Problems

2009 Paper 2 Q10

D: 1600.0 B: 1500.0

momentum collisions coefficient of restitution simultaneous collisions conservation of momentum two-dimensional constraint equations

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Four particles $P_1$, $P_2$, $P_3$ and $P_4$, of masses $m_1$, $m_2$, $m_3$ and $m_4$, respectively, are arranged on smooth horizontal axes as shown in the diagram. Initially, $P_2$ and $P_3$ are stationary, and both $P_1$ and $P_4$ are moving towards $O$ with speed $u$. Then $P_1$ and $P_2$ collide, at the same moment as $P_4$ and $P_3$ collide. Subsequently, $P_2$ and $P_3$ collide at $O$, as do $P_1$ and $P_4$ some time later. The coefficient of restitution between each pair of particles is~$e$, and $e>0$. Show that initially $P_2$ and $P_3$ are equidistant from $O$.

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2009 Paper 3 Q11

D: 1700.0 B: 1500.1

variable force momentum variable mass Newton's second law differential equation conservation of momentum dust cloud comet separable ODE limiting behaviour

A comet in deep space picks up mass as it travels through a large stationary dust cloud. It is subject to a gravitational force of magnitude $M\!f$ acting in the direction of its motion. When it entered the cloud, the comet had mass $M$ and speed $V$. After a time $t$, it has travelled a distance $x$ through the cloud, its mass is $M(1+bx)$, where~$b$ is a positive constant, and its speed is $v$.

In the case when $f=0$, write down an equation relating $V$, $x$, $v$ and $b$. Hence find an expression for $x$ in terms of $b$, $V$ and $t$.
In the case when $f$ is a non-zero constant, use Newton's second law in the form \[ \text{force} = \text{rate of change of momentum} \] to show that \[ v = \frac{ft+V}{1+bx}\,. \] Hence find an expression for $x$ in terms of $b$, $V$, $f$ and $t$. Show that it is possible, if $b$, $V$ and $f$ are suitably chosen, for the comet to move with constant speed. Show also that, if the comet does not move with constant speed, its speed tends to a constant as $t\to\infty$.

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2007 Paper 2 Q9

D: 1600.0 B: 1500.0

projectiles collision coefficient of restitution Newton's experimental law conservation of momentum cone normal impact optimisation

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.

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

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2006 Paper 2 Q10

D: 1600.0 B: 1500.0

momentum collisions coefficient of restitution Newton's law of restitution conservation of momentum inequality three particles successive collisions

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 $\frac14$.

Find the range of values of $k$ for which $A$ and $B$ collide for a second time.
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 $\dfrac{60d}{13u}\,$.

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2006 Paper 3 Q11

D: 1700.0 B: 1516.0

momentum collisions coefficient of restitution energy loss lift and counterweight pulley impulsive force connected particles conservation of momentum free fall

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.

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2005 Paper 1 Q10

D: 1500.0 B: 1484.0

momentum collisions coefficient of restitution Newton's law of restitution conservation of momentum collinear particles sequential collisions algebraic manipulation

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:

$\displaystyle \frac ab = \frac bc = e\,$;
$\displaystyle \frac ba = \frac cb = e\,$.

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2005 Paper 3 Q9

D: 1700.0 B: 1484.0

momentum collisions coefficient of restitution Newton's law of restitution conservation of momentum wall rebound geometric series equal time intervals two-particle collision recurrence

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.

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2002 Paper 1 Q11

D: 1500.0 B: 1484.0

momentum elastic collision oblique collision conservation of energy conservation of momentum trigonometric identity two-dimensional collision geometric proof

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A particle $P_1$ of mass $m$ collides with a particle $P_2$ of mass $km$ which is at rest. No energy is lost in the collision. The direction of motion of $P_1$ and $P_2$ after the collision make non-zero angles of $\theta$ and $\phi$, respectively, with the direction of motion of $P_1$ before the collision, as shown. Show that \[ \sin^2\theta + k\sin^2\phi = k\sin^2(\theta+\phi) \;. \] Show that, if the angle between the particles after the collision is a right angle, then $k=1\,$.

