Problems

Two particles move on a smooth horizontal table and collide. The masses of the particles are \(m\) and \(M\). Their velocities before the collision are \(u{\bf i}\) and \(v{\bf i}\,\), respectively, where \(\bf i\) is a unit vector and \(u>v\). Their velocities after the collision are \(p{\bf i}\) and \(q{\bf i}\,\), respectively. The coefficient of restitution between the two particles is \(e\), where \(e<1\).

1. Show that the loss of kinetic energy due to the collision is \[ \tfrac12 m (u-p)(u-v)(1-e)\,, \] and deduce that \(u\ge p\).
2. Given that each particle loses the same (non-zero) amount of kinetic energy in the collision, show that \[ u+v+p+q=0\,, \] and that, if \(m\ne M\), \[ e= \frac{(M+3m)u + (3M+m)v}{(M-m)(u-v)}\,. \]

\(\,\)

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.

Show Solution

---

Solution:

+ - ↺

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 solid right circular cone, of mass \(M\), has semi-vertical angle \(\alpha\) and smooth surfaces. It stands with its base on a smooth horizontal table. A particle of mass \(m\) is projected so that it strikes the curved surface of the cone at speed \(u\). The coefficient of restitution between the particle and the cone is \(e\). The impact has no rotational effect on the cone and the cone has no vertical velocity after the impact.

1. The particle strikes the cone in the direction of the normal at the point of impact. Explain why the trajectory of the particle immediately after the impact is parallel to the normal to the surface of the cone. Find an expression, in terms of \(M\), \(m\), \(\alpha\), \(e\) and \(u\), for the speed at which the cone slides along the table immediately after impact.
2. If instead the particle falls vertically onto the cone, show that the speed \(w\) at which the cone slides along the table immediately after impact is given by \[ w= \frac{mu(1+e)\sin\alpha\cos\alpha}{M+m\cos^2\alpha}\,. \] Show also that the value of \(\alpha\) for which \(w\) is greatest is given by \[ \cos \alpha = \sqrt{ \frac{M}{2M+m}}\ . \]

Three particles, \(A\), \(B\) and \(C\), of masses \(m\), \(km\) and \(3m\) respectively, are initially at rest lying in a straight line on a smooth horizontal surface. Then \(A\) is projected towards \(B\) at speed \(u\). After the collision, \(B\) collides with \(C\). The coefficient of restitution between \(A\) and \(B\) is \(\frac12\) and the coefficient of restitution between \(B\) and \(C\) is \(\frac14\).

1. Find the range of values of \(k\) for which \(A\) and \(B\) collide for a second time.
2. Given that \(k=1\) and that \(B\) and \(C\) are initially a distance \(d\) apart, show that the time that elapses between the two collisions of \(A\) and \(B\) is \(\dfrac{60d}{13u}\,\).

A lift of mass \(M\) and its counterweight of mass \(M\) are connected by a light inextensible cable which passes over a fixed frictionless pulley. The lift is constrained to move vertically between smooth guides. The distance between the floor and the ceiling of the lift is \(h\). Initially, the lift is at rest, and the distance between the top of the lift and the pulley is greater than \(h\). A small tile of mass \(m\) becomes detached from the ceiling of the lift and falls to the floor of the lift. Show that the speed of the tile just before the impact is \[ \sqrt{\frac{(2M-m)gh \;}{M}}\;. \] The coefficient of restitution between the tile and the floor of the lift is \(e\). Given that the magnitude of the impulsive force on the lift due to tension in the cable is equal to the magnitude of the impulsive force on the counterweight due to tension in the cable, show that the loss of energy of the system due to the impact is \(mgh(1-e^2)\). Comment on this result.

Three collinear, non-touching particles \(A\), \(B\) and \(C\) have masses \(a\), \(b\) and \(c\), respectively, and are at rest on a smooth horizontal surface. The particle \(A\) is given an initial velocity \(u\) towards~\(B\). These particles collide, giving \(B\) a velocity \(v\) towards \(C\). These two particles then collide, giving \(C\) a velocity \(w\). The coefficient of restitution is \(e\) in both collisions. Determine an expression for \(v\), and show that \[ \displaystyle w = \frac {abu \l 1+e \r^2}{\l a + b \r \l b+c \r}\;. \] Determine the final velocities of each of the three particles in the cases:

1. \(\displaystyle \frac ab = \frac bc = e\,\);
2. \(\displaystyle \frac ba = \frac cb = e\,\).

