Edexcel M5 (Mechanics 5)

At time \(t = 0\), a particle \(P\) of mass \(3\) kg is at rest at the point \(A\) with position vector \((j - 3k)\) m. Two constant forces \(\mathbf{F}_1\) and \(\mathbf{F}_2\) then act on the particle \(P\) and it passes through the point \(B\) with position vector \((8i - 3j + 5k)\) m. Given that \(\mathbf{F}_1 = (4i - 2j + 5k)\) N and \(\mathbf{F}_2 = (8i - 4j + 7k)\) N and that \(\mathbf{F}_1\) and \(\mathbf{F}_2\) are the only two forces acting on \(P\), find the velocity of \(P\) as it passes through \(B\), giving your answer as a vector. [7]

At time \(t = 0\), the position vector of a particle \(P\) is \(-3j\) m. At time \(t\) seconds, the position vector of \(P\) is \(\mathbf{r}\) metres and the velocity of \(P\) is \(\mathbf{v}\) m s\(^{-1}\). Given that $$\mathbf{v} - 2\mathbf{r} = 4e^t \mathbf{j},$$ find the time when \(P\) passes through the origin. [7]

A particle moves from the point \(A\) with position vector \((3i - j + 3k)\) m to the point \(B\) with position vector \((i - 2j - 4k)\) m under the action of the force \((2i - 3j - k)\) N. Find the work done by the force. [4]

A particle \(P\) moves in a plane such that its position vector \(\mathbf{r}\) metres at time \(t\) seconds \((t > 0)\) satisfies the differential equation $$\frac{d\mathbf{r}}{dt} - \frac{2}{t}\mathbf{r} = 4i$$ When \(t = 1\), the particle is at the point with position vector \((i + j)\) m. Find \(\mathbf{r}\) in terms of \(t\). [9]

A particle moves in a plane in such a way that its position vector \(\mathbf{r}\) metres at time \(t\) seconds satisfies the differential equation $$\frac{d^2\mathbf{r}}{dt^2} - 2\frac{d\mathbf{r}}{dt} = 0$$ When \(t = 0\), the particle is at the origin and is moving with velocity \((4i + 2j)\) m s\(^{-1}\). Find \(\mathbf{r}\) in terms of \(t\). [7]

At time \(t\) seconds, the position vector of a particle \(P\) is \(\mathbf{r}\) metres, where \(\mathbf{r}\) satisfies the vector differential equation $$\frac{d^2\mathbf{r}}{dt^2} + 4\mathbf{r} = e^{2t} \mathbf{j}.$$ When \(t = 0\), \(P\) has position vector \((i + j)\) m and velocity \(2i\) m s\(^{-1}\). Find an expression for \(\mathbf{r}\) in terms of \(t\). [11]

\includegraphics{figure_2} **Figure 1** A uniform circular disc has mass \(4m\), centre \(O\) and radius \(4a\). The line \(POQ\) is a diameter of the disc. A circular hole of radius \(2a\) is made in the disc with the centre of the hole at the point \(R\) on \(PQ\) where \(QR = 5a\), as shown in Figure 1. The resulting lamina is free to rotate about a fixed smooth horizontal axis \(L\) which passes through \(Q\) and is perpendicular to the plane of the lamina.

1. Show that the moment of inertia of the lamina about \(L\) is \(69ma^2\). [7]

The lamina is hanging at rest with \(P\) vertically below \(Q\) when it is given an angular velocity \(\Omega\). Given that the lamina turns through an angle \(\frac{2\pi}{3}\) before it first comes to instantaneous rest,

1. find \(\Omega\) in terms of \(g\) and \(a\). [6]

A particle \(P\) moves in the \(x\)-\(y\) plane so that its position vector \(\mathbf{r}\) metres at time \(t\) seconds satisfies the differential equation $$\frac{d^2\mathbf{r}}{dt^2} - 4\mathbf{r} = -3e^t\mathbf{j}$$ When \(t = 0\), the particle is at the origin and is moving with velocity \((2i + j)\) ms\(^{-1}\). Find \(\mathbf{r}\) in terms of \(t\). [10]

A rocket, with initial mass 1500 kg, including 600 kg of fuel, is launched vertically upwards from rest. The rocket burns fuel at a rate of 15 kg s\(^{-1}\) and the burnt fuel is ejected vertically downwards with a speed of 1000 m s\(^{-1}\) relative to the rocket. At time \(t\) seconds after launch \((t \leqslant 40)\) the rocket has mass \(m\) kg and velocity \(v\) m s\(^{-1}\).

