Problems

The origin \(O\) of coordinates lies on a smooth horizontal table and the \(x\)- and \(y\)-axes lie in the plane of the table. A smooth sphere \(A\) of mass \(m\) and radius \(r\) is at rest on the table with its lowest point at the origin. A second smooth sphere \(B\) has the same mass and radius and also lies on the table. Its lowest point has \(y\)-coordinate \(2r\sin\alpha\), where \(\alpha\) is an acute angle, and large positive \(x\)-coordinate. Sphere \(B\) is now projected parallel to the \(x\)-axis, with speed \(u\), so that it strikes sphere \(A\). The coefficient of restitution in this collision is \(\frac{1}{3}\).

1. Show that, after the collision, sphere \(B\) moves with velocity \[\begin{pmatrix} -\frac{1}{3}u\bigl(1 + 2\sin^2\alpha\bigr) \\ \frac{2}{3}u\sin\alpha\cos\alpha \end{pmatrix}.\]
2. Show further that the lowest point of sphere \(B\) crosses the \(y\)-axis at the point \((0, Y)\), where \(Y = 2r(\cos\alpha\tan\beta + \sin\alpha)\) and \[\tan\beta = \frac{2\sin\alpha\cos\alpha}{1 + 2\sin^2\alpha}.\]

3. Show that, if \(h > Y + 2r\sec\beta\), sphere \(B\) will not strike sphere \(C\) in its motion after the collision with sphere \(A\).
4. Show that \(Y < 2r\sec\beta\). Hence show that sphere \(B\) will not strike sphere \(C\) for any value of \(\alpha\), if \(h > \dfrac{8r}{\sqrt{3}}\).

1. Two particles \(A\) and \(B\), of masses \(m\) and \(km\) respectively, lie at rest on a smooth horizontal surface. The coefficient of restitution between the particles is \(e\), where \(0 < e < 1\). Particle \(A\) is then projected directly towards particle \(B\) with speed \(u\). Let \(v_1\) and \(v_2\) be the velocities of particles \(A\) and \(B\), respectively, after the collision, in the direction of the initial velocity of \(A\). Show that \(v_1 = \alpha u\) and \(v_2 = \beta u\), where \(\alpha = \dfrac{1 - ke}{k+1}\) and \(\beta = \dfrac{1+e}{k+1}\). Particle \(B\) strikes a vertical wall which is perpendicular to its direction of motion and a distance \(D\) from the point of collision with \(A\), and rebounds. The coefficient of restitution between particle \(B\) and the wall is also \(e\). Show that, if \(A\) and \(B\) collide for a second time at a point \(\frac{1}{2}D\) from the wall, then \[ k = \frac{1+e-e^2}{e(2e+1)}\,. \]
2. Three particles \(A\), \(B\) and \(C\), of masses \(m\), \(km\) and \(k^2m\) respectively, lie at rest on a smooth horizontal surface in a straight line, with \(B\) between \(A\) and \(C\). A vertical wall is perpendicular to this line and lies on the side of \(C\) away from \(A\) and \(B\). The distance between \(B\) and \(C\) is equal to \(d\) and the distance between \(C\) and the wall is equal to \(3d\). The coefficient of restitution between each pair of particles, and between particle \(C\) and the wall, is \(e\), where \(0 < e < 1\). Particle \(A\) is then projected directly towards particle \(B\) with speed \(u\). Show that, if all three particles collide simultaneously at a point \(\frac{3}{2}d\) from the wall, then \(e = \frac{1}{2}\).

An equilateral triangle \(ABC\) has sides of length \(a\). The points \(P\), \(Q\) and \(R\) lie on the sides \(BC\), \(CA\) and \(AB\), respectively, such that the length \(BP\) is \(x\) and \(QR\) is parallel to \(CB\). Show that \[ (\sqrt{3}\cot\phi + 1)(\sqrt{3}\cot\theta + 1)x = 4(a - x), \] where \(\theta = \angle CPQ\) and \(\phi = \angle BRP\). A horizontal triangular frame with sides of length \(a\) and vertices \(A\), \(B\) and \(C\) is fixed on a smooth horizontal table. A small ball is placed at a point \(P\) inside the frame, in contact with side \(BC\) at a distance \(x\) from \(B\). It is struck so that it moves round the triangle \(PQR\) described above, bouncing off the frame at \(Q\) and then \(R\) before returning to point \(P\). The frame is smooth and the coefficient of restitution between the ball and the frame is \(e\). Show that \[ x = \frac{ae}{1 + e}. \] Show further that if the ball continues to move round \(PQR\) after returning to \(P\), then \(e = 1\).

