科目9231-further-mathematicsPaper 3 — Further Mechanics考前速记本页总览考前速记 Recommended考前 15 分钟阅读,重点回忆公式和易错点。 核心公式 Projectile Motion x=ucosθ⋅ty=usinθ⋅t−12gt2y=xtanθ−gx22u2cos2θ\begin{aligned} x &= u\cos\theta \cdot t \\ y &= u\sin\theta \cdot t - \frac{1}{2}gt^2 \\ y &= x\tan\theta - \frac{gx^2}{2u^2\cos^2\theta} \end{aligned}xyy=ucosθ⋅t=usinθ⋅t−21gt2=xtanθ−2u2cos2θgx2 Equilibrium of Rigid Body 力矩 M=r×F\mathbf{M} = \mathbf{r} \times \mathbf{F}M=r×F,大小 M=FdM = FdM=Fd 平衡条件:∑Fx=0, ∑Fy=0, ∑M=0\sum F_x = 0,\ \sum F_y = 0,\ \sum M = 0∑Fx=0, ∑Fy=0, ∑M=0 Circular Motion a=v2r=rω2,F=mv2ra = \frac{v^2}{r} = r\omega^2,\quad F = m\frac{v^2}{r}a=rv2=rω2,F=mrv2 Hooke's Law T=λxL,EPE=λx22LT = \frac{\lambda x}{L},\quad \text{EPE} = \frac{\lambda x^2}{2L}T=Lλx,EPE=2Lλx2 Linear Motion Under Variable Force F=ma=mdvdt=mvdvdxF = ma = m\frac{dv}{dt} = mv\frac{dv}{dx}F=ma=mdtdv=mvdxdv Momentum m1u1+m2u2=m1v1+m2v2,e=v2−v1u1−u2m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2,\quad e = \frac{v_2 - v_1}{u_1 - u_2}m1u1+m2u2=m1v1+m2v2,e=u1−u2v2−v1 易错清单 问题注意事项抛体最高点vy=0v_y = 0vy=0,但 vxv_xvx 不为零刚体力矩使用垂直距离(不是斜边)圆周运动最顶点速度最小值对应 mgmgmg 提供向心力弹性绳松弛T=0T = 0T=0 当 x≤0x \leq 0x≤0,EPE = 0变力方向注意正负号与运动方向的关系斜碰分别沿法线和切线方向处理