Motion in a Plane

In uniform circular motion, the speed is constant but the direction of velocity changes continuously; since velocity is a vector, it is varying. The centripetal acceleration has constant magnitude $\frac{v^2}{r}$ but its direction (towards the centre) keeps changing, so acceleration is also varying. So the particle has varying velocity and varying acceleration.

$h_{max}=\frac{u^2\sin^2\theta}{2g}=\frac{280\times280}{2\times9.8}\times\frac{1}{4}=1000\ m$

$H=\frac{u^2\sin^2\theta}{2g}=\frac{(20)^2\times(1/2)^2}{2\times10}=5$ m.

Given $T_A=T_B=T$. $\omega_A=\frac{2\pi}{T_A}$, $\omega_B=\frac{2\pi}{T_B}$ $\frac{\omega_A}{\omega_B}=\frac{T_B}{T_A}=\frac{T}{T}=1$ Ratio $=1:1$

$x = 5t - 2t^2$, $y = 10t$ $v_x = \frac{dx}{dt} = 5 - 4t$, $v_y = \frac{dy}{dt} = 10$ $a_x = \frac{dv_x}{dt} = -4$, $a_y = \frac{dv_y}{dt} = 0$ Acceleration of the particle at t = 2 s is $-4\ m/s^2$.

$|\vec{P} + \vec{Q}| = |\vec{P} - \vec{Q}|$
Squaring: $P^2 + Q^2 + 2PQ\cos\theta = P^2 + Q^2 - 2PQ\cos\theta$
$\Rightarrow 4PQ\cos\theta = 0 \Rightarrow \cos\theta = 0$
$\theta = \frac{\pi}{2} = 90^\circ$

$\vec v_{av} = \frac{\Delta\vec r}{\Delta t} = \frac{(13-2)\hat i + (14-3)\hat j}{5 - 0} = \frac{11}{5}(\hat i + \hat j)$

B is at the same level as A. The horizontal component of velocity stays the same and the vertical component is reversed.
So the velocity at B is $2\hat i - 3\hat j$ m/s.

Maximum height $H = \frac{u^2\sin^2\theta}{2g}$; range $R = \frac{2u^2\sin\theta\cos\theta}{g}$
H = R: $\frac{u^2\sin^2\theta}{2g} = \frac{2u^2\sin\theta\cos\theta}{g}$
$\sin\theta = 4\cos\theta \Rightarrow \tan\theta = 4$
$\theta = \tan^{-1}(4)$

$\vec v = \vec u + \vec at$
$= (2\hat i + 3\hat j) + (0.3\hat i + 0.2\hat j)\times10$
$= (2 + 3)\hat i + (3 + 2)\hat j = 5\hat i + 5\hat j$
$|\vec v| = \sqrt{5^2 + 5^2} = 5\sqrt2$ units

R = 5 cm = $5\times10^{-2}$ m, T = $0.2\pi$ s.
Angular speed $\omega = \frac{2\pi}{T} = \frac{2\pi}{0.2\pi} = 10$ rad/s
For constant speed the only acceleration is centripetal: $a = \omega^2R$
$a = (10)^2\times5\times10^{-2} = 100\times0.05$
a = 5 m/$s^2$

Initial velocity $\vec{v_i} = 30\hat{i}$ m/s (east); final velocity $\vec{v_f} = 40\hat{j}$ m/s (north).
Change in velocity $\Delta\vec{v} = \vec{v_f} - \vec{v_i} = 40\hat{j} - 30\hat{i}$
$|\Delta\vec{v}| = \sqrt{40^2 + 30^2} = \sqrt{1600 + 900} = \sqrt{2500} = 50$ m/s
Average acceleration $= \frac{|\Delta\vec{v}|}{t} = \frac{50}{10}$
= 5 m/$s^2$

Range of a projectile: $R = \frac{u^2\sin2\theta}{g}$
The range is maximum when $\sin2\theta = 1$, i.e., $\theta = 45^\circ$.
$R_{max} = \frac{u^2}{g}$
$R_{max} = \frac{(20)^2}{10} = \frac{400}{10}$
$R_{max} = 40$ m

Let the initial speed be $v_0$ and the angle of projection be $\theta$.
At the maximum height the vertical component of velocity is zero; only the horizontal component remains: $v = v_0\cos\theta$
Given $v = \frac{v_0}{2}$: $\frac{v_0}{2} = v_0\cos\theta$
$\cos\theta = \frac{1}{2}$
$\theta = 60^\circ$

$\frac{x}{a} = \sin\omega t$ and $\frac{y}{a} = \cos\omega t$
Squaring and adding to eliminate t: $\frac{x^2}{a^2} + \frac{y^2}{a^2} = \sin^2\omega t + \cos^2\omega t = 1$
$x^2 + y^2 = a^2$
This is the equation of a circle of radius a centred at the origin.
Hence the particle follows a circular path.

By the triangle law, if two vectors are drawn head to tail, their sum is the vector drawn from the tail of the first to the head of the second.
Using the magnitudes and directions shown in the figure, place the tail of $\vec{e}$ at the head of $\vec{d}$.
The vector from the tail of $\vec{d}$ to the head of $\vec{e}$ then has the same magnitude and direction as $\vec{f}$.
The other combinations ($\vec{b} + \vec{e}$, $\vec{b} + \vec{c}$, $\vec{d} + \vec{c}$) give resultants that do not coincide with $\vec{f}$ in direction.
Hence $\vec{d} + \vec{e} = \vec{f}$.

