OCR MEI Further Mechanics Major (Further Mechanics Major) 2019 June

Three forces represented by the vectors \(-4\mathbf{i} + \mathbf{j} + 2\mathbf{j}\) and \(k\mathbf{i} - 2\mathbf{j}\) act at the points with coordinates \((0, 0)\), \((3, 0)\) and \((0, 4)\) respectively.

1. Given that the three forces form a couple, find the value of \(k\). [2]
2. Find the magnitude and direction of the couple. [3]

The Reynolds number, \(R\), is an important dimensionless quantity in fluid dynamics; it can be used to predict flow patterns when a fluid is in motion relative to a surface. The Reynolds number is defined as $$R = \frac{\rho ul}{\mu},$$ where \(\rho\) is the density of the fluid, \(u\) is the velocity of the fluid relative to the surface, \(l\) is the distance travelled by the fluid and \(\mu\) is the viscosity of the fluid. Find the dimensions of \(\mu\). [4]

A ball of mass \(2\)kg is moving with velocity \((3\mathbf{i} - 2\mathbf{j})\)ms\(^{-1}\) when it is struck by a bat. The impulse on the ball is \((-8\mathbf{i} + 10\mathbf{j})\)Ns.

1. Find the speed of the ball immediately after the impact. [4]
2. State one modelling assumption you have used in answering part (a). [1]

\includegraphics{figure_4} Fig. 4 shows a uniform lamina ABCDE such that ABDE is a rectangle and BCD is an isosceles triangle. AB = 5a, AE = 4a and BC = CD. The point F is the midpoint of BD and FC = a.

1. Find, in terms of \(a\), the exact distance of the centre of mass of the lamina from AE. [4]

The lamina is freely suspended from B and hangs in equilibrium.

1. Find the angle between AB and the downward vertical. [2]

A particle P of mass 4 kilograms moves in such a way that its position vector at time \(t\) seconds is \(\mathbf{r}\) metres, where $$\mathbf{r} = 3t\mathbf{i} + 2e^{-3t}\mathbf{j}.$$

1. Find the initial kinetic energy of P. [4]
2. Find the time when the acceleration of P is 2 metres per second squared. [3]

\includegraphics{figure_6} The rim of a smooth hemispherical bowl is a circle of centre O and radius \(a\). The bowl is fixed with its rim horizontal and uppermost. A particle P of mass \(m\) is released from rest at a point A on the rim as shown in Fig. 6. When P reaches the lowest point of the bowl it collides directly with a stationary particle Q of mass \(\frac{1}{2}m\). After the collision Q just reaches the rim of the bowl. Find the coefficient of restitution between P and Q. [7]

In this question you must show detailed reasoning. \includegraphics{figure_7} Fig. 7 shows the curve with equation \(y = \frac{2}{3}\ln x\). The region R, shown shaded in Fig. 7, is bounded by the curve and the lines \(x = 0\), \(y = 0\) and \(y = \ln 2\). A uniform solid of revolution is formed by rotating the region R completely about the \(y\)-axis. Find the exact \(x\)-coordinate of the centre of mass of the solid. [8]

A car of mass 800kg travels up a line of greatest slope of a straight road inclined at \(5°\) to the horizontal. The power developed by the car is constant and equal to 25kW. The resistance to the motion of the car is constant and equal to 750N. The car passes through a point A on the road with speed \(7\)ms\(^{-1}\).

1. Find
   [5]

The car later passes through a point B on the road where AB = 131m. The time taken to travel from A to B is 10.4s.

1. Calculate the speed of the car at B. [6]

\includegraphics{figure_9} A particle P of mass \(m\) is joined to a fixed point O by a light inextensible string of length \(l\). P is released from rest with the string taut and making an acute angle \(\alpha\) with the downward vertical, as shown in Fig. 9. At a time \(t\) after P is released the string makes an angle \(\theta\) with the downward vertical and the tension in the string is \(T\). Angles \(\alpha\) and \(\theta\) are measured in radians.

1. Show that $$\left(\frac{\mathrm{d}\theta}{\mathrm{d}t}\right)^2 = \frac{2g}{l}\cos\theta + k_1,$$ where \(k_1\) is a constant to be determined in terms of \(g\), \(l\) and \(\alpha\). [4]
2. Show that $$T = 3mg\cos\theta + k_2,$$ where \(k_2\) is a constant to be determined in terms of \(m\), \(g\) and \(\alpha\). [3]

It is given that \(\alpha\) is small enough for \(\alpha^2\) to be negligible.

