M1 · Mechanics 1

Mechanics 1 (M1, paper WME01/01, 1 hour 30 minutes, 75 marks, first assessment June 2019) is the applied unit of International AS Mathematics: P1 and P2 are assumed. Take $g = 9.8\ \mathrm{m\,s^{-2}}$ unless told otherwise, and give inexact answers to 2 or 3 significant figures. This reference covers modelling assumptions, vectors in mechanics, constant-acceleration kinematics, Newton's laws with connected particles, momentum and impulse, friction, statics and moments.

Must-know for M1 (not in the formula booklet — memorise):

The words in a stem are instructions. A particle 粒子 has mass but no size; a lamina 薄片 has area but no thickness; a rod 杆 has length but no thickness; a rigid body 刚体 keeps its shape. Light means massless (a light rope transmits tension unchanged), uniform means the centre of mass sits at the midpoint, an inextensible string 不可伸长绳 keeps both ends moving with the same speed, and a smooth surface 光滑表面 versus a rough surface 粗糙表面 decide whether friction acts. A light smooth pulley 滑轮 changes a tension's direction, not its size. State the assumption you use — "smooth, so no friction" — because the marks are in the modelling as much as the algebra.

Displacements, velocities, accelerations and forces are vectors: $(3\mathbf i - 4\mathbf j)$ has magnitude $5$ and direction $\tan^{-1}(-4/3)$ from the positive $x$-axis. Resolve into components, work component-wise, and combine at the end. The resultant 合力 of $\begin{pmatrix}a_1\\b_1\end{pmatrix} + \begin{pmatrix}a_2\\b_2\end{pmatrix}$ is $\begin{pmatrix}a_1+a_2\\b_1+b_2\end{pmatrix}$. With $\mathbf i$ east and $\mathbf j$ north, a bearing 方位角 is measured clockwise from north: the vector $-5\mathbf i - 12\mathbf j$ points south-west, at $\tan^{-1}(5/12) = 22.6°$ west of due south, so its bearing is $180 + 23 = 203°$.

Worked example — position vectors and bearings (WME01 June 2025 Q2). Particle A has $\mathbf r_A(2) = (-5\mathbf i + 16\mathbf j)$ and $\mathbf r_A(5) = (10\mathbf i + 4\mathbf j)$. Velocity $= \dfrac{\Delta\mathbf r}{\Delta t} = \dfrac{15\mathbf i - 12\mathbf j}3 = 5\mathbf i - 4\mathbf j$, so $\mathbf r_A(t) = (-5\mathbf i + 16\mathbf j) + (t-2)(5\mathbf i - 4\mathbf j) = (5t-15)\mathbf i + (24-4t)\mathbf j$. With B at $\mathbf s_B = -5\mathbf i + 7\mathbf j + t(2\mathbf i - 3\mathbf j)$, at $t = 5$: $\overrightarrow{AB} = (5\mathbf i - 8\mathbf j) - (10\mathbf i + 4\mathbf j) = -5\mathbf i - 12\mathbf j$, bearing $203°$.

Constant acceleration means the suvat formulae apply once per phase; the discipline is signs. Choose positive, write $s, u, v, a, t$ with signs for each phase, and pick the formula that avoids the unknown you do not want. Graphs: on a velocity–time graph the gradient is acceleration and the area is displacement (area below the axis counts negative); a speed–time graph cannot go negative, so upward motion then downward motion appears as a V.

Worked example — up and down (WME01 June 2025 Q3). A ball is projected vertically up from A, 12 m above the ground at B, and takes 4 s to reach B. Taking up positive with $A$ as origin, $s = -12$: $-12 = 4U - 4.9(16)$, so $U = 16.6\ \mathrm{m\,s^{-1}}$ (2 s.f. 17). Speed at the ground: $v^2 = 16.6^2 + 2(9.8)(12) = 510.8$, so $v = 22.6\ \mathrm{m\,s^{-1}}$. The speed–time graph is a V: falling from 16.6 to zero at the top (after $16.6/9.8 = 1.7$ s), then rising to 22.6 at $t = 4$ — the downward leg is longer and steeper.

Newton's second law, applied one way along a line: $F_{\text{net}} = ma$ with every force resolved and signed. For two connected particles on a light inextensible string, write an equation of motion for EACH body (same tension, same speed magnitude), then eliminate $T$. On an inclined plane resolve along and perpendicular to the slope; $\tan\theta = 3/4$ means $\sin\theta = 3/5$, $\cos\theta = 4/5$.

Worked example — two ropes on a barge (WME01 June 2025 Q4). Forces of 800 N at $25°$ above the line of motion and 500 N at $\alpha$ below it pull a barge straight along the canal. Perpendicular to the motion there is no acceleration: $800\sin25° = 500\sin\alpha$, so $\sin\alpha = 0.676$ and $\alpha = 43°$. Along the motion, with mass 15000 kg and resistance 750 N: $800\cos25° + 500\cos\alpha - 750 = 15000a$, giving $a = 0.023\ \mathrm{m\,s^{-2}}$ — three-decimal answers are expected when the numbers are this small.