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Solution:

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\begin{align*} && \frac{x}{\sin \phi} &= \frac{PR}{\sin 60^{\circ}} \\ && \frac{a-x}{\sin (120^{\circ}-\theta)} &= \frac{QP}{\sin 60^{\circ}} \\ && \frac{PR}{\sin \theta} &= \frac{QP}{\sin(120^{\circ}-\phi)} \\ \\ \Rightarrow && PR &= \frac{\sqrt3}2 \frac{x}{\sin \phi} \\ && QP &= \frac{\sqrt3}2 \frac{a-x}{\sin(120^{\circ}-\theta)} \\ \Rightarrow && \frac{\sqrt3}2 \frac{x}{\sin \phi} \frac{1}{\sin \theta} &= \frac{\sqrt3}2 \frac{a-x}{\sin(120^{\circ}-\theta)} \frac{1}{\sin(120^{\circ}-\phi)} \\ \Rightarrow && a-x &= \frac{(\frac{\sqrt3}2 \cos \phi + \frac12 \sin \phi)(\frac{\sqrt3}2 \cos \theta + \frac12 \sin \theta)}{\sin \phi \sin \theta} x \\ \Rightarrow && 4(a-x) &= (\sqrt3 \cot \phi + 1)(\sqrt 3 \cot \theta + 1) x \end{align*}

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Notice that \(e = \frac{\tan 60^{\circ}}{\tan (120^{\circ}-\theta)} = \frac{\tan \phi}{\tan 60^{\circ}}\) or \begin{align*} && \tan \phi &= \sqrt 3 e \\ && \frac{\sqrt3}{e} &= \tan(120^{\circ} - \theta) \\ &&&= \frac{-\sqrt3 - \tan \theta}{1 - \sqrt3 \tan \theta} \\ &&&= \frac{\sqrt3 \cot \theta +1}{\sqrt3-\cot \theta} \\ \Rightarrow && \sqrt3(-e-1)\cot \theta &= e-3 \\ \Rightarrow && \cot\theta &= \frac{3-e}{\sqrt{3}(1+e)} \\ \\ \Rightarrow && 4(a-x) &= (\sqrt3 \cot \phi + 1)(\sqrt 3 \cot \theta + 1) x \\ &&&= \left (\sqrt3 \frac{1}{\sqrt3 e} + 1 \right) \left (\sqrt 3\frac{3-e}{\sqrt{3}(1+e)}+1 \right) x \\ &&&= \frac{1+e}{e}\frac{3-e+1+e}{1+e} x \\ \Rightarrow && (a-x) &= \frac{1}{e}x \\ \Rightarrow && a &= \frac{1+e}{e}x \\ \Rightarrow && x &= \frac{ae}{1+e} \end{align*} The ball will continue to move around \(PQR\) if \(e \tan(120^{\circ} - \phi) = \tan \theta\) ie \begin{align*} && e \frac{-\sqrt3-\tan \phi}{1-\sqrt3 \tan \phi} &= \tan \theta \\ \Rightarrow && e \frac{\sqrt3 + \sqrt3 e}{3e-1} &= \frac{\sqrt3(1+e)}{3-e} \\ \Rightarrow && \frac{e}{3e-1} &= \frac{1}{3-e} \tag{\(e \neq -1\)} \\ \Rightarrow && 3e-e^2 &= 3e-1 \\ \Rightarrow && e^2 &= 1 \\ \Rightarrow && e &= 1 \end{align*}

Two small beads, \(A\) and \(B\), of the same mass, are threaded onto a vertical wire on which they slide without friction, and which is fixed to the ground at \(P\). They are released simultaneously from rest, \(A\) from a height of \(8h\) above \(P\) and \(B\) from a height of \(17h\) above \(P\). When \(A\) reaches the ground for the first time, it is moving with speed \( V\). It then rebounds with coefficient of restitution \(\frac{1}{2}\) and subsequently collides with \(B\) at height \(H\) above \(P\). Show that \(H= \frac{15}8h\) and find, in terms of \(g\) and \(h\), the speeds \(u_A\) and \(u_B\) of the two beads just before the collision. When \(A\) reaches the ground for the second time, it is again moving with speed \( V\). Determine the coefficient of restitution between the two beads.