1. Find, in terms of \(m\) and \(g\), the approximate tension in the string. [2]
2. Show that the motion of P is approximately simple harmonic. [3]

A particle P, of mass \(m\), moves on a rough horizontal table. P is attracted towards a fixed point O on the table by a force of magnitude \(\frac{kmg}{x^2}\), where \(x\) is the distance OP. The coefficient of friction between P and the table is \(\mu\). P is initially projected in a direction directly away from O. The velocity of P is first zero at a point A which is a distance \(a\) from O.

1. Show that the velocity \(v\) of P, when P is moving away from O, satisfies the differential equation $$\frac{\mathrm{d}}{\mathrm{d}x}(v^2) + \frac{2kg}{x^2} + 2\mu g = 0.$$ [3]
2. Verify that $$v^2 = 2gk\left(\frac{1}{x} - \frac{1}{a}\right) + 2\mu g(a-x).$$ [3]
3. Find, in terms of \(k\) and \(a\), the range of values of \(\mu\) for which P remains at A. [2]

Two uniform smooth spheres A and B have equal radii and are moving on a smooth horizontal surface. The mass of A is 0.2kg and the mass of B is 0.6kg. The spheres collide obliquely. When the spheres collide the line joining their centres is parallel to \(\mathbf{i}\). Immediately before the collision the velocity of A is \(\mathbf{u}_A\)ms\(^{-1}\) and the velocity of B is \(\mathbf{u}_B\)ms\(^{-1}\). The coefficient of restitution between A and B is 0.5. Immediately after the collision the velocity of A is \((-4\mathbf{i} + 2\mathbf{j})\)ms\(^{-1}\) and the velocity of B is \((2\mathbf{i} + 3\mathbf{j})\)ms\(^{-1}\).

1. Find \(\mathbf{u}_A\) and \(\mathbf{u}_B\). [7]

After the collision B collides with a smooth vertical wall which is parallel to \(\mathbf{j}\). The loss in kinetic energy of B caused by the collision with the wall is 1.152J.

1. Find the coefficient of restitution between B and the wall. [3]
2. Find the angle through which the direction of motion of B is deflected as a result of the collision with the wall. [4]

\includegraphics{figure_12} The ends of a light inextensible string are fixed to two points A and B in the same vertical line, with A above B. The string passes through a small smooth ring of mass \(m\). The ring is fastened to the string at a point P. When the string is taut the angle APB is a right angle, the angle BAP is \(\theta\) and the perpendicular distance of P from AB is \(r\). The ring moves in a horizontal circle with constant angular velocity \(\omega\) and the string taut as shown in Fig. 12.

1. By resolving horizontally and vertically, show that the tension in the part of the string BP is \(m(r\omega^2\cos\theta - g\sin\theta)\). [6]
2. Find a similar expression, in terms of \(r\), \(\omega\), \(m\), \(g\) and \(\theta\), for the tension in the part of the string AP. [2]

It is given that AB = 5a and AP = 4a.

1. Show that \(16a\omega^2 > 5g\). [3]

The ring is now free to move on the string but remains in the same position on the string as before. The string remains taut and the ring continues to move in a horizontal circle.

1. Find the period of the motion of the ring, giving your answer in terms of \(a\), \(g\) and \(\pi\). [5]

\includegraphics{figure_13} A step-ladder has two sides AB and AC, each of length \(4a\). Side AB has weight \(W\) and its centre of mass is at the half-way point; side AC is light. The step-ladder is smoothly hinged at A and the two parts of the step-ladder, AB and AC, are connected by a light taut rope DE, where D is on AB, E is on AC and AD = AE = \(a\). A man of weight \(4W\) stands at a point F on AB, where BF = \(x\). The system is in equilibrium with B and C on a smooth horizontal floor and the sides AB and AC are each at an angle \(\theta\) to the vertical, as shown in Fig. 13.

1. By taking moments about A for side AB of the step-ladder and then for side AC of the step-ladder show that the tension in the rope is $$W\left(1 + \frac{2x}{a}\right)\tan\theta.$$ [7]

The rope is elastic with natural length \(\frac{1}{2}a\) and modulus of elasticity \(W\).

1. Show that the condition for equilibrium is that $$x = \frac{1}{2}a(8\cos\theta - \cot\theta - 1).$$ [5]

In this question you must show detailed reasoning.

1. Hence determine, in terms of \(a\), the maximum value of \(x\) for which equilibrium is possible. [5]

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