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2001 Paper 2 Q11

D: 1600.0 B: 1484.0

projectiles two-stage missile explosion conservation of momentum quadratic equation range inequality optimisation

A two-stage missile is projected from a point $A$ on the ground with horizontal and vertical velocity components $u$ and $v$, respectively. When it reaches the highest point of its trajectory an internal explosion causes it to break up into two fragments. Immediately after this explosion one of these fragments, $P$, begins to move vertically upwards with speed $v_e$, but retains the previous horizontal velocity. Show that $P$ will hit the ground at a distance $R$ from $A$ given by $$ \frac{gR}u = v+v_e + \sqrt{v_e^2 +v^2}\, . $$ It is required that the range $R$ should be greater than a certain distance $D$ (where $D> 2uv/g$). Show that this requirement is satisfied if \[ v_e> \frac{gD}{2u}\left( \frac{gD-2uv}{gD-uv}\right). \] \noindent[{\sl The effect of air resistance is to be neglected.}]

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2000 Paper 1 Q10

D: 1516.0 B: 1500.0

momentum collisions coefficient of restitution conservation of momentum inequality energy optimisation three particles

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.

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2000 Paper 3 Q9

D: 1700.0 B: 1500.0

momentum elastic collision kinetic energy oblique collision two-dimensional collision vector dot product inequality conservation of energy conservation of momentum

Two small discs of masses $m$ and $\mu m$ lie on a smooth horizontal surface. The disc of mass $\mu m$ is at rest, and the disc of mass $m$ is projected towards it with velocity $\mathbf{u}$. After the collision, the disc of mass $\mu m$ moves in the direction given by unit vector $\mathbf{n}$. The collision is perfectly elastic.

Show that the speed of the disc of mass $\mu m$ after the collision is \ \ $ \dfrac {2\mathbf{u} \cdot \mathbf{n}}{1+\mu}. $
Given that the two discs have equal kinetic energy after the collision, find an expression for the cosine of the angle between $\bf n$ and $\bf u$ and show that $3-\sqrt8\le \mu \le 3+\sqrt8$.

Solution:

In the direction of $\mathbf{n}$, Conservation of momentum gives: $m \mathbf{u} \cdot \mathbf{n} = m v_m + \mu m v_{\mu m}$ Newton's experimental law gives: $\frac{\mathbf{u} \cdot \mathbf{n}}{v_{\mu m} - v_m} = 1$ Therefore \begin{align*} && \mathbf{u} \cdot \mathbf{n} &= v_m + \mu v_{\mu m} \\ && \mathbf{u} \cdot \mathbf{n} &= v_{\mu m} - v_m \\ \Rightarrow && 2 \mathbf{u} \cdot \mathbf{n} &= (1 + \mu)v_{\mu m} \\ \Rightarrow && v_{\mu m} &= \frac{2 \mathbf{u} \cdot \mathbf{n}}{1 + \mu} \\ \end{align*}
Kinetic energy after the collision for the second mass is: $\frac12 m \mu \frac{4 (\mathbf{u} \cdot \mathbf{n})^2 }{(1 + \mu)^2}$ For the first mass the final speed (in the direction $\mathbf{n}$ is: $\mathbf{u} \cdot \mathbf{n}- \frac{2 \mathbf{u} \cdot \mathbf{n}}{1 + \mu} = \frac{(\mu - 1) \mathbf{u} \cdot \mathbf{n}}{1 + \mu}$ It's velocity perpendicular to $\mathbf{n}$ is unchanged, which is $\mathbf{u} - (\mathbf{u} \cdot \mathbf{n}) \mathbf{n}$, so it's speed squared is $\mathbf{u}\cdot \mathbf{u} - (\mathbf{u} \cdot \mathbf{n})^2$ Therefore the total kinetic energy is: $\frac12 m \frac{(\mu - 1)^2 (\mathbf{u} \cdot \mathbf{n})^2}{(1 + \mu)} + \frac12 m (\mathbf{u}\cdot \mathbf{u} - (\mathbf{u} \cdot \mathbf{n})^2)$ Therefore since the kinetic energies are equal we have: \begin{align*} && \frac12 m \frac{(\mu - 1)^2 (\mathbf{u} \cdot \mathbf{n})^2}{(1 + \mu)} + \frac12 m (\mathbf{u}\cdot \mathbf{u} - (\mathbf{u} \cdot \mathbf{n})^2) &= \frac12 m \mu \frac{4 (\mathbf{u} \cdot \mathbf{n})^2 }{(1 + \mu)^2} \\ \Rightarrow && \frac{(\mu - 1)^2 (\mathbf{u} \cdot \mathbf{n})^2}{(1 + \mu)} + (\mathbf{u}\cdot \mathbf{u} - (\mathbf{u} \cdot \mathbf{n})^2) &= \mu \frac{4 (\mathbf{u} \cdot \mathbf{n})^2 }{(1 + \mu)^2} \\ \Rightarrow && (\mathbf{u} \cdot \mathbf{n})^2 \l 1 + \frac{4\mu}{(1+ \mu)^2} - \frac{(1-\mu)^2}{(1+ \mu)^2} \r &= (\mathbf{u} \cdot \mathbf{u}) \\ \Rightarrow && (\mathbf{u} \cdot \mathbf{n})^2 \l \frac{(1 + \mu)^2 + 4\mu - (1-\mu)^2}{(1+ \mu)^2} \r &= (\mathbf{u} \cdot \mathbf{u}) \\ \Rightarrow && (\mathbf{u} \cdot \mathbf{n})^2 \l \frac{8\mu}{(1+ \mu)^2} \r &= (\mathbf{u} \cdot \mathbf{u}) \\ \Rightarrow && \cos^2 \theta &=\frac{(1 + \mu)^2}{8\mu} \\ \end{align*} We need \begin{align*} && \frac{(1 + \mu)^2}{8\mu} & \leq 1 \\ \Rightarrow && 1 +2 \mu + \mu^2 \leq 8 \mu \\ \Rightarrow && 1 - 6 \mu + \mu^2 \leq 0 \end{align*} This quadratic has roots at $3 \pm \sqrt{8}$ and therefore our quadratic inequality is satisfied if: $\boxed{3 - \sqrt{8} \leq \mu \leq 3 + \sqrt{8}}$

