Edexcel M5 (Mechanics 5) 2012 June

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 $$\frac{d\mathbf{r}}{dt} - \frac{2}{t}\mathbf{r} = 4\mathbf{i}$$ When \(t = 1\), the particle is at the point with position vector \((\mathbf{i} + \mathbf{j})\) m. Find \(\mathbf{r}\) in terms of \(t\). [9]

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

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

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]

The points \(P\) and \(Q\) have position vectors \(4\mathbf{i} - 6\mathbf{j} - 12\mathbf{k}\) and \(2\mathbf{i} + 4\mathbf{j} + 4\mathbf{k}\) 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]
2. Show that the resultant of \(\mathbf{F}_1\), \(\mathbf{F}_2\) and \(\mathbf{F}_3\) is \((2\mathbf{i} - 10\mathbf{j} - 16\mathbf{k})\) N. [2]
3. 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]

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]
2. 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]