CAIE Further Paper 3 (Further Paper 3) 2023 November

One end of a light inextensible string of length \(a\) is attached to a fixed point \(O\). The other end of the string is attached to a particle of mass \(m\). The string is taut and makes an angle \(\theta\) with the downward vertical through \(O\), where \(\cos \theta = \frac{2}{3}\). The particle moves in a horizontal circle with speed \(v\). Find \(v\) in terms of \(a\) and \(g\). [4]

A particle \(P\) of mass \(0.5\) kg moves in a straight line. At time \(t\) s the velocity of \(P\) is \(v\) m s\(^{-1}\) and its displacement from a fixed point \(O\) on the line is \(x\) m. The only forces acting on \(P\) are a force of magnitude \(\frac{150}{(x+1)^2}\) N in the direction of increasing displacement and a resistive force of magnitude \(\frac{450}{(x+1)^3}\) N. When \(t = 0\), \(x = 0\) and \(v = 20\). Find \(v\) in terms of \(x\), giving your answer in the form \(v = \frac{Ax + B}{(x + 1)}\), where \(A\) and \(B\) are constants to be determined. [6]

\includegraphics{figure_3} A uniform lamina is in the form of an isosceles triangle \(ABC\) in which \(AC = 2a\) and angle \(ABC = 90°\). The point \(D\) on \(AB\) is such that the ratio \(DB : AB = 1 : k\). The point \(E\) on \(CB\) is such that \(DE\) is parallel to \(AC\). The triangle \(DBE\) is removed from the lamina (see diagram).

1. Find, in terms of \(k\), the distance of the centre of mass of the remaining lamina \(ADEC\) from the midpoint of \(AC\). [4]

When the lamina \(ADEC\) is freely suspended from the vertex \(A\), the edge \(AC\) makes an angle \(\theta\) with the downward vertical, where \(\tan \theta = \frac{2}{15}\).

1. Find the value of \(k\). [3]

\includegraphics{figure_4} Two smooth vertical walls meet at right angles. The smooth sphere \(A\), with mass \(m\), is at rest on a smooth horizontal surface and is at a distance \(d\) from each wall. An identical smooth sphere \(B\) is moving on the horizontal surface with speed \(u\) at an angle \(\theta\) with the line of centres when the spheres collide (see diagram). After the collision, the spheres take the same time to reach a wall. The coefficient of restitution between the spheres is \(\frac{1}{2}\).

1. Find the value of \(\tan \theta\). [4]
2. Find the percentage loss in the total kinetic energy of the spheres as a result of this collision. [3]

\includegraphics{figure_5} A bead of mass \(m\) moves on a smooth circular wire, with centre \(O\) and radius \(a\), in a vertical plane. The bead has speed \(v_A\) when it is at the point \(A\) where \(OA\) makes an angle \(\alpha\) with the downward vertical through \(O\), and \(\cos \alpha = \frac{2}{3}\). Subsequently the bead has speed \(v_B\) at the point \(B\), where \(OB\) makes an angle \(\theta\) with the upward vertical through \(O\). Angle \(AOB\) is a right angle (see diagram). The reaction of the wire on the bead at \(B\) is in the direction \(OB\) and has magnitude equal to \(\frac{1}{6}\) of the magnitude of the reaction when the bead is at \(A\).

1. Find, in terms of \(m\) and \(g\), the magnitude of the reaction at \(B\). [6]
2. Given that \(v_A = \sqrt{kag}\), find the value of \(k\). [2]

A particle \(P\) is projected with speed \(u\) at an angle \(\alpha\) 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. Derive the equation of the trajectory of \(P\) in the form $$y = x \tan \alpha - \frac{gx^2}{2u^2} \sec^2 \alpha.$$ [3]

During its flight, \(P\) must clear an obstacle of height \(h\) m that is at a horizontal distance of \(32\) m from the point of projection. When \(u = 40\sqrt{2}\) m s\(^{-1}\), \(P\) just clears the obstacle. When \(u = 40\) m s\(^{-1}\), \(P\) only achieves \(80\%\) of the height required to clear the obstacle.

1. Find the two possible values of \(h\). [6]

\includegraphics{figure_7} A particle \(P\) of mass \(m\) is attached to one end of a light rod of length \(3a\). The other end of the rod is able to pivot smoothly about the fixed point \(A\). The particle is also attached to one end of a light spring of natural length \(a\) and modulus of elasticity \(kmg\). The other end of the spring is attached to a fixed point \(B\). The points \(A\) and \(B\) are in a horizontal line, a distance \(5a\) apart, and these two points and the rod are in a vertical plane. Initially, \(P\) is held in equilibrium by a vertical force \(F\) with the stretched length of the spring equal to \(4a\) (see diagram). The particle is released from rest in this position and has a speed of \(\frac{6}{5}\sqrt{2ag}\) when the rod becomes horizontal.

1. Find the value of \(k\). [5]
2. Find \(F\) in terms of \(m\) and \(g\). [2]
3. Find, in terms of \(m\) and \(g\), the tension in the rod immediately before it is released. [2]