CAIE Further Paper 3 (Further Paper 3) 2024 June

\includegraphics{figure_1} Two smooth uniform spheres \(A\) and \(B\) of equal radii have masses \(m\) and \(2m\) respectively. The two spheres are moving on a smooth horizontal surface when they collide with speeds \(u\) and \(\frac{1}{2}u\) respectively. Immediately before the collision, \(A\)'s direction of motion is along the line of centres, and \(B\)'s direction of motion makes an angle \(\theta\) with the line of centres (see diagram). As a result of the collision, the direction of motion of \(A\) is reversed and its speed is reduced to \(\frac{1}{4}u\). The direction of motion of \(B\) again makes an angle \(\theta\) with the line of centres, but on the opposite side of the line of centres. The speed of \(B\) is unchanged. Find the value of the coefficient of restitution between the spheres. [4]

A particle \(P\) of mass \(m\) is attached to one end of a light elastic string of natural length \(a\) and modulus of elasticity \(2mg\). A particle \(Q\) of mass \(km\) is attached to the other end of the string. Particle \(P\) lies on a smooth horizontal table. The string passes through a small smooth hole \(H\) in the table and then passes through a small smooth hole \(H\) in the table. Particle \(P\) moves in a horizontal circle on the surface of the table with constant speed \(\sqrt{\frac{1}{3}ga}\). Particle \(Q\) hangs in equilibrium vertically below the hole with \(HQ = \frac{1}{4}a\).

1. Find, in terms of \(a\), the extension in the string. [4]
2. Find the value of \(k\). [2]

A particle \(P\) 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 \(O\). When the particle is hanging vertically below \(O\), it is projected horizontally with speed \(u\) so that it begins to move along a circular path. When \(P\) is at the lowest point of its motion, the tension in the string is \(T\). When \(OP\) makes an angle \(\theta\) with the upward vertical, the tension in the string is \(S\).

1. Show that \(S = T - 3mg(1 + \cos\theta)\). [5]
2. Given that \(u = \sqrt{4ag}\), find the value of \(\cos\theta\) when the string goes slack. [2]

\includegraphics{figure_4} A light spring of natural length \(a\) and modulus of elasticity \(kmg\) is attached to a fixed point \(O\) on a smooth plane inclined to the horizontal at an angle \(\theta\), where \(\sin\theta = \frac{1}{4}\). A particle of mass \(m\) is attached to the lower end of the spring and is held at the point \(A\) on the plane, where \(OA = 2a\) and \(OA\) is along a line of greatest slope of the plane (see diagram). The particle is released from rest and is moving with speed \(V\) when it passes through the point \(B\) on the plane, where \(OB = \frac{3}{2}a\). The speed of the particle is \(\frac{1}{3}V\) when it passes through the point \(C\) on the plane, where \(OC = \frac{3}{4}a\). Find the value of \(k\). [7]

\includegraphics{figure_5} A uniform lamina is in the form of a triangle \(OBC\), with \(OC = 18a\), \(OB = 24a\) and angle \(COB = 90°\). The point \(A\) on \(OB\) is such that \(OA = x\) (see diagram). The triangle \(OAC\) is removed from the lamina.

1. Find, in terms of \(a\) and \(x\), the distance of the centre of mass of the remaining object \(ABC\) from \(OC\). [3]

The object \(ABC\) is suspended from \(C\). In its equilibrium position, the side \(AB\) makes an angle \(\theta\) with the vertical, where \(\tan\theta = \frac{8}{5}\).

1. Find \(x\) in terms of \(a\). [4]

A particle \(P\) is projected with speed \(u\text{ ms}^{-1}\) at an angle \(\theta\) above the horizontal from a point \(O\) and moves freely under gravity. After 5 seconds the speed of \(P\) is \(\frac{3}{4}u\).

1. Show that \(\frac{7}{16}u^2 - 100u\sin\theta + 2500 = 0\). [3]
2. It is given that the velocity of \(P\) after 5 seconds is perpendicular to the initial velocity. Find, in either order, the value of \(u\) and the value of \(\sin\theta\). [5]

A parachutist of mass \(m\) kg opens his parachute when he is moving vertically downwards with a speed of \(50\text{ ms}^{-1}\). At time \(t\) s after opening his parachute, he has fallen a distance \(x\) m from the point where he opened his parachute, and his speed is \(v\text{ ms}^{-1}\). The forces acting on him are his weight and a resistive force of magnitude \(mv\) N.

1. Find an expression for \(v\) in terms of \(t\). [6]
2. Find an expression for \(x\) in terms of \(t\). [3]
3. Find the distance that the parachutist has fallen, since opening his parachute, when his speed is \(15\text{ ms}^{-1}\). [2]