CAIE Further Paper 3 (Further Paper 3) 2021 June

\includegraphics{figure_1} A uniform lamina \(ABCD\) consists of two isosceles triangles \(ABD\) and \(BCD\). The diagonals of \(ABCD\) meet at the point \(O\). The length of \(AO\) is \(3a\), the length of \(OC\) is \(6a\) and the length of \(BD\) is \(16a\) (see diagram). Find the distance of the centre of mass of the lamina from \(DB\). [3]

One end of a light elastic string of natural length \(0.8\) m and modulus of elasticity \(36\) N is attached to a fixed point \(O\) on a smooth plane. The plane is inclined at an angle \(\alpha\) to the horizontal, where \(\sin \alpha = \frac{3}{5}\). A particle \(P\) of mass \(2\) kg is attached to the other end of the string. The string lies along a line of greatest slope of the plane with the particle below the level of \(O\). The particle is projected with speed \(\sqrt{2}\) m s\(^{-1}\) directly down the plane from the position where \(OP\) is equal to the natural length of the string. Find the maximum extension of the string during the subsequent motion. [5]

\includegraphics{figure_3} Particles \(A\) and \(B\), of masses \(3m\) and \(m\) respectively, are connected by a light inextensible string of length \(a\) that passes through a fixed smooth ring \(R\). Particle \(B\) hangs in equilibrium vertically below the ring. Particle \(A\) moves in horizontal circles on a smooth horizontal surface with speed \(\frac{2}{3}\sqrt{ga}\). The angle between \(AR\) and \(BR\) is \(\theta\) (see diagram). The normal reaction between \(A\) and the surface is \(\frac{15}{2}mg\).

1. Find \(\cos \theta\). [3]

1. Find, in terms of \(a\), the distance of \(B\) below the ring. [3]

\includegraphics{figure_4} A particle of mass \(m\) is attached to one end of a light inextensible string of length \(a\). The other end of the string is attached to a fixed point \(O\). The particle is initially held with the string taut at the point \(A\), where \(OA\) makes an angle \(\theta\) with the downward vertical through \(O\). The particle is then projected with speed \(u\) perpendicular to \(OA\) and begins to move upwards in part of a vertical circle. The string goes slack when the particle is at the point \(B\) where angle \(AOB\) is a right angle. The speed of the particle when it is at \(B\) is \(\frac{1}{2}u\) (see diagram). Find the tension in the string at \(A\), giving your answer in terms of \(m\) and \(g\). [8]

A particle \(P\) of mass \(m\) kg is projected vertically upwards from a point \(O\), with speed \(20\) m s\(^{-1}\), and moves under gravity. There is a resistive force of magnitude \(2mv\) N, where \(v\) m s\(^{-1}\) is the speed of \(P\) at time \(t\) s after projection.

1. Find an expression for \(v\) in terms of \(t\), while \(P\) is moving upwards. [6]

The displacement of \(P\) from \(O\) is \(x\) m at time \(t\) s.

1. Find an expression for \(x\) in terms of \(t\), while \(P\) is moving upwards. [2]
2. Find, correct to 3 significant figures, the greatest height above \(O\) reached by \(P\). [2]

\includegraphics{figure_6} Two uniform smooth spheres \(A\) and \(B\) of equal radii have masses \(m\) and \(km\) respectively. Sphere \(A\) is moving with speed \(u\) on a smooth horizontal surface when it collides with sphere \(B\) which is at rest. Immediately before the collision, \(A\)'s direction of motion makes an angle \(\theta\) with the line of centres (see diagram). The coefficient of restitution between the spheres is \(\frac{1}{3}\).

1. Show that the speed of \(B\) after the collision is \(\frac{4u \cos \theta}{3(1 + k)}\). [3]

70% of the total kinetic energy of the spheres is lost as a result of the collision.

1. Given that \(\tan \theta = \frac{1}{3}\), find the value of \(k\). [6]

A particle \(P\) is projected with speed \(u\) at an angle \(\theta\) above the horizontal from a point \(O\) on a horizontal plane and moves freely under gravity. The horizontal and vertical displacements of \(P\) from \(O\) at a subsequent time \(t\) are denoted by \(x\) and \(y\) respectively.

1. Use the equation of the trajectory given in the List of formulae (MF19), together with the condition \(y = 0\), to establish an expression for the range \(R\) in terms of \(u\), \(\theta\) and \(g\). [2]
2. Deduce an expression for the maximum height \(H\), in terms of \(u\), \(\theta\) and \(g\). [2]

It is given that \(R = \frac{4H}{\sqrt{3}}\).

1. Show that \(\theta = 60°\). [1]

It is given also that \(u = \sqrt{40}\) m s\(^{-1}\).

1. Find, by differentiating the equation of the trajectory or otherwise, the set of values of \(x\) for which the direction of motion makes an angle of less than \(45°\) with the horizontal. [4]