Two particles, A and B, move without friction along a horizontal line which is perpendicular to a vertical wall. The coefficient of restitution between the two particles is \(e\) and the coefficient of restitution between particle B and the wall is also \(e\), where \( 0< e < 1\). The mass of particle~A is \(4em\) (with \(m > 0\)), and the mass of particle B is \((1-e)^2m\)\,. Initially, A is moving towards the wall with speed \((1-e)v\) (where \(v > 0\)) and B is moving away from the wall and towards A with speed \(2ev\). The two particles collide at a distance \(d\) from the wall. Find the speeds of A and B after the collision. When B strikes the wall, it rebounds along the same line. Show that a second collision will take place, at a distance \(de\) from the wall. Deduce that further collisions will take place. Find the distance from the wall at which the \(n\)th collision takes place, and show that the times between successive collisions are equal.

Two thin discs, each of radius \(r\) and mass \(m\), are held on a rough horizontal surface with their centres a distance \(6r\) apart. A thin light elastic band, of natural length \(2\pi r\) and modulus \(\dfrac{\pi mg}{12}\), is wrapped once round the discs, its straight sections being parallel. The contact between the elastic band and the discs is smooth. The coefficient of static friction between each disc and the horizontal surface is \(\mu\), and each disc experiences a force due to friction equal to \(\mu mg\) when it is sliding. The discs are released simultaneously. If the discs collide, they rebound and a half of their total kinetic energy is lost in the collision.

1. Show that the discs start sliding, but come to rest before colliding, if and only if \mbox{\(\frac23 <\mu <1\)}.
2. Show that, if the discs collide at least once, their total kinetic energy just before the first collision is \(\frac43 mgr(2-3\mu)\).
3. Show that if \(\frac 4 9 > \mu^2 >\frac{5}{27}\) the discs come to rest exactly once after the first collision.

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 \(\frac14 {m u^2 \big( 1 - e^2 \big) }\,\).

Show Solution

---

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*}

+ - ↺

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 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.

Two particles \(A\) and \(B\) of masses \(m\) and \(km\), respectively, are at rest on a smooth horizontal surface. The direction of the line passing through \(A\) and \(B\) is perpendicular to a vertical wall which is on the other side of \(B\) from \(A\). The particle \(A\) is now set in motion towards \(B\) with speed \(u\). The coefficient of restitution between \(A\) and \(B\) is \(e_1\) and between \(B\) and the wall is \(e_2\). Show that there will be a second collision between \(A\) and \(B\) provided $$ k< \frac {1+e_2(1+e_1)} {e_1}\;. $$ Show that, if \(e_1=\frac13\), \(e_2=\frac12\) and \(k<5\), then the kinetic energy of \(A\) and \(B\) immediately after \(B\) rebounds from the wall is greater than \(mu^2/27\).

Show Solution

---

Solution: First collision:

