Problems

Points \(A\) and \(B\) are at the same height and a distance \(\sqrt{2}r\) apart. Two small, spherical particles of equal mass, \(P\) and \(Q\), are suspended from \(A\) and \(B\), respectively, by light inextensible strings of length \(r\). Each particle individually may move freely around and inside a circle centred at the point of suspension. The particles are projected simultaneously from points which are a distance \(r\) vertically below their points of suspension, directly towards each other and each with speed \(u\). When the particles collide, the coefficient of restitution in the collision is \(e\).

1. Show that, immediately after the collision, the horizontal component of each particle's velocity has magnitude \(\frac{1}{2}ev\sqrt{2}\), where \(v^2 = u^2 - gr(2 - \sqrt{2})\) and write down the vertical component in terms of \(v\).
2. Show that the strings will become taut again at a time \(t\) after the collision, where \(t\) is a non-zero root of the equation \[(r - evt)^2 + \left(-r + vt - \frac{1}{2}\sqrt{2}gt^2\right)^2 = 2r^2.\]
3. Show that, in terms of the dimensionless variables \[z = \frac{vt}{r} \quad \text{and} \quad c = \frac{\sqrt{2}v^2}{rg}\] this equation becomes \[\left(\frac{z}{c}\right)^3 - 2\left(\frac{z}{c}\right)^2 + \left(\frac{2}{c} + 1 + e^2\right)\left(\frac{z}{c}\right) - \frac{2}{c}(1 + e) = 0.\]
4. Show that, if this equation has three equal non-zero roots, \(e = \frac{1}{3}\) and \(v^2 = \frac{9}{2}\sqrt{2}rg\). Explain briefly why, in this case, no energy is lost when the string becomes taut.
5. In the case described in (iv), the particles have speed \(U\) when they again reach the points of their motion vertically below their points of suspension. Find \(U^2\) in terms of \(r\) and \(g\).

Show Solution

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Solution:

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1. Assuming the particles have mass \(m\), and speed \(v\) just before collision, then \begin{align*} \text{COE}: && \underbrace{\frac12 m u^2}_{\text{initial kinetic energy}} + \underbrace{0}_{\text{initial GPE}} &= \underbrace{\frac12m v^2}_{\text{kinetic energy just before collision}} + \underbrace{mgr\left(1-\frac1{\sqrt{2}}\right)}_{\text{GPE just before collision}} \\ \Rightarrow && v^2 &= u^2 - gr(2-\sqrt{2}) \end{align*} Therefore the particles has velocity \(\frac{\sqrt{2}}2v \binom{\pm 1}{1}\) before the collision. By symmetry, the impulse between the particles will be horizontal, so the vertical velocities will be unchanged at \(\frac{\sqrt{2}}{2}v\). By conservation of momentum (or symmetry) the particles will have equal but opposite velocities after the collision (say \(w\)) satisfying: \[ e = \frac{2w}{2\frac{\sqrt{2}}{2}v} \] ie \(w = \frac{\sqrt{2}}2 e v\) as required.
2. Once the particles have rebounded, they will be projectiles whilst the strings are slack. If we consider the left-most point \(A = (0,0)\) then the particles colide at \(\left ( \frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right)\) and the position at time \(t\) after the collision (before the string goes slack) will be: \begin{align*} \mathbf{x}_t &= \frac{\sqrt{2}}{2}r\binom{1}{-1} + \frac{\sqrt{2}}{2} vt \binom{-e}{1} + \frac12 gt^2 \binom{0}{-1} \end{align*} The string will go taught when \(|\mathbf{x}_t|^2 = r^2\), ie \begin{align*} && r^2 &= \left ( \frac{\sqrt{2}}{2} r - \frac{\sqrt{2}}{2}evt \right)^2 + \left (-\frac{\sqrt{2}}{2} r + \frac{\sqrt{2}}{2}vt -\frac12 gt^2 \right)^2 \\ \Rightarrow && r^2 &= \frac12 \left (r - evt \right)^2 + \frac12 \left (-r+vt - \frac{\sqrt{2}}{2}gt^2 \right)^2 \\ \Rightarrow && 2r^2 &= \left (r - evt \right)^2 + \left (-r+vt - \frac{\sqrt{2}}{2}gt^2 \right)^2 \\ \end{align*} as required.
3. Suppose \(z = \frac{vt}{r}\), \(c = \frac{\sqrt{2}v^2}{rg}\), then \begin{align*} && 2r^2 &= \left (r - evt \right)^2 + \left (-r+vt - \frac{\sqrt{2}}{2}gt^2 \right)^2 \\ \Leftrightarrow && 2 &= \left (1 - e\frac{vt}{r} \right)^2 + \left (-1 + \frac{vt}{r}- \frac{\sqrt{2}}{2} \frac{gt^2}{r} \right)^2 \\ \Leftrightarrow && 2 &= \left (1 - ez \right)^2 + \left (-1 +z- \frac{v^2t^2}{r^2} \frac{gr}{\sqrt{2}v^2}\right)^2 \\ \Leftrightarrow && 2 &= \left (1 - ez \right)^2 + \left (-1 +z- \frac{z^2}{c} \right)^2 \\ \Leftrightarrow && 2 &= 1-2ez + e^2z^2 + 1 + z^2 +\frac{z^4}{c^2} - 2z-2\frac{z^3}{c}+2\frac{z^2}{c} \\ \Leftrightarrow && 0 &= z(-2e-2) + z^2(e^2+1 + \frac{2}{c}) + z^3(-\frac{2}{c}) + z^4 \frac{1}{c^2} \\ \underbrace{\Leftrightarrow}_{z \neq 0} && 0 &= \left ( \frac{z}{c} \right)^3 - 2\left ( \frac{z}{c} \right)^2 + \left ( \frac{z}{c} \right) (1 + e^2 + \frac{2}{c} ) - \frac{2}{c}(1+e) \end{align*} as required, (where on the last step we divide by \(z/c\)).
4. If a cubic has \(3\) equal, non-zero roots then it must have the form \((z-a)^3 = z^3 -3az^2 + 3a^2 z -a^3 = 0\), so \(3a = 2\), and so the expansion must be \(\left ( \frac{z}{c} \right)^3 - 2\left ( \frac{z}{c} \right)^2 + \frac{4}{3}\left ( \frac{z}{c} \right) - \frac{8}{27} = 0\) \begin{align*} && \frac{2}{c}(1+e) &= \frac{8}{27} \\ \Rightarrow && \frac{2}{c} &= \frac{8}{27} \frac{1}{1+e} \\ && 1 + e^2 + \frac{2}{c} &= \frac43 \\ \Rightarrow && e^2 + \frac{8}{27(1+e)} &= \frac{1}{3} \\ \Rightarrow && 27(1+e)e^2+8 &= 9(1+e) \\ \Rightarrow && 27e^3 + 27e^2-9e-1 &= 0 \\ \Rightarrow && (3e-1)(9e^2+12e+1) &= 0 \end{align*} The only (positive) root is \(e = \frac13\), therefore \(e = \frac13\). We must also have \begin{align*} && \frac{2}{c} \frac43 &= \frac{8}{27} \\ \Rightarrow && c &= 9 \\ \Rightarrow && \frac{\sqrt{2}v^2}{rg} &= 9 \\ \Rightarrow && v^2 &= \frac{9\sqrt{2}rg}{2} \end{align*} as required. If we consider the path of the particle acting as a projectile, iff the path is tangent to the circle then there will be exactly one solution for \(z/c\) and (importantly) it will be a repeated root. Therefore the particle rejoins the circle at a tangent and the tension is acting perpendicularly to the direction of motion (ie no energy loss).
5. Since the only energy lost is lost in the collision, we can apply conservation of energy again: \begin{align*} \text{COE:} && \frac12 m U^2 &= \frac12 m \frac12v^2(1+e^2) + mgr\left (1 - \frac1{\sqrt{2}} \right) \\ \Rightarrow && U^2 &= \frac12 \frac{9 \sqrt{2}}{2}gr(1+\frac19) + gr(2 - \sqrt{2}) \\ &&&= \left (\frac{5\sqrt{2}}{2}+2 - \sqrt{2} \right)gr \\ &&&= \left (\frac{4+3\sqrt{2}}{2} \right)gr \end{align*}

In this question, the \(x\)- and \(y\)-axes are horizontal and the \(z\)-axis is vertically upwards.

1. A particle \(P_\alpha\) is projected from the origin with speed \(u\) at an acute angle \(\alpha\) above the positive \(x\)-axis. The curve \(E\) is given by \(z = A - Bx^2\) and \(y = 0\). If \(E\) and the trajectory of \(P_\alpha\) touch exactly once, show that \[u^2 - 2gA = u^2(1 - 4AB)\cos^2\alpha\,.\] \(E\) and the trajectory of \(P_\alpha\) touch exactly once for all \(\alpha\) with \(0 < \alpha < \frac{1}{2}\pi\). Write down the values of \(A\) and \(B\) in terms of \(u\) and \(g\).

