科目：进阶数学 Further Mathematics (9231)
年份：2023 冬季
试卷：Paper 32
总页数：14
章节：9231 Further Mathematics
1. 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]2 A particle $P$ of mass 0.5 kg moves in a straight line. At time $t$ is the velocity of $P$ is $v \text{ m/s}^{-1}$ and its displacement from a fixed point $O$ on the line is $x\text{ m}$. The only forces acting on $P$ are a force of magnitude $\frac{150}{(x+1)^2}\text{ N}$ in the direction of increasing displacement and a resistive force of magnitude $\frac{450}{(x+1)^3}\text{ N}$. 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]A uniform lamina is in the form of an isosceles triangle $ABC$ in which $AC = 2a$ and angle $ABC = 90^\circ$. 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). (a) 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 $\tan \theta = \frac{2}{18}$.
(b) Find the value of $k$. [3]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 $\frac{1}{2}$.
(a) Find the value of $\tan \theta$. [4](b) Find the percentage loss in the total kinetic energy of the spheres as a result of this collision. [3]
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 $\cos \alpha = \frac{3}{2}$. 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 $A$.
(a) Find, in terms of $m$ and $g$, the magnitude of the reaction at $B$. [6](b) Given that $v_A = \sqrt{kag}$, find the value of $k$. [2]6 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.
(a) 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 $u = 40\sqrt{2} \text{ ms}^{-1}$, $P$ just clears the obstacle. When $u = 40 \text{ ms}^{-1}$, $P$ only achieves 80% of the height required to clear the obstacle.
(b) Find the two possible values of $h$.
[6]11 © UCLES 2023 9231/32/O/N/23 [Turn over]
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.
(a) Find the value of $k$. [5](b) Find $F$ in terms of $m$ and $g$. [2] (c) Find, in terms of $m$ and $g$, the tension in the rod immediately before it is released. [2]
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