Motion in a Plane

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

At the highest point the projectile is at height H and at horizontal distance $\frac{R}{2}$ from the point of projection.
$H = \frac{u^2\sin^2\theta}{2g}$ and $R = \frac{2u^2\sin\theta\cos\theta}{g}$
Elevation angle $\phi$: $\tan\phi = \frac{H}{R/2} = \frac{u^2\sin^2\theta/2g}{u^2\sin\theta\cos\theta/g}$
$\tan\phi = \frac{\tan\theta}{2}$
With $\theta = 45^\circ$: $\tan\phi = \frac{\tan45^\circ}{2} = \frac{1}{2}$
$\phi = \tan^{-1}\frac{1}{2}$

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

Only the vertical component of velocity changes during projectile motion, so the change in momentum is in the vertical direction only.
Horizontal component of velocity: $v_x = v\cos 45^\circ$. It stays constant throughout the flight, so $\Delta p_x = 0$.
Vertical component at the time of projection: $v_{iy} = +v\sin 45^\circ$ (upward).
Vertical component when the particle lands on the same level: $v_{fy} = -v\sin 45^\circ$ (downward).
Initial vertical momentum: $p_{iy} = mv\sin 45^\circ$.
Final vertical momentum: $p_{fy} = -mv\sin 45^\circ$.
Change in momentum: $\Delta \vec{p} = (p_{fy} - p_{iy})\,\hat{j} = (-mv\sin 45^\circ - mv\sin 45^\circ)\,\hat{j} = -2mv\sin 45^\circ\,\hat{j}$
$|\Delta \vec{p}| = 2mv \times \dfrac{1}{\sqrt{2}} = \sqrt{2}\,mv$
The negative sign only shows that the change in momentum is directed vertically downward (along $-y$).
So the magnitude of the change in momentum is $\sqrt{2}\,mv$.

The 4 N and 2 N forces are opposite to each other, so their net is $4 - 2 = 2$ N along the direction of the 4 N force.
Now the figure reduces to: 2 N at $30^\circ$ to the +y axis (towards -x) and 1 N at $60^\circ$ to the +x axis.
Horizontal component of the 2 N force along the -ve x-direction: $x_1 = 2\sin 30^\circ = 1$ N
Horizontal component of the 1 N force along the +ve x-direction: $x_2 = 1\cos 60^\circ = \dfrac{1}{2}$ N
Net horizontal force $= 1 - \dfrac{1}{2} = \dfrac{1}{2}$ N along the -ve x-direction.
For the resultant to be only along the y-direction, the net x-component must be zero.
The smallest force that does this is one that just cancels the x-component: $\dfrac{1}{2}$ N along the +ve x-direction.
So the minimum additional force needed is 0.5 N.

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.

Let the two boys meet at point C after time $t$. The boy from B runs along BC, perpendicular to AB, and the boy from A runs straight along AC.
BC $= v_1 t$ and AC $= vt$.
Triangle ABC is right-angled at B, so $AB^2 = AC^2 - BC^2$
$a^2 = v^2t^2 - v_1^2t^2 = t^2(v^2 - v_1^2)$
$t^2 = \dfrac{a^2}{v^2 - v_1^2}$
$t = \sqrt{\dfrac{a^2}{v^2 - v_1^2}}$

Given $|\vec{A} \times \vec{B}| = \sqrt{3}\,\vec{A}\cdot\vec{B}$
$\Rightarrow AB\sin\theta = \sqrt{3}AB\cos\theta$
$\Rightarrow \tan\theta = \sqrt{3}$
$\theta = 60^\circ$
Therefore $|\vec{A} + \vec{B}| = \sqrt{A^2 + B^2 + 2AB\cos\theta}$
$= \sqrt{A^2 + B^2 + 2AB\cos 60^\circ} = \sqrt{A^2 + B^2 + 2AB \times \dfrac{1}{2}}$
$= (A^2 + B^2 + AB)^{1/2}$

The tangential acceleration is constant, $\dfrac{dv}{dt} = a$ (say), so the equations of uniformly accelerated motion apply along the circular path.
Use $v^2 = u^2 + 2as$, where s is the distance travelled along the circle.
Distance in two revolutions: $s = 2 \times 2\pi r = 2 \times 2\pi \times \dfrac{20}{\pi} = 80$ m
$u = 0$ (motion begins from rest), $v = 80$ m/s
$80^2 = 0 + 2 \times a \times 80$
$a = \dfrac{6400}{160} = 40$ m s$^{-2}$
So the tangential acceleration is 40 m s$^{-2}$.

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).

In half a rotation, the wheel rolls forward a distance equal to half its circumference: $\pi R$.
The point P, initially touching the ground, reaches the top of the wheel, i.e. a height 2R above the ground.

So P has a horizontal displacement $PM = \pi R$ and a vertical displacement $QM = 2R$.
Displacement $PQ = \sqrt{(PM)^2 + (QM)^2} = \sqrt{(\pi R)^2 + (2R)^2}$
With R = 1 m: $PQ = \sqrt{\pi^2 + 4}$ m