}{(1-e)x-\frac{1}{x}}}{1+\frac{2}{(1-e)x^2}} \\ &&&= \frac{2x-(1-e)x}{(1-e)x^2+2} \\ &&&= \frac{(1+e)x}{(1-e)x^2+2} \\ \\ \frac{\d}{\d t}: &&&= \frac{(1+e)((1-e)x^2+2)-2(1+e)(1-e)x^2}{\sim} \\ &&&= \frac{2(1+e)-(1+e)(1-e)x^2}{\sim}\\ \frac{\d}{\d t} = 0: &&0 &= 2(1+e)-(1+e)(1-e)x^2 \\ \Rightarrow && x &= \pm \sqrt{\frac{2}{1-e}} \\ \\ \Rightarrow && \tan \theta &= \frac{\pm(1+e)\sqrt{\frac{2}{1-e}}}{2+2} \\ &&&= \pm \frac{\sqrt{2}(1+e)}{4\sqrt{1-e}} \\ \Rightarrow && \cot^2 \theta &= \frac{8(1-e)}{(1+e)^2} \\ \Rightarrow && \cosec^2 \theta &= \frac{8(1-e)}{(1+e)^2} + 1 \\ &&&= \frac{8-8e+1+2e+e^2}{(1+e)^2} \\ &&&= \frac{9-6e+e^2}{(1+e)^2} \\ &&&= \frac{(3-e)^2}{(1+e)^2} \\ \Rightarrow && \theta &= \sin^{-1} \left ( \frac{1+e}{3-e}\right) \end{align*}

\(\,\) \vspace{-1cm}

+ - ↺

Ice snooker is played on a rectangular horizontal table, of length \(L\) and width \(B\), on which a small disc (the puck) slides without friction. The table is bounded by smooth vertical walls (the cushions) and the coefficient of restitution between the puck and any cushion is \(e\). If the puck is hit so that it bounces off two adjacent cushions, show that its final path (after two bounces) is parallel to its original path. The puck rests against the cushion at a point which divides the side of length \(L\) in the ratio \(z:1\). Show that it is possible, whatever \(z\), to hit the puck so that it bounces off the three other cushions in succession clockwise and returns to the spot at which it started. By considering these paths as \(z\) varies, explain briefly why there are two different ways in which, starting at any point away from the cushions, it is possible to perform a shot in which the puck bounces off all four cushions in succession clockwise and returns to its starting point.

Show Solution

---

Solution:

+ - ↺

The puck sets off at some velocity \(\displaystyle \binom{u_x}{u_y}\), after the first bounce off the wall parallel to the \(y\)-axis, it has velocity \(\displaystyle \binom{-eu_x}{u_y}\). After it bounces off the wall parallel to the \(x\)-axis, it has velocity \(\displaystyle \binom{-eu_x}{-eu_y}\) which is clearly parallel to the original velocity.

+ - ↺

If the puck bounces off 3 walls and returns to the same point the shape formed must be a parallelogram. We need to hit the point on the opposite side which is in a ratio of \(1:z\), but this must be possible if we aim towards the side further away.

+ - ↺

For a fixed path, as \(z\) increases we generate more parallelograms which cross ours (on two of the legs) twice. As they move the full length it will cover the full leg of the parallogram. Similarly going the other way will cover the other leg of the parallelogram. Therefore from every point there are two circuits round the table