2. Describe the set of points which can be hit by particles from the explosion, explaining your answer.
3. Show that, at a time \(t\) after the explosion, the particles lie on a sphere whose centre and radius you should find.
4. Another particle \(Q\) is projected horizontally from the point \((0, 0, A)\) with speed \(u\) in the positive \(x\) direction. Show that, at all times, \(Q\) lies on the curve \(E\).
5. Show that for particles \(Q\) and \(P_\alpha\) to collide, \(Q\) must be projected a time \(\dfrac{u(1-\cos\alpha)}{g\sin\alpha}\) after the explosion.

Point \(A\) is a distance \(h\) above ground level and point \(N\) is directly below \(A\) at ground level. Point \(B\) is also at ground level, a distance \(d\) horizontally from \(N\). The angle of elevation of \(A\) from \(B\) is \(\beta\). A particle is projected horizontally from \(A\), with initial speed \(V\). A second particle is projected from \(B\) with speed \(U\) at an acute angle \(\theta\) above the horizontal. The horizontal components of the velocities of the two particles are in opposite directions. The two particles are projected simultaneously, in the vertical plane through \(A\), \(N\) and \(B\). Given that the two particles collide, show that \[d\sin\theta - h\cos\theta = \frac{Vh}{U}\] and also that

1. \(\theta > \beta\);
2. \(U\sin\theta \geqslant \sqrt{\dfrac{gh}{2}}\);
3. \(\dfrac{U}{V} > \sin\beta\).

1. Two particles move on a smooth horizontal surface. The positions, in Cartesian coordinates, of the particles at time \(t\) are \((a+ut\cos\alpha \,,\, ut\sin\alpha)\) and \((vt\cos\beta\,,\, b+vt\sin\beta )\), where \(a\), \(b\), \(u\) and \(v\) are positive constants, \(\alpha\) and \(\beta\) are constant acute angles, and \(t\ge0\). Given that the two particles collide, show that \[ u \sin(\theta+\alpha) = v\sin(\theta +\beta)\,, \] where \(\theta \) is the acute angle satisfying \(\tan\theta = \dfrac b a\).
2. A gun is placed on the top of a vertical tower of height \(b\) which stands on horizontal ground. The gun fires a bullet with speed \(v\) and (acute) angle of elevation \(\beta\). Simultaneously, a target is projected from a point on the ground a horizontal distance \(a\) from the foot of the tower. The target is projected with speed \(u\) and (acute) angle of elevation \(\alpha\), in a direction directly away from the tower. Given that the target is hit before it reaches the ground, show that \[ 2u\sin\alpha (u\sin\alpha - v\sin\beta) > bg\,. \] Explain, with reference to part (i), why the target can only be hit if \(\alpha > \beta\).

Two particles, \(A\) and \(B\), are projected simultaneously towards each other from two points which are a distance \(d\) apart in a horizontal plane. Particle \(A\) has mass \(m\) and is projected at speed \(u\) at angle \(\alpha\) above the horizontal. Particle \(B\) has mass \(M\) and is projected at speed \(v\) at angle \(\beta\) above the horizontal. The trajectories of the two particles lie in the same vertical plane. The particles collide directly when each is at its point of greatest height above the plane. Given that both \(A\) and \(B\) return to their starting points, and that momentum is conserved in the collision, show that \[ m\cot \alpha = M \cot \beta\,. \] Show further that the collision occurs at a point which is a horizontal distance \(b\) from the point of projection of \(A\) where \[ b= \frac{Md}{m+M}\, , \] and find, in terms of \(b\) and \(\alpha\), the height above the horizontal plane at which the collision occurs.

A uniform rod \(AB\) has mass \(M\) and length \(2a\). The point \(P\) lies on the rod a distance \(a-x\) from~\(A\). Show that the moment of inertia of the rod about an axis through \(P\) and perpendicular to the rod is \[ \tfrac13 M(a^2 +3x^2)\,. \] The rod is free to rotate, in a horizontal plane, about a fixed vertical axis through \(P\). Initially the rod is at rest. The end \(B\) is struck by a particle of mass \(m\) moving horizontally with speed \(u\) in a direction perpendicular to the rod. The coefficient of restitution between the rod and the particle is \(e\). Show that the angular velocity of the rod immediately after impact is \[ \frac{3mu(1+e)(a+x)}{M(a^2+3x^2) +3m(a+x)^2}\,. \] In the case \(m=2M\), find the value of \(x\) for which the angular velocity is greatest and show that this angular velocity is \(u(1+e)/a\,\).