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Solution: \begin{align*} && v^2 &= u^2 +2as \\ \Rightarrow && V^2 &= 2 g \cdot (8h)\\ \Rightarrow && V &=4\sqrt{hg}\\ \end{align*} When the first particle collides with the ground, the second particle is at \(9h\) traveling with speed \(V\), the first particle is at \(0\) traveling (upwards) with speed \(\tfrac12 V\). For a collision we need: \begin{align*} && \underbrace{\frac12 V t- \frac12 g t^2}_{\text{position of A}} &= \underbrace{9h - Vt - \frac12 gt^2}_{\text{position of B}} \\ \Rightarrow && \frac32Vt &= 9h \\ \Rightarrow && t &= \frac{6h}{V} \\ \\ && \underbrace{\frac12 V t- \frac12 g t^2}_{\text{position of A}} &= \frac12 V \frac{6h}{V} - \frac12 g t^2 \\ &&&= 3h - \frac12 g\frac{36h^2}{16hg} \\ &&&= 3h - \frac{9}{8}h \\ &&&= \frac{15}{8}h \end{align*} Just before the collision, \(A\) will be moving with velocity (taking upwards as positive) \begin{align*} && u_A &= \frac12 V-gt \\ &&&= 2\sqrt{hg}-g \frac{6h}{V} \\ &&&= 2\sqrt{hg} - g \frac{6h}{4\sqrt{hg}} \\ &&&= 2\sqrt{hg}-\frac32\sqrt{hg} \\ &&&= \frac12 \sqrt{hg} \end{align*} Similarly, for \(B\). \begin{align*} && u_B &= -V -gt \\ &&&= -4\sqrt{hg} - \frac32\sqrt{hg} \\ &&&= -\frac{11}{2}\sqrt{hg} \end{align*} Considering \(A\), to figure out \(v_A\). \begin{align*} && v^2 &= u^2 + 2as \\ && V^2 &= v_A^2 + 2g\frac{15}{8}h \\ && 16hg &= v_A^2 + \frac{15}{4}gh \\ \Rightarrow && v_A^2 &= \frac{49}{4}gh \\ \Rightarrow && v_A &= -\frac{7}{2}\sqrt{gh} \end{align*}

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To keep things clean, lets use units of \(\sqrt{hg}\) so we don't need to focus on that for now: \begin{align*} \text{COM}: && \frac12 - \frac{11}{2} &= -\frac{7}{2}+v_B \\ \Rightarrow && v_B& =-\frac{3}{2} \\ \text{NEL}: && e &= \frac{2}{6} = \frac13 \end{align*}

Two parallel vertical barriers are fixed a distance \(d\) apart on horizontal ice. A small ice hockey puck moves on the ice backwards and forwards between the barriers, in the direction perpendicular to the barriers, colliding with each in turn. The coefficient of friction between the puck and the ice is \(\mu\) and the coefficient of restitution between the puck and each of the barriers is \(r\). The puck starts at one of the barriers, moving with speed \(v\) towards the other barrier. Show that \[ v_{i+1}^2 - r^2 v_i^2 = - 2 r^2 \mu gd\, \] where \(v_i\) is the speed of the puck just after its \(i\)th collision. The puck comes to rest against one of the barriers after traversing the gap between them \(n\) times. In the case \(r\ne1\), express \(n\) in terms of \(r\) and \(k\), where \(k= \dfrac{v^2}{2\mu g d}\,\). If \(r=\e^{-1}\) (where \(\e\) is the base of natural logarithms) show that \[ n = \tfrac12 \ln\big(1+k(\e^2-1)\big)\,. \] Give an expression for \(n\) in the case \(r=1\).

The lengths of the sides of a rectangular billiards table \(ABCD\) are given by \(AB = DC = a\) and \(AD=BC = 2b\). There are small pockets at the midpoints \(M\) and \(N\) of the sides \(AD\) and \(BC\), respectively. The sides of the table may be taken as smooth vertical walls. A small ball is projected along the table from the corner \(A\). It strikes the side \(BC\) at \(X\), then the side \(DC\) at \(Y\) and then goes directly into the pocket at \(M\). The angles \(BAX\), \(CXY\) and \(DY\!M\) are \(\alpha\), \(\beta\) and \(\gamma\) respectively. On each stage of its path, the ball moves with constant speed in a straight line, the speeds being \(u\), \(v\) and \(w\) respectively. The coefficient of restitution between the ball and the sides is \(e\), where \(e>0\).