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1998 Paper 1 Q9

D: 1500.0 B: 1484.0

SHM simple pendulum Newton's experimental law coefficient of restitution collision conservation of momentum successive collisions equal pendulums period

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 $B$.

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1998 Paper 2 Q10

D: 1600.0 B: 1500.0

mechanics wedge motion on a slope conservation of momentum energy conservation Newton's laws differential equation smooth surface particle on inclined plane

A wedge of mass $M$ rests on a smooth horizontal surface. The face of the wedge is a smooth plane inclined at an angle $\alpha$ to the horizontal. A particle of mass $m$ slides down the face of the wedge, starting from rest. At a later time $t$, the speed $V$ of the wedge, the speed $v$ of the particle and the angle $\beta$ of the velocity of the particle below the horizontal are as shown in the diagram.

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\noindent Let $y$ be the vertical distance descended by the particle. Derive the following results, stating in (ii) and (iii) the mechanical principles you use:

$V\sin\alpha=v\sin(\beta-\alpha)$;
$\tan\beta=(1+m/M)\tan\alpha$;
$2gy=v^2(M+m\cos^2\beta)/M$.

Write down a differential equation for $y$ and hence show that $$y={gMt^2\sin^2\beta \over 2\,(M+m\cos^2\beta)}.$$

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1997 Paper 2 Q10

D: 1600.0 B: 1500.0

momentum collision coalescence elastic bounce explosion conservation of momentum energy cylinder two-dimensional motion geometric constraints

\noindent{\it In this question the effect of gravity is to be neglected.} A small body of mass $M$ is moving with velocity $v$ along the axis of a long, smooth, fixed, circular cylinder of radius $L$. An internal explosion splits the body into two spherical fragments, with masses $qM$ and $(1-q)M$, where $q\le\frac{1}{2}$. After bouncing perfectly elastically off the cylinder (one bounce each) the fragments collide and coalesce at a point $\frac{1}{2}L$ from the axis. Show that $q=\frac{3}{ 8}$. The collision occurs at a time $5L/v$ after the explosion. Find the energy imparted to the fragments by the explosion, and find the velocity after coalescence.

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1996 Paper 1 Q10

D: 1500.0 B: 1516.0

momentum conservation of momentum kinetic energy explosion optimisation relative speed inequality

A spaceship of mass $M$ is at rest. It separates into two parts in an explosion in which the total kinetic energy released is $E$. Immediately after the explosion the two parts have masses $m_{1}$ and $m_{2}$ and speeds $v_{1}$ and $v_{2}$ respectively. Show that the minimum possible relative speed $v_{1}+v_{2}$ of the two parts of the spaceship after the explosion is $(8E/M)^{1/2}.$

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