Particles \(P\) and \(Q\), each of mass \(m\), lie initially at rest a distance \(a\) apart on a smooth horizontal plane. They are connected by a light elastic string of natural length \(a\) and modulus of elasticity \(\frac12 m a \omega^2\), where \(\omega\) is a constant. Then \(P\) receives an impulse which gives it a velocity \(u\) directly away from \(Q\). Show that when the string next returns to length \(a\), the particles have travelled a distance \(\frac12 \pi u/\omega\,\), and find the speed of each particle. Find also the total time between the impulse and the subsequent collision of the particles.

Two points \(A\) and \(B\) lie on horizontal ground. A particle \(P_1\) is projected from \(A\) towards \(B\) at an acute angle of elevation \(\alpha\) and simultaneously a particle \(P_2\) is projected from \(B\) towards \(A\) at an acute angle of elevation \(\beta\). Given that the two particles collide in the air a horizontal distance \(b\) from \(B\), and that the collision occurs after \(P_1\) has attained its maximum height \(h\), show that \[ 2h \cot\beta < b < 4h \cot\beta \hphantom{\,,} \] and \[ 2h \cot\alpha < a < 4h \cot\alpha \,, \] where \(a\) is the horizontal distance from \(A\) to the point of collision.

Two particles \(P\) and \(Q\) are projected simultaneously from points \(O\) and \(D\), respectively, where~\(D\) is a distance \(d\) directly above \(O\). The initial speed of \(P\) is \(V\) and its angle of projection {\em above} the horizontal is \(\alpha\). The initial speed of \(Q\) is \(kV\), where \(k>1\), and its angle of projection {\em below} the horizontal is \(\beta\). The particles collide at time \(T\) after projection. Show that \(\cos\alpha = k\cos\beta\) and that \(T\) satisfies the equation \[ (k^2-1)V^2T^2 +2dVT\sin\alpha -d^2 =0\,. \] Given that the particles collide when \(P\) reaches its maximum height, find an expression for~\(\sin^2\alpha\) in terms of \(g\), \(d\), \(k\) and \(V\), and deduce that \[ gd\le (1+k)V^2\,. \]

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 \[ \cos \alpha = \sqrt{ \frac{M}{2M+m}}\ . \]

The points \(A\) and \(B\) are \(180\) metres apart and lie on horizontal ground. A missile is launched from \(A\) at speed of \(100\,\)m\,s\(^{-1}\) and at an acute angle of elevation to the line \(AB\) of \(\arcsin \frac35\). A time \(T\) seconds later, an anti-missile missile is launched from \(B\), at speed of \(200\,\)m\,s\(^{-1}\) and at an acute angle of elevation to the line \(BA\) of \(\arcsin \frac45\). The motion of both missiles takes place in the vertical plane containing \(A\) and \(B\), and the missiles collide. Taking \(g =10\,\)m\,s\(^{-2}\) and ignoring air resistance, find \(T\). \noindent [Note that \(\arcsin \frac35\) is another notation for \(\sin^{-1} \frac35\,\).]

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.

In this question take \(g = 10 ms^{-2}.\) The point \(A\) lies on a fixed rough plane inclined at \(30^{\circ}\) to the horizontal and \(\ell\) is the line of greatest slope through \(A\). A particle \(P\) is projected up \(\ell\) from \(A\) with initial speed \(6\)ms\(^{-1}\). A time \(T\) seconds later, a particle \(Q\) is projected from \(A\) up \(\ell\), also with speed \(6\)ms\(^{-1}\). The coefficient of friction between each particle and the plane is \(1/(5\sqrt{3})\,\) and the mass of each particle is \(4\)kg.

1. Given that \(T<1+\sqrt{3/2}\), show that the particles collide at a time \((3-\sqrt6)T+1\) seconds after \(P\) is projected.
2. In the case \(T=1+\sqrt{2/3}\,\), determine the energy lost due to friction from the instant at which \(P\) is projected to the time of the collision.

Point \(B\) is a distance \(d\) due south of point \(A\) on a horizontal plane. Particle \(P\) is at rest at \(B\) at \(t=0\), when it begins to move with constant acceleration \(a\) in a straight line with fixed bearing~\(\beta\,\). Particle \(Q\) is projected from point \(A\) at \(t=0\) and moves in a straight line with constant speed \(v\,\). Show that if the direction of projection of \(Q\) can be chosen so that \(Q\) strikes \(P\), then \[ v^2 \ge ad \l 1 - \cos \beta \r\;. \] Show further that if \(v^2 >ad(1-\cos\beta)\) then the direction of projection of \(Q\) can be chosen so that \(Q\) strikes \(P\) before \(P\) has moved a distance \(d\,\).

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 \(B\).