1. Show that \(\tan\alpha \tan \beta = e\) and find \(\gamma\) in terms of \(\alpha\).
2. Show that \(\displaystyle \tan\alpha = \frac {(1+2e)b} {(1+e)a}\) and deduce that the shot is possible whatever the value of \(e\).
3. Find an expression in terms of \(e\) for the fraction of the kinetic energy of the ball that is lost during the motion.

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Solution:

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1. The initial velocity is \(u = \binom{u \cos\alpha}{u \sin \alpha}\), therefore \(v = \binom{v_x}{u \sin \alpha}\). Newton's experimental law tells us \(v_x = -e u_x = -eu \cos\alpha\), therefore \(v = \binom{-eu \cos \alpha}{u \sin \alpha} = \binom{-v \sin \beta}{v\cos \beta} \Rightarrow -\tan \beta = -e \cot \alpha \Rightarrow \tan \alpha \tan \beta = e\). There is nothing special about the result here, and so it must also be the case that \(\tan \beta \tan \gamma = e \Rightarrow \tan \gamma = \tan \alpha\)
2. \(\tan \alpha = \frac{XB}{BA}\) so the point \(X\) is at \((a, \tan \alpha a)\). The point \(Y\) satisfies \(\tan \beta = \frac{CY}{CX} = \frac{CY}{2b - \tan \alpha a}\) so the point \(Y\) is \((a-(2b - a \tan \alpha)\tan \beta,2b) = (a - 2b\tan \beta + ea, 2b) = ((1+e)a-2b\tan \beta, 2b)\). Finally, the point \(M\) is the midpoint, so \(\tan \gamma = \frac{DM}{DY}\) so \(M\) is the point \((0, 2b - ((1+e)a-2b\tan \beta)\tan \gamma) = (0, 2b - (1+e)a \tan \gamma - 2b e) = (0, (2b(1-e) - (1+e)a \tan \gamma)\), but \(M\) is the point \((0, b)\), ie \begin{align*} && b &= 2b(1-e) - (1+e)a \tan \gamma \\ \Rightarrow && b+2eb &= (1+e)a \tan \gamma \\ \Rightarrow && \tan \gamma &= \frac{(1+2e)b}{(1+e)a} \\ \Rightarrow && \tan \alpha &= \frac{(1+2e)b}{(1+e)a} \end{align*} Since \( \frac{(1+2e)b}{(1+e)a} = \frac{b}{a} + \frac{e}{1+e}b\) which is clearly an increasing function of \(e\) on \([0,1]\), so \(\tan \alpha \in \left [\frac{b}{a}, \frac{3b}{2a} \right]\) which are all all angles which place \(X\) in sensible places, therefore we can always hit the middle pocket. (Except \(e = 0\), where we would put the ball in \(N\), but we are given \(e > 0\)).
3. After the first collision the velocity is \(\binom{-eu \cos \alpha}{u \sin \alpha}\) after the second collision the velocity is \(\binom{-eu \cos \alpha}{-eu \sin \alpha}\). Initial kinetic energy is therefore \(\frac12 m u^2\) and final kinetic energy is \(\frac12 m e^2u^2\) therefore the fraction lost is \(\frac{\frac12 m u^2 - \frac12 m e^2u^2}{\frac12 m u^2} = 1-e^2\)

A smooth plane is inclined at an angle \(\alpha\) to the horizontal. \(A\) and \(B\) are two points a distance \(d\) apart on a line of greatest slope of the plane, with \(B\) higher than \(A\). A particle is projected up the plane from \(A\) towards \(B\) with initial speed \(u\), and simultaneously another particle is released from rest at \(B\,\). Show that they collide after a time \(\displaystyle {d /u}\,\). The coefficient of restitution between the two particles is \(e\) and both particles have mass \(m\,\). Show that the loss of kinetic energy in the collision is \(\frac14 {m u^2 \big( 1 - e^2 \big) }\,\).