+ - ↺

Since the \(e = e_1\), the speed of approach is \(u\) the speed of separation will be \(e_1u\) and so \(v_B = v_A + e_1u\). \begin{align*} \text{COM}: && mu &= mv_A + km(v_A + e_1u) \\ \Rightarrow && v_A(1+k) &= u(1-ke_1) \\ \Rightarrow && v_A &= \frac{1-ke_1}{1+k} u \\ && v_B &= \frac{1-ke_1}{1+k} u + e_1 u \\ &&&= \frac{1-ke_1 + e_1+ke_1}{1+k}u \\ &&&= \frac{1+e_1}{1+k}u \end{align*} Once the ball rebounds from the wall it will have velocity (still taking towards the wall as +ve) of \(-\frac{1+e_1}{1+k}e_2u\). There will be another collision if it is travelling faster than \(A\), ie if: \begin{align*} -\frac{1+e_1}{1+k}e_2u &< \frac{1-ke_1}{1+k} u \\ \Leftrightarrow && 0 &< (1-ke_1) + (1+e_1)e_2 \\ \Leftrightarrow && ke_1 &< 1 +e_2 (1+e_1) \\ \Leftrightarrow && k &< \frac{1 +e_2 (1+e_1)}{e_1} \\ \end{align*} If \(e_1 = \frac13, e_2 = \frac12\), then \(v_A = \frac{1-\frac13k}{1+k}u = \frac{3-k}{3(1+k)}u\) and \(v_B = \frac{4}{3(1+k)}u\). Therefore \begin{align*} && \text{total k.e.} &= \underbrace{\frac12 m v_A^2}_{\text{k.e. of }A} + \underbrace{\frac12 (km) (e_2 v_B)^2}_{\text{k.e. of }B} \\ &&&= \frac12 m \frac{(3-k)^2}{9(1+k)^2}u^2 + \frac12 km \frac14 \frac{16}{9(1+k)^2}u^2 \\ &&&= \frac12mu^2 \frac{1}{9(1+k)^2}\left ( (3-k)^2+4k \right) \\ &&&= \frac12mu^2 \frac{1}{9(1+k)^2}\left ( 9-2k+k^2 \right) \\ &&&= \frac{mu^2}{18} \frac{9-2k+k^2}{1+2k+k^2} \end{align*} We wish to minimize this as a function of \(k\). \begin{align*} \frac{\d}{\d k} \left ( \frac{9-2k+k^2}{1+2k+k^2}\right) &= \frac{(1+k)^2(2k-2)-2(1+k)(k^2-2k+9)}{(1+k)^4} \\ &= \frac{2(k^2-1) - 2(k^2-2k+9)}{(1+k)^3} \\ &= \frac{2(2k-10)}{(1+k)^3} \end{align*} Therefore the minimum will be when \(k = 5\) can't be a maximum by considering \(k \to 0\). This value is \(\frac{2}{3}\) and therefore \(\frac{mu^2}{18} \frac{2}{3} = \frac{mu^2}{27}\) is the smallest energy (which isn't quite achievable since \(k < 5\).

Three particles \(P_1\), \(P_2\) and \(P_3\) of masses \(m_{1}\), \(m_{2}\) and \(m_{3}\) respectively lie at rest in a straight line on a smooth horizontal table. \(P_1\) is projected with speed \(v\) towards \(P_2\) and brought to rest by the collision. After \(P_2\) collides with \(P_3\), the latter moves forward with speed \(v\). The coefficients of restitution in the first and second collisions are \(e\) and \(e'\), respectively. Show that \[ e'= \frac{m_{2}+m_{3}-m_{1}}{m_{1}}. \] Show that \(2m_1\ge m_2 +m_3\ge m_1\) for such collisions to be possible. If \(m_1\), \(m_3\) and \(v\) are fixed, find, in terms of \(m_1\), \(m_3\) and \(v\), the largest and smallest possible values for the final energy of the system.

\(N\) particles \(P_1\), \(P_2\), \(P_3\), \(\ldots\), \(P_N\) with masses \(m\), \(qm\), \(q^2m\), \(\ldots\) , \({q^{N-1}}m\), respectively, are at rest at distinct points along a straight line in gravity-free space. The particle \(P_1\) is set in motion towards \(P_2\) with velocity \(V\) and in every subsequent impact the coefficient of restitution is \(e\), where \(0 < e < 1\). Show that after the first impact the velocities of \(P_1\) and \(P_2\) are $$ {\left({{1-eq}\over{1+q}}\right)}V \mbox{ \ \ \ and \ \ \ } {\left({{1+e}\over{1+q}}\right)}V, $$ respectively. Show that if \(q \le e\), then there are exactly \(N-1\) impacts and that if \(q=e\), then the total loss of kinetic energy after all impacts have occurred is equal to $$ {1\over 2}{me}{\left(1-e^{N-1}\right)}{V^2}. $$

Two small spheres \(A\) and \(B\) of equal mass \(m\) are suspended in contact by two light inextensible strings of equal length so that the strings are vertical and the line of centres is horizontal. The coefficient of restitution between the spheres is \(e\). The sphere \(A\) is drawn aside through a very small distance in the plane of the strings and allowed to fall back and collide with the other sphere \(B\), its speed on impact being \(u\). Explain briefly why the succeeding collisions will all occur at the lowest point. (Hint: Consider the periods of the two pendulums involved.) Show that the speed of sphere \(A\) immediately after the second impact is \(\frac{1}{2}u(1+e^{2})\) and find the speed, then, of sphere \(B\).