Worked example — pulley, start to finish (WME01 June 2025 Q8). Balls P (mass $kM$) and Q (mass $3M$, $k < 3$) hang over a fixed light smooth pulley and accelerate at $\tfrac g5$. Equations of motion: $T - kMg = kM\cdot\tfrac g5$ (P up) and $3Mg - T = 3M\cdot\tfrac g5$ (Q down). Eliminating $T$: $3Mg - kMg = \tfrac{g}5(3M + kM)$, so $5(3-k) = 3+k$ and $k = 2$. The string pulls the pulley down with $2T$: with $M = 0.5$ kg, $T = 1(9.8 + 1.96) = 11.76$ N, so the force is $23.5$ N. When Q lands after falling 2.5 m its speed is $\sqrt{2\cdot\frac g5\cdot2.5} = \sqrt{g} = 3.13$; if it rebounds at $0.4\ \mathrm{m\,s^{-1}}$, the impulse on Q is $1.5(3.13 + 0.4) = 5.3$ N s. P then coasts upward against gravity alone for $\tfrac{\sqrt g}g = 0.32$ s after a $5/\sqrt g = 1.60$ s descent phase: total time to rest $\approx 1.9$ s.

Momentum $mv$ and impulse 冲量 $I = mv - mu$ are signed along the line. Newton's third law makes the impulse exerted on each of two colliding bodies equal in magnitude and opposite in direction — use one of them, not both. For a direct collision, conservation of momentum $m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$ needs consistent signs; restitution is NOT in M1.

Worked example — impulse on both bodies (WME01 June 2025 Q1). P (mass $km$, speed $5u$) catches Q (mass $3m$, speed $2u$); afterwards P still moves forward at $2u$ and Q at $v$. The impulse on Q is $4.5mu$: $3m(v - 2u) = 4.5mu$, so $v = 3.5u$. The impulse on P is equal and opposite: $km(5u - 2u) = 4.5mu$, so $k = 1.5$.

On a rough surface the contact force splits into the normal reaction 法向反力 $R$ (perpendicular) and friction $F$ (along, opposing motion or its tendency). Moving: $F = \mu R$ with $\mu$ the coefficient of friction 摩擦系数. In equilibrium $F \le \mu R$, with equality at limiting equilibrium — the phrase "on the point of moving" licenses $F = \mu R$.

Worked example — a package on a rough slope (WME01 June 2025 Q6). With $\tan\theta = \tfrac34$ and $\mu = \tfrac14$: perpendicular, $R = mg\cos\theta = \tfrac45mg$; along (up positive, decelerating): $-mg\sin\theta - \tfrac14R = ma$, so $a = -g(\tfrac35 + \tfrac15) = -\tfrac45g$ — the deceleration is $\tfrac45g$, and the answer must stay as an exact fraction of $g$. Held at rest on the slope by a force $X$ up the plane (mass 2 kg): $X + \tfrac14(2g\cdot\tfrac45) = 2g\cdot\tfrac35$, so $X = \tfrac45g = 7.84$ N.

A particle is in equilibrium 平衡 when the resultant force is zero: resolve in two perpendicular directions and set both totals to zero, or draw a closed vector triangle. Weight $mg$ down, normal reaction perpendicular to the surface, tension 张力 along a string away from the body (thrust along a rod pushes), friction along the surface opposing slipping. On rough ground at rest, $F$ is whatever closes the polygon, capped at $\mu R$.

Worked example — forces in equilibrium (original). A particle of weight 50 N hangs from two light strings at $40°$ and $60°$ to the vertical. Resolving horizontally: $T_1\sin40° = T_2\sin60°$; vertically: $T_1\cos40° + T_2\cos60° = 50$. From the first, $T_2 = 0.742T_1$; then $T_1(0.766 + 0.371) = 50$, so $T_1 = 44$ N and $T_2 = 33$ N (2 s.f.).

The moment of a force 力矩 about a point is force × perpendicular distance, with a sign per sense. A rigid body in equilibrium has zero resultant moment about ANY point — so pick the pivot that kills the most unknowns. "On the point of tilting about X" means the reaction elsewhere has dropped to zero and moments are taken about X.

Worked example — a rod on a ledge (WME01 June 2025 Q5). A non-uniform rod AB of length $3a$ rests on a horizontal ledge with the edge at C, $AC = 2a$, centre of mass $d$ from A. An upward force $P$ at B tilts it about A: moments about A give $3aP = Wd$. A downward force $1.25P$ at B tilts it about C: $1.25Pa = W(2a - d)$. Substituting $P = \tfrac{Wd}{3a}$ (in units of $a$): $1.25\cdot\tfrac d3 = 2 - d$, so $1.25d = 6a - 3d$, giving $4.25d = 6a$ and $d = \tfrac{24a}{17}$. The lesson: each tilting case has its own pivot, and only the weight's moment survives unchanged between them.

• Name the modelling assumptions in play (particle, light rod, inextensible string, smooth/rough) and say what each buys you.
• Work with position and velocity vectors component-wise; convert directions to bearings; find relative positions at a common time.
• Solve one-phase and multi-phase suvat problems with signs fixed per phase; read gradients and areas off motion graphs; sketch the speed-time V for up-then-down motion.
• Apply $F = ma$ per body along the line of motion; handle connected particles over pulleys with paired equations; find the force on a pulley as $2T$.
• Use impulse-momentum on ONE body of a collision pair; conserve momentum with consistent signs.
• Split contact forces into $R$ and $F$; apply $F = \mu R$ when moving or limiting; resolve on inclined planes using $\sin\theta = 3/5$, $\cos\theta = 4/5$ for $\tan\theta = 3/4$.
• Close force polygons for equilibrium in two directions; cap static friction at $\mu R$.
• Take moments about the pivot that removes unknowns; switch pivots when the tilting point changes.