1. Show that $$\frac{dv}{dt} + \frac{1000}{m} \frac{dm}{dt} = -9.8$$ [5]

1. Find \(v\) at time \(t\), \(0 \leqslant t \leqslant 40\) [5]

Three forces \(\mathbf{F}_1 = (3i - j + k)\) N, \(\mathbf{F}_2 = (2i - k)\) N, and \(\mathbf{F}_3\) act on a rigid body. The force \(\mathbf{F}_1\) acts through the point with position vector \((i + 2j + k)\) m, the force \(\mathbf{F}_2\) acts through the point with position vector \((i - 2j)\) m and the force \(\mathbf{F}_3\) acts through the point with position vector \((i + j + k)\) m. Given that the system \(\mathbf{F}_1\), \(\mathbf{F}_2\) and \(\mathbf{F}_3\) reduces to a couple \(\mathbf{G}\),

1. find \(\mathbf{G}\). [6]

The line of action of \(\mathbf{F}_3\) is changed so that the system \(\mathbf{F}_1\), \(\mathbf{F}_2\) and \(\mathbf{F}_3\) now reduces to a couple \((6i + 8j + 2k)\) N m.

1. Find an equation of the new line of action of \(\mathbf{F}_3\), giving your answer in the form \(\mathbf{r} = \mathbf{a} + t\mathbf{b}\), where \(\mathbf{a}\) and \(\mathbf{b}\) are constant vectors. [5]

A spaceship is moving in a straight line in deep space and needs to increase its speed. This is done by ejecting fuel backwards from the spaceship at a constant speed \(c\) relative to the spaceship. When the speed of the spaceship is \(v\), its mass is \(m\).

1. Show that, while the spaceship is ejecting fuel, $$\frac{dv}{dm} = -\frac{c}{m}.$$ [5]

The initial mass of the spaceship is \(m_0\) and at time \(t\) the mass of the spaceship is given by \(m = m_0(1 - kt)\), where \(k\) is a positive constant.

1. Find the acceleration of the spaceship at time \(t\). [4]

A uniform lamina \(ABC\) of mass \(m\) is in the shape of an isosceles triangle with \(AB = AC = 5a\) and \(BC = 8a\).

1. Show, using integration, that the moment of inertia of the lamina about an axis through \(A\), parallel to \(BC\), is \(\frac{9}{2}ma^2\). [6]

The foot of the perpendicular from \(A\) to \(BC\) is \(D\). The lamina is free to rotate in a vertical plane about a fixed smooth horizontal axis which passes through \(D\) and is perpendicular to the plane of the lamina. The lamina is released from rest when \(DA\) makes an angle \(\alpha\) with the downward vertical. It is given that the moment of inertia of the lamina about an axis through \(D\), perpendicular to \(BC\) and in the plane of the lamina, is \(\frac{8}{3}ma^2\).

1. Find the angular acceleration of the lamina when \(DA\) makes an angle \(\theta\) with the downward vertical. [8]

Given that \(\alpha\) is small,

1. find an approximate value for the period of oscillation of the lamina about the vertical. [2]

A rocket propels itself by its engine ejecting burnt fuel. Initially the rocket has total mass \(M\), of which a mass \(kM\), \(k < 1\), is fuel. The rocket is at rest when its engine is started. The burnt fuel is ejected with constant speed \(c\), relative to the rocket, in a direction opposite to that of the rocket's motion. Assuming that there are no external forces, find the speed of the rocket when all its fuel has been burnt. [7]

A uniform rod \(PQ\), of mass \(m\) and length \(3a\), is free to rotate about a fixed smooth horizontal axis \(L\), which passes through the end \(P\) of the rod and is perpendicular to the rod. The rod hangs at rest in equilibrium with \(Q\) vertically below \(P\). One end of a light inextensible string of length \(2a\) is attached to the rod at \(P\) and the other end is attached to a particle of mass \(3m\). The particle is held with the string taut, and horizontal and perpendicular to \(L\), and is then released. After colliding, the particle sticks to the rod forming a body \(B\).

