Edexcel M5 (Mechanics 5) 2009 June

1. At time \(t = 0\), a particle \(P\) of mass 3 kg is at rest at the point \(A\) with position vector \(( \mathbf { j } - 3 \mathbf { k } ) \mathrm { m }\). 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 \(( 8 \mathbf { i } - 3 \mathbf { j } + 5 \mathbf { k } ) \mathrm { m }\).

Given that \(\mathbf { F } _ { 1 } = ( 4 \mathbf { i } - 2 \mathbf { j } + 5 \mathbf { k } ) \mathrm { N }\) and \(\mathbf { F } _ { 2 } = ( 8 \mathbf { i } - 4 \mathbf { j } + 7 \mathbf { k } ) \mathrm { N }\) 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.

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

1. 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,
   $$\frac { \mathrm { d } v } { \mathrm {~d} m } = - \frac { c } { m } .$$ The initial mass of the spaceship is \(m _ { 0 }\) and at time \(t\) the mass of the spaceship is given by \(m = m _ { 0 } ( 1 - k t )\), where \(k\) is a positive constant.
2. Find the acceleration of the spaceship at time \(t\).

4. \begin{figure}[h] \end{figure} A uniform lamina of mass \(M\) is in the shape of a right-angled triangle \(O A B\). The angle \(O A B\) is \(90 ^ { \circ } , O A = a\) and \(A B = 2 a\), as shown in Figure 1.

1. Prove, using integration, that the moment of inertia of the lamina \(O A B\) about the edge \(O A\) is \(\frac { 2 } { 3 } M a ^ { 2 }\).
   (You may assume without proof that the moment of inertia of a uniform rod of mass \(m\) and length \(2 l\) about an axis through one end and perpendicular to the rod is \(\frac { 4 } { 3 } m l ^ { 2 }\).) The lamina \(O A B\) is free to rotate about a fixed smooth horizontal axis along the edge \(O A\) 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 \(120 ^ { \circ }\),
2. find an expression for \(J\), in terms of \(M , a\) and \(g\).

1. Two forces \(\mathbf { F } _ { 1 } = ( 2 \mathbf { i } + \mathbf { j } ) \mathrm { N }\) and \(\mathbf { F } _ { 2 } = ( - 2 \mathbf { j } - \mathbf { k } ) \mathrm { N }\) act on a rigid body. The force \(\mathbf { F } _ { 1 }\) acts at the point with position vector \(\mathbf { r } _ { 1 } = ( 3 \mathbf { i } + \mathbf { j } + \mathbf { k } ) \mathrm { m }\) and the force \(\mathbf { F } _ { 2 }\) acts at the point with position vector \(\mathbf { r } _ { 2 } = ( \mathbf { i } - 2 \mathbf { j } ) \mathrm { m }\). 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 \(\mathbf { F } _ { 3 }\).
   2. Find a vector equation of the line of action of \(\mathbf { F } _ { 3 }\).
   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.
2. Find the magnitude of this couple.

1. A pendulum consists of a uniform rod \(A B\), of length \(4 a\) and mass \(2 m\), whose end \(A\) is rigidly attached to the centre \(O\) of a uniform square lamina \(P Q R S\), of mass \(4 m\) and side \(a\). The \(\operatorname { rod } A B\) 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 \(A B\) and parallel to the edge \(P Q\) of the square.
   1. Show that the moment of inertia of the pendulum about \(L\) is \(75 m a ^ { 2 }\).
   The pendulum is released from rest when \(B A\) makes an angle \(\alpha\) with the downward vertical through \(B\), where \(\tan \alpha = \frac { 7 } { 24 }\). When \(B A\) makes an angle \(\theta\) with the downward vertical through \(B\), the magnitude of the component, in the direction \(A B\), of the force exerted by the axis \(L\) on the pendulum is \(X\).
2. Find an expression for \(X\) in terms of \(m , g\) and \(\theta\). Using the approximation \(\theta \approx \sin \theta\),
3. find an estimate of the time for the pendulum to rotate through an angle \(\alpha\) from its initial rest position.