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Solution: We can `ignore' the fact that they are both accelerating, because the acceleration is the same for both object so it will "cancel" out. Therefore the time taken is the same as if the object has to travel distance \(d\) at speed \(u\), ie \(d/u\). \begin{align*} && u_A &= u - g \frac{d}{u} \\ && u_B &= -g\frac{d}{u} \end{align*}

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The speed of approach is \(u\), therefore the speed of separation is \(eu\), in particular \(v_B = v_A + eu\) \begin{align*} \text{COM}: && m\left (u-g\frac{d}{u} \right)+m\left (-g\frac{d}{u} \right) &= mv_A + m(v_A + eu) \\ \Rightarrow && 2v_A &= u - 2g\frac{d}{u}-eu \\ \Rightarrow && v_A &= \frac12 (1-e)u - \frac{gd}{u} \\ \Rightarrow && v_B &= \frac12 (1+e)u - \frac{gd}{u} \\ \\ && \text{initial k.e.} &= \frac12 m \left (u-g\frac{d}{u} \right)^2 + \frac12 m \left (-g\frac{d}{u} \right)^2 \\ &&&= \frac12m \left (u^2 -2gd + \frac{2g^2d^2}{u^2} \right) \\ && \text{final k.e.} &= \frac12 m \left ( \frac12 (1-e)u - \frac{gd}{u}\right)^2 + \frac12 m \left ( \frac12 (1+e)u - \frac{gd}{u}\right)^2 \\ &&&= \frac12 m \left (\frac14 \left ( (1-e)^2+(1+e)^2\right)u^2 - gd \left ((1-e)+(1+e) \right) +\frac{2g^2d^2}{u^2}\right) \\ &&&= \frac12 m \left (\frac12(1+e^2)u^2-2gd+ \frac{2g^2d^2}{u^2}\right) \\ \Rightarrow && \text{loss k.e.} &= \frac12m \left ( u^2 - \frac12(1+e^2)u^2\right) \\ &&&= \frac14mu^2(1-e^2) \end{align*}

A bicycle pump consists of a cylinder and a piston. The piston is pushed in with steady speed~\(u\). A particle of air moves to and fro between the piston and the end of the cylinder, colliding perfectly elastically with the piston and the end of the cylinder, and always moving parallel with the axis of the cylinder. Initially, the particle is moving towards the piston at speed \(v\). Show that the speed, \(v_n\), of the particle just after the \(n\)th collision with the piston is given by \(v_n=v+2nu\). Let \(d_n\) be the distance between the piston and the end of the cylinder at the \(n\)th collision, and let \(t_n\) be the time between the \(n\)th and \((n+1)\)th collisions. Express \(d_n - d_{n+1}\) in terms of \(u\) and \(t_n\), and show that \[ d_{n+1} = \frac{v+(2n-1)u}{v+(2n+1)u} \, d_n \;. \] Express \(d_n\) in terms of \(d_1\), \(u\), \(v\) and \(n\). In the case \(v=u\), show that \(ut_n = \displaystyle \frac {d_1} {n(n+1)}\). %%%%%Verify that \(\sum\limits_1^\infty t_n = d/u\).

A particle moves on a smooth triangular horizontal surface \(AOB\) with angle \(AOB = 30^\circ\). The surface is bounded by two vertical walls \(OA\) and \(OB\) and the coefficient of restitution between the particle and the walls is \(e\), where \(e < 1\). The particle, which is initially at point \(P\) on the surface and moving with velocity \(u_1\), strikes the wall \(OA\) at \(M_1\), with angle \(PM_1A = \theta\), and rebounds, with velocity \(v_1\), to strike the wall \(OB\) at \(N_1\), with angle \(M_1N_1B = \theta\). Find \(e\) and \(\displaystyle {v_1 \over u_1}\) in terms of \(\theta\). The motion continues, with the particle striking side \(OA\) at \(M_2\), \(M_3\), \( \ldots \) and striking side \(OB\) at \(N_2\), \(N_3\), \(\ldots \). Show that, if \(\theta < 60^\circ\,\), the particle reaches \(O\) in a finite time.

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 \(B\).

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A smooth particle \(P_{1}\) is projected from a point \(O\) on the horizontal floor of a room with has a horizontal ceiling at a height \(h\) above the floor. The speed of projection is \(\sqrt{8gh}\) and the direction of projection makes an acute angle \(\alpha\) with the horizontal. The particle strikes the ceiling and rebounds, the impact being perfectly elastic. Show that for this to happen \(\alpha\) must be at least \(\frac{1}{6}\pi\) and that the range on the floor is then \[ 8h\cos\alpha\left(2\sin\alpha-\sqrt{4\sin^{2}\alpha-1}\right). \] Another particle \(P_{2}\) is projected from \(O\) with the same velocity as \(P_{1}\) but its impact with the ceiling is perfectly inelastic. Find the difference \(D\) between the ranges of \(P_{1}\) and \(P_{2}\) on the floor and show that, as \(\alpha\) varies, \(D\) has a maximum value when \(\alpha=\frac{1}{4}\pi.\)

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 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 \(125U/127.\)