1. Show that the moment of inertia of \(B\) about \(L\) is \(15ma^2\). [2]

1. Show that \(B\) first comes to instantaneous rest after it has turned through an angle \(\arccos\left(\frac{9}{25}\right)\). [10]

\includegraphics{figure_4} **Figure 1** A uniform lamina of mass \(M\) is in the shape of a right-angled triangle \(OAB\). The angle \(OAB\) is \(90°\), \(OA = a\) and \(AB = 2a\), as shown in Figure 1.

1. Prove, using integration, that the moment of inertia of the lamina \(OAB\) about the edge \(OA\) is \(\frac{8}{3}Ma^2\). (You may assume without proof that the moment of inertia of a uniform rod of mass \(m\) and length \(2l\) about an axis through one end and perpendicular to the rod is \(\frac{4}{3}ml^2\).) [6]

The lamina \(OAB\) is free to rotate about a fixed smooth horizontal axis along the edge \(OA\) and hangs at rest with \(B\) vertically below \(A\). The lamina is then given a horizontal impulse of magnitude \(J\). The impulse is applied to the lamina at the point \(B\), in a direction which is perpendicular to the plane of the lamina. Given that the lamina first comes to instantaneous rest after rotating through an angle of \(120°\),

1. find an expression for \(J\), in terms of \(M\), \(a\) and \(g\). [7]

Two forces \(\mathbf{F}_1 = (i + 2j + 3k)\) N and \(\mathbf{F}_2 = (3i + j + 2k)\) N act on a rigid body. The force \(\mathbf{F}_1\) acts through the point with position vector \((2i + k)\) m and the force \(\mathbf{F}_2\) acts through the point with position vector \((j + 2k)\) m.

1. If the two forces are equivalent to a single force \(\mathbf{R}\), find
   1. \(\mathbf{R}\), [2]
   2. a vector equation of the line of action of \(\mathbf{R}\), in the form \(\mathbf{r} = \mathbf{a} + \lambda \mathbf{b}\). [6]

1. If the two forces are equivalent to a single force acting through the point with position vector \((i + 2j + k)\) m together with a couple of moment \(\mathbf{G}\), find the magnitude of \(\mathbf{G}\). [5]

Two forces \(\mathbf{F}_1 = (3i + k)\) N and \(\mathbf{F}_2 = (4i + j - k)\) N act on a rigid body. The force \(\mathbf{F}_1\) acts at the point with position vector \((2i - j + 3k)\) m and the force \(\mathbf{F}_2\) acts at the point with position vector \((-3i + 2k)\) m. The two forces are equivalent to a single force \(\mathbf{R}\) acting at the point with position vector \((i + 2j + k)\) m together with a couple of moment \(\mathbf{G}\). Find,

1. \(\mathbf{R}\), [2]
2. \(\mathbf{G}\). [4]

A third force \(\mathbf{F}_3\) is now added to the system. The force \(\mathbf{F}_3\) acts at the point with position vector \((2i - k)\) m and the three forces \(\mathbf{F}_1\), \(\mathbf{F}_2\) and \(\mathbf{F}_3\) are equivalent to a couple.

1. Find the magnitude of the couple. [6]

A body consists of a uniform plane circular disc, of radius \(r\) and mass \(2m\), with a particle of mass \(3m\) attached to the circumference of the disc at the point \(P\). The line \(PQ\) is a diameter of the disc. The body is free to rotate in a vertical plane about a fixed smooth horizontal axis, \(L\), which is perpendicular to the plane of the disc and passes through \(Q\). The body is held with \(QP\) making an angle \(\beta\) with the downward vertical through \(Q\), where \(\sin \beta = 0.25\), and released from rest. Find the magnitude of the component, perpendicular to \(PQ\), of the force acting on the body at \(Q\) at the instant when it is released. [You may assume that the moment of inertia of the body about \(L\) is \(15mr^2\).] [6]

Two forces \(\mathbf{F}_1 = (2i + j)\) N and \(\mathbf{F}_2 = (-2j - k)\) N act on a rigid body. The force \(\mathbf{F}_1\) acts at the point with position vector \(\mathbf{r}_1 = (3i + j + k)\) m and the force \(\mathbf{F}_2\) acts at the point with position vector \(\mathbf{r}_2 = (i - 2j)\) m. A third force \(\mathbf{F}_3\) acts on the body such that \(\mathbf{F}_1\), \(\mathbf{F}_2\) and \(\mathbf{F}_3\) are in equilibrium.