$\vec{v} = \vec{u} + \vec{a}t$
$\vec{v} = (3\hat{i} + 4\hat{j}) + (0.4\hat{i} + 0.3\hat{j})\times10$
$\vec{v} = (3 + 4)\hat{i} + (4 + 3)\hat{j} = 7\hat{i} + 7\hat{j}$
Speed $= |\vec{v}| = \sqrt{7^2 + 7^2} = \sqrt{98}$
$= 7\sqrt{2}$ units

Let $\theta$ be the angle which the path of the particle makes with the x-axis.
The particle moves along the straight line from (0, 0) to $(\sqrt{3}, 3)$.
$\tan\theta = \dfrac{y}{x} = \dfrac{3}{\sqrt{3}}$
$\tan\theta = \sqrt{3}$
$\theta = \tan^{-1}(\sqrt{3}) = 60^\circ$

$|\vec{A} \times \vec{B}| = AB\sin\theta$ and $\vec{A}\cdot\vec{B} = AB\cos\theta$
Given $|\vec{A} \times \vec{B}| = \sqrt{3}\,(\vec{A}\cdot\vec{B})$
$\Rightarrow AB\sin\theta = \sqrt{3}\,AB\cos\theta$
$\Rightarrow \tan\theta = \sqrt{3}$
$\Rightarrow \theta = 60^\circ$

Horizontal range: $R = \dfrac{u^2\sin 2\alpha}{g}$
$R_1 = \dfrac{u^2\sin 2(45^\circ - \theta)}{g} = \dfrac{u^2\sin(90^\circ - 2\theta)}{g} = \dfrac{u^2\cos 2\theta}{g}$
$R_2 = \dfrac{u^2\sin 2(45^\circ + \theta)}{g} = \dfrac{u^2\sin(90^\circ + 2\theta)}{g} = \dfrac{u^2\cos 2\theta}{g}$
$\dfrac{R_1}{R_2} = \dfrac{\cos 2\theta}{\cos 2\theta} = 1$
In other words, $(45^\circ - \theta)$ and $(45^\circ + \theta)$ are complementary angles (their sum is $90^\circ$), and for complementary angles of projection the range is the same.
So the ranges are in the ratio 1 : 1.

$|\vec{A} + \vec{B}| = |\vec{A} - \vec{B}| \Rightarrow |\vec{A} + \vec{B}|^2 = |\vec{A} - \vec{B}|^2$
$|\vec{A} + \vec{B}|^2 = A^2 + B^2 + 2AB\cos\theta$
$|\vec{A} - \vec{B}|^2 = A^2 + B^2 - 2AB\cos\theta$
$\Rightarrow A^2 + B^2 + 2AB\cos\theta = A^2 + B^2 - 2AB\cos\theta$
$\Rightarrow 4AB\cos\theta = 0 \Rightarrow \cos\theta = 0$
$\Rightarrow \theta = 90^\circ$

Frequency: $n = \dfrac{22}{44} = \dfrac{1}{2}$ revolution per second
Angular speed: $\omega = 2\pi n = 2\pi \times \dfrac{1}{2} = \pi$ rad/s
Centripetal acceleration: $a_r = \omega^2 r = (2\pi n)^2 r$
$a_r = (\pi)^2 \times 1 = \pi^2$ m s$^{-2}$
The speed is constant, so the tangential acceleration is zero: $a_T = 0$.
So the acceleration is $\pi^2$ m s$^{-2}$, directed along the radius towards the centre.

Two vectors are perpendicular when their scalar (dot) product is zero.
$(2\hat{i} + 3\hat{j} + 8\hat{k})\cdot(-4\hat{i} + 4\hat{j} + \alpha\hat{k}) = 0$
$2 \times (-4) + 3 \times 4 + 8 \times \alpha = 0$
$-8 + 12 + 8\alpha = 0$
$8\alpha = -4$
$\alpha = -\dfrac{1}{2}$

$\vec{B} \times \vec{A} = BA\sin\theta\,\hat{n}$, where $\hat{n}$ is a unit vector perpendicular to the plane containing $\vec{A}$ and $\vec{B}$.
$(\vec{B} \times \vec{A})\cdot\vec{A} = (BA\sin\theta)(\hat{n}\cdot\vec{A})$
Since $\hat{n}$ is perpendicular to $\vec{A}$, $\hat{n}\cdot\vec{A} = A\cos 90^\circ = 0$.
$\Rightarrow (\vec{B} \times \vec{A})\cdot\vec{A} = 0$
In general, a scalar triple product with two identical vectors is always zero.

Two vectors are perpendicular if their dot product is zero.
$(\vec{A} + \vec{B})\cdot(\vec{A} - \vec{B}) = 0$
$\Rightarrow A^2 - \vec{A}\cdot\vec{B} + \vec{B}\cdot\vec{A} - B^2 = 0$
Since $\vec{A}\cdot\vec{B} = \vec{B}\cdot\vec{A}$, the middle terms cancel: $A^2 - B^2 = 0$
$\Rightarrow A = B$
So the forces are equal in magnitude. They need not be equal as vectors, since their directions can be different (if they were equal vectors, their difference would be zero).