Motion in a Plane formulas

Class 11 physics formula sheet for NEET and JEE: the key equations of NCERT chapter 3, the special cases questions are built on, and diagrams where they help.

35 formulas5 sectionsClass 11 · Chapter 32 of 5 sections free

By Sreeraj P, M.Sc Physics · 10+ years teaching NEET and JEE

Most used formulasOther formulas and cases

1Vectors

$$R=\sqrt{P^2+Q^2+2PQ\cos\theta},\qquad \tan\alpha=\frac{Q\sin\theta}{P+Q\cos\theta}$$

$$\begin{array}{l}\displaystyle R=\sqrt{P^2+Q^2+2PQ\cos\theta}\\[6pt]\displaystyle \tan\alpha=\frac{Q\sin\theta}{P+Q\cos\theta}\end{array}$$

$\alpha$ measured from $\vec P$; resultant leans towards the larger vector. $|P-Q|\le R\le P+Q$. Difference: $|\vec P-\vec Q|=\sqrt{P^2+Q^2-2PQ\cos\theta}$.

Equal vectors $P$ at angle $\theta$	Result
Resultant / difference	$2P\cos\frac\theta2$ (along bisector) / $2P\sin\frac\theta2$
$\theta=60^\circ,90^\circ,120^\circ$	$R=\sqrt3P,\ \sqrt2P,\ P$
$\|\vec A+\vec B\|=n\|\vec A-\vec B\|$	$\theta=2\tan^{-1}\dfrac1n$
$n$ equal forces, each at $2\pi/n$ to the previous	resultant 0 (closed polygon)
Speed $v$ turned through $\theta$ (circular motion)	$\|\Delta\vec v\|=2v\sin\frac\theta2$

$$\begin{array}{l}\displaystyle \hat A=\frac{\vec A}{A},\quad A=\sqrt{A_x^2+A_y^2+A_z^2}\\[5pt]\displaystyle \cos^2\alpha+\cos^2\beta+\cos^2\gamma=1\end{array}$$

$$\begin{array}{l}\displaystyle \hat A=\frac{\vec A}{A}\\[6pt]\displaystyle A=\sqrt{A_x^2+A_y^2+A_z^2}\\[6pt]\displaystyle \cos^2\alpha+\cos^2\beta+\cos^2\gamma=1\end{array}$$

Direction cosines $\cos\alpha=A_x/A$ etc. ($\sin^2$ sum $=2$). Rotating axes changes components, not magnitude.

$$\vec A\cdot\vec B=AB\cos\theta=A_xB_x+A_yB_y+A_zB_z$$

Commutative; zero if perpendicular. Angle: $\cos\theta=\dfrac{\vec A\cdot\vec B}{AB}$. Component of $\vec A$ along $\vec B$: $\vec A\cdot\hat B$ (vector: $(\vec A\cdot\hat B)\hat B$). Work $\vec F\cdot\vec s$, power $\vec F\cdot\vec v$, flux $\vec B\cdot\vec A$.

$$\vec A\times\vec B=AB\sin\theta\,\hat n=\begin{vmatrix}\hat i&\hat j&\hat k\\A_x&A_y&A_z\\B_x&B_y&B_z\end{vmatrix}$$

$\vec A\times\vec B=-\vec B\times\vec A$; zero if parallel ($A_x/B_x=A_y/B_y=A_z/B_z$). $\hat i\times\hat j=\hat k$ (cyclic). Unit normal $\dfrac{\vec A\times\vec B}{|\vec A\times\vec B|}$. Triangle area $\tfrac12|\vec A\times\vec B|$; parallelogram $|\vec A\times\vec B|$ (from diagonals: $\tfrac12|\vec d_1\times\vec d_2|$). $|\vec A\times\vec B|^2+(\vec A\cdot\vec B)^2=A^2B^2$.

$$\frac{P}{\sin\alpha}=\frac{Q}{\sin\beta}=\frac{R}{\sin\gamma}$$

Lami's theorem: three concurrent forces in equilibrium, each angle opposite the force ($\alpha$ between $Q$ and $R$). Diagonals of a parallelogram: $R^2+S^2=2(P^2+Q^2)$. Angle between body diagonals of a cube: $\cos^{-1}\frac13$.

2Motion in 2-D and relative velocity

$$\vec v=\frac{d\vec r}{dt},\quad \vec a=\frac{d\vec v}{dt},\quad \vec v=\vec u+\vec at,\quad \vec r=\vec r_0+\vec ut+\tfrac12\vec at^2$$

$$\begin{array}{l}\displaystyle \vec v=\frac{d\vec r}{dt}\\[6pt]\displaystyle \vec a=\frac{d\vec v}{dt}\\[6pt]\displaystyle \vec v=\vec u+\vec at\\[6pt]\displaystyle \vec r=\vec r_0+\vec ut+\tfrac12\vec at^2\end{array}$$

Apply separately to $x$ and $y$. Speed rises if angle between $\vec v$ and $\vec a$ is acute, falls if obtuse, constant if 90°.

Collision of two particles: $\vec r_1-\vec r_2$ must be parallel to $\vec v_2-\vec v_1$. Bead sliding along any chord from the top of a vertical circle: $t=2\sqrt{R/g}$ (same for all chords).
Rod sliding on wall and floor: $v_B=v_A\cot\theta$ (differentiate $x^2+y^2=l^2$).

$$\vec v_{AB}=\vec v_A-\vec v_B,\qquad d_{\min}=d\sin\phi,\ \ t=\frac{d\cos\phi}{v_{\text{rel}}}$$

$$\begin{array}{l}\displaystyle \vec v_{AB}=\vec v_A-\vec v_B\\[6pt]\displaystyle d_{\min}=d\sin\phi,\ \ t=\frac{d\cos\phi}{v_{\text{rel}}}\end{array}$$

Closest approach: separation $d$, $\phi$ = angle between line joining them and $\vec v_{\text{rel}}$. Escalator: walk time $t_1$, ride time $t_2$ → both: $\dfrac{t_1t_2}{t_1+t_2}$.

$$\vec v_{RM}=\vec v_R-\vec v_M,\qquad \tan\theta=\frac{v_M}{v_R}\ (\text{umbrella from vertical})$$

$$\begin{array}{l}\displaystyle \vec v_{RM}=\vec v_R-\vec v_M\\[6pt]\displaystyle \tan\theta=\frac{v_M}{v_R}\ (\text{umbrella from vertical})\end{array}$$

Rain falling vertically, man walking: tilt umbrella forward. With wind: add wind to rain first. Rain appears vertical at speed $v_1$ and at 45° at $v_2$: horizontal part of rain $=v_1$, vertical part $=v_2-v_1$.

3 more sections and 19 formulas in the full chapter

3River crossing1 case table · 1 diagram
4Projectile motion11 formulas · 1 case table · 1 diagram
5Circular motion (kinematics)8 formulas

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