1. Find the magnitude of \(\mathbf{F}_3\). [4]

1. Find a vector equation of the line of action of \(\mathbf{F}_3\). [8]

The force \(\mathbf{F}_3\) is replaced by a fourth force \(\mathbf{F}_4\), acting through the origin \(O\), such that \(\mathbf{F}_1\), \(\mathbf{F}_2\) and \(\mathbf{F}_4\) are equivalent to a couple.

1. Find the magnitude of this couple. [4]

A raindrop falls vertically under gravity through a cloud. In a model of the motion the raindrop is assumed to be spherical at all times and the cloud is assumed to consist of stationary water particles. At time \(t = 0\), the raindrop is at rest and has radius \(a\). As the raindrop falls, water particles from the cloud condense onto it and the radius of the raindrop is assumed to increase at a constant rate \(\lambda\). At time \(t\) the speed of the raindrop is \(v\).

1. Show that $$\frac{dv}{dt} + \frac{3\lambda v}{(\lambda t + a)} = g.$$ [8]

1. Find the speed of the raindrop when its radius is \(3a\). [7]

A uniform rod \(PQ\), of mass \(m\) and length \(2a\), is made to rotate in a vertical plane with constant angular speed \(\sqrt{\frac{g}{a}}\) about a fixed smooth horizontal axis through the end \(P\) of the rod. Show that, when the rod is inclined at an angle \(\theta\) to the downward vertical, the magnitude of the force exerted on the axis by the rod is \(2mg|\cos(\frac{1}{2}\theta)|\). [8]

The points \(P\) and \(Q\) have position vectors \(4i - 6j - 12k\) and \(2i + 4j + 4k\) respectively, relative to a fixed origin \(O\). Three forces, \(\mathbf{F}_1\), \(\mathbf{F}_2\) and \(\mathbf{F}_3\), act along \(\overrightarrow{OP}\), \(\overrightarrow{OQ}\) and \(\overrightarrow{QP}\) respectively, and have magnitudes \(7\) N, \(3\) N and \(3\sqrt{10}\) N respectively.

1. Express \(\mathbf{F}_1\), \(\mathbf{F}_2\) and \(\mathbf{F}_3\) in vector form. [3]

1. Show that the resultant of \(\mathbf{F}_1\), \(\mathbf{F}_2\) and \(\mathbf{F}_3\) is \((2i - 10j - 16k)\) N. [2]

1. Find a vector equation of the line of action of this resultant, giving your answer in the form \(\mathbf{r} = \mathbf{a} + \lambda\mathbf{b}\), where \(\mathbf{a}\) and \(\mathbf{b}\) are constant vectors and \(\lambda\) is a parameter. [5]

A pendulum consists of a uniform rod \(AB\), of length \(4a\) and mass \(2m\), whose end \(A\) is rigidly attached to the centre \(O\) of a uniform square lamina \(PQRS\), of mass \(4m\) and side \(a\). The rod \(AB\) is perpendicular to the plane of the lamina. The pendulum is free to rotate about a fixed smooth horizontal axis \(L\) which passes through \(B\). The axis \(L\) is perpendicular to \(AB\) and parallel to the edge \(PQ\) of the square.

1. Show that the moment of inertia of the pendulum about \(L\) is \(75ma^2\). [4]

The pendulum is released from rest when \(BA\) makes an angle \(\alpha\) with the downward vertical through \(B\), where \(\tan \alpha = \frac{3}{4}\). When \(BA\) makes an angle \(\theta\) with the downward vertical through \(B\), the magnitude of the component, in the direction \(AB\), of the force exerted by the axis \(L\) on the pendulum is \(X\).

1. Find an expression for \(X\) in terms of \(m\), \(g\) and \(\theta\). [9]

Using the approximation \(\theta \approx \sin \theta\),

1. find an estimate of the time for the pendulum to rotate through an angle \(\alpha\) from its initial rest position. [6]

A uniform circular disc has mass \(m\), centre \(O\) and radius \(2a\). It is free to rotate about a fixed smooth horizontal axis \(L\) which lies in the same plane as the disc and which is tangential to the disc at the point \(A\). The disc is hanging at rest in equilibrium with \(O\) vertically below \(A\) when it is struck at \(O\) by a particle of mass \(m\). Immediately before the impact the particle is moving perpendicular to the plane of the disc with speed \(3\sqrt{ag}\). The particle adheres to the disc at \(O\).

1. Find the angular speed of the disc immediately after the impact. [5]

1. Find the magnitude of the force exerted on the disc by the axis immediately after the impact. [6]

A uniform rod \(AB\) of mass \(4m\) is free to rotate in a vertical plane about a fixed smooth horizontal axis, \(L\), through \(A\). The rod is hanging vertically at rest when it is struck at its end \(B\) by a particle of mass \(m\). The particle is moving with speed \(u\), in a direction which is horizontal and perpendicular to \(L\), and after striking the rod it rebounds in the opposite direction with speed \(v\). The coefficient of restitution between the particle and the rod is 1. Show that \(u = 7v\). [7]

A uniform circular pulley, of mass \(4m\) and radius \(r\), is free to rotate about a fixed smooth horizontal axis which passes through the centre of the pulley and is perpendicular to the plane of the pulley. A light inextensible string passes over the pulley and has a particle of mass \(2m\) attached to one end and a particle of mass \(3m\) attached to the other end. The particles hang with the string vertical and taut on each side of the pulley. The rim of the pulley is sufficiently rough to prevent the string slipping. The system is released from rest.

1. Find the angular acceleration of the pulley. [8]

When the angular speed of the pulley is \(\Omega\), the string breaks and a constant braking couple of magnitude \(G\) is applied to the pulley which brings it to rest.

1. Find an expression for the angle turned through by the pulley from the instant when the string breaks to the instant when the pulley first comes to rest. [4]

Prove, using integration, that the moment of inertia of a uniform solid right circular cone, of mass \(M\) and base radius \(a\), about its axis is \(\frac{3}{10}Ma^2\). [You may assume, without proof, that the moment of inertia of a uniform circular disc, of mass \(m\) and radius \(r\), about an axis through its centre and perpendicular to its plane is \(\frac{1}{2}mr^2\).] [10]

1. A uniform lamina of mass \(m\) is in the shape of a triangle \(ABC\). The perpendicular distance of \(C\) from the line \(AB\) is \(h\). Prove, using integration, that the moment of inertia of the lamina about \(AB\) is \(\frac{1}{6}mh^2\). [7]

1. Deduce the radius of gyration of a uniform square lamina of side \(2a\), about a diagonal. [3]

The points \(X\) and \(Y\) are the mid-points of the sides \(RQ\) and \(RS\) respectively of a square \(PQRS\) of side \(2a\). A uniform lamina of mass \(M\) is in the shape of \(PQXYS\).

1. Show that the moment of inertia of this lamina about \(XY\) is \(\frac{79}{84}Ma^2\). [6]

A pendulum consists of a uniform rod \(PQ\), of mass \(3m\) and length \(2a\), which is rigidly fixed at its end \(Q\) to the centre of a uniform circular disc of mass \(m\) and radius \(a\). The rod is perpendicular to the plane of the disc. The pendulum is free to rotate about a fixed smooth horizontal axis \(L\) which passes through the end \(P\) of the rod and is perpendicular to the rod.

1. Show that the moment of inertia of the pendulum about \(L\) is \(\frac{33}{4}ma^2\). [5]

The pendulum is released from rest in the position where \(PQ\) makes an angle \(\alpha\) with the downward vertical. At time \(t\), \(PQ\) makes an angle \(\theta\) with the downward vertical.

1. Show that the angular speed, \(\dot{\theta}\), of the pendulum satisfies $$\dot{\theta}^2 = \frac{40g(\cos \theta - \cos \alpha)}{33a}$$ [4]

1. Hence, or otherwise, find the angular acceleration of the pendulum. [3]

Given that \(\alpha = \frac{\pi}{20}\) and that \(PQ\) has length \(\frac{8}{33}\) m,

1. find, to 3 significant figures, an approximate value for the angular speed of the pendulum 0.2 s after it has been released from rest. [5]