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.

Initial velocity $=-v\hat{j}$, final velocity $=v\hat{i}$ Change in velocity $=v\hat{i}-(-v\hat{j})=v(\hat{i}+\hat{j})$ Momentum gain is along $\hat{i}+\hat{j}$ $\Rightarrow$ force experienced is along $\hat{i}+\hat{j}$ $\Rightarrow$ force experienced is in North-East direction.

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

$\vec V=\dfrac{d\vec r}{dt}=4\hat i+4t\hat j+0\hat k$. At $t=1$ s, $\vec V=4\hat i+4\hat j$. $|\vec V|=\sqrt{4^2+4^2}=4\sqrt2$. $\tan\alpha=\dfrac{4}{4}=1\Rightarrow\alpha=45^\circ$.

At the topmost point of its trajectory, the particle has only the horizontal component of velocity: $V_{top}=v\cos\theta=20\times\cos60^\circ=20\times\dfrac{1}{2}=10$ m/s.

In uniform circular motion, the direction of both velocity and the (centripetal) acceleration changes continuously (even though their magnitudes stay constant), so neither $\vec v$ nor $\vec a$ is a constant vector.

At the highest point, only the horizontal component of velocity remains. Angle with the horizontal $\theta = 90^\circ - 60^\circ = 30^\circ$ $u_x = u\cos\theta = 10\cos 30^\circ = 5\sqrt{3}\ ms^{-1}$

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

$\vec{F}\cdot\vec{r}=2\times3+1\times2+(-1)(-2)=10$. $\vec{F}\times\vec{r}=\hat{i}(-2+2)-\hat{j}(-4+3)+\hat{k}(4-3)=\hat{j}+\hat{k}$, so $|\vec{F}\times\vec{r}|=\sqrt{2}$.

Speed in circular motion: $T = \frac{2\pi R}{v} \Rightarrow v = \frac{2\pi R}{T}$ $H_{max} = \frac{v^2\sin^2\theta}{2g} = \frac{4\pi^2R^2\sin^2\theta}{2gT^2} = \frac{2\pi^2R^2\sin^2\theta}{gT^2} = 4R$ $\sin\theta = \left(\frac{2gT^2}{\pi^2R}\right)^{1/2}$ $\theta = \sin^{-1}\left(\frac{2gT^2}{\pi^2R}\right)^{1/2}$

Velocity of the car at t = 4 s: $v = 0 + 5 \times 4 = 20$ m/s, so the ball has horizontal velocity $v_x = 20$ m/s. After being dropped, the only acceleration of the ball is due to gravity: a = g = 10 $m/s^2$. At t = 6 s (2 s after release): $v_y = 0 + 10 \times 2 = 20$ m/s (downward) $v = \sqrt{v_x^2 + v_y^2} = \sqrt{20^2 + 20^2} = 20\sqrt{2}$ m/s

$\vec{V}_{SG}=\vec{V}_{SR}+\vec{V}_{RG}$, with $V_{SR}=20\ m/s$ and $V_{RG}=10\ m/s$ (due east). For the shortest path, the resultant velocity must be due north, so the swimmer must head at angle $\theta$ west of north such that $\sin\theta=\frac{V_{RG}}{V_{SR}}=\frac{10}{20}=\frac{1}{2}$ $\theta=30^\circ$ west

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 $=1:1$

Relative horizontal speed $=25+25=50$ m/s, so $t=\frac{100}{50}=2$ s. Vertical fall $=\frac{1}{2}\times10\times2^2=20$ m, so they collide at a height of $200-20=180$ m.

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

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

$\vec{v} = \frac{d\vec{r}}{dt} = \omega(-\sin\omega t\,\hat{x} + \cos\omega t\,\hat{y})$ ...(ii)
$\vec{a} = \frac{d\vec{v}}{dt} = -\omega^2(\cos\omega t\,\hat{x} + \sin\omega t\,\hat{y}) = -\omega^2\vec{r}$ ...(iii)
So the acceleration is antiparallel to r, i.e. directed towards the origin.
$\vec{v}\cdot\vec{r} = \omega(-\sin\omega t\cos\omega t + \sin\omega t\cos\omega t) = 0$, so the velocity is perpendicular to r.

Hence the velocity is perpendicular to r and the acceleration is directed towards the origin.

$\frac{1}{2}mv^2 = 8\times10^{-4}$ J $\Rightarrow v^2 = \frac{2\times8\times10^{-4}}{10\times10^{-3}} = \frac{16}{100} \Rightarrow v = 0.4$ m/s
Distance in two revolutions $s = 2\times2\pi R = 4\pi R$
$v^2 = u^2 + 2as \Rightarrow \left(\frac{4}{10}\right)^2 = 0 + 2a\times4\pi\times6.4\times10^{-2}$
$a = \frac{16}{100\times8\pi\times6.4\times10^{-2}} = \frac{160}{8\times3.14\times64} \approx 0.1\ m/s^2$

Centripetal component: $a\cos30^\circ = a_c = \frac{v^2}{R}$
$\Rightarrow v^2 = aR\times\frac{\sqrt{3}}{2} = 15\times2.5\times\frac{\sqrt{3}}{2} = 32.47$
$\Rightarrow v \approx 5.7$ m/s

$\vec v_A = 10(-\hat i)$ km/h and $\vec v_B = 10\hat j$ km/h.
Velocity of B relative to A: $\vec v_{BA} = 10\hat j + 10\hat i$, so $|\vec v_{BA}| = \sqrt{10^2 + 10^2} = 10\sqrt{2}$ km/h, directed at $45^\circ$.

Relative to A, B moves along BC; the distance is least when B reaches the foot C of the perpendicular from A.
$BC = 100\cos 45^\circ = 50\sqrt{2}$ km
$t = \frac{50\sqrt{2}}{10\sqrt{2}} = 5$ h

Orthogonal: $\vec A\cdot\vec B = 0$
$\cos\omega t\cos\frac{\omega t}{2} + \sin\omega t\sin\frac{\omega t}{2} = 0 \Rightarrow \cos\left(\omega t - \frac{\omega t}{2}\right) = \cos\frac{\omega t}{2} = 0$
$\frac{\omega t}{2} = \frac{\pi}{2} \Rightarrow t = \frac{\pi}{\omega}$

$x^2 + y^2 = 16$, so the path is a circle of radius 4 m, and the acceleration is centripetal (along $-\vec R$) with magnitude $\frac{v^2}{R}$.
$\vec v = \frac{d\vec R}{dt} = 8\pi\cos 2\pi t\,\hat i - 8\pi\sin 2\pi t\,\hat j$
$|\vec v| = 8\pi$ m/s, not 8 m/s — so that statement is wrong.

For collision, the velocity of B relative to A must point along the position of A relative to B.
Direction of relative position of A w.r.t. B: $\frac{\vec r_1 - \vec r_2}{|\vec r_1 - \vec r_2|}$
Direction of velocity of B w.r.t. A: $\frac{\vec u_2 - \vec u_1}{|\vec u_2 - \vec u_1|}$
Condition: $\frac{\vec r_1 - \vec r_2}{|\vec r_1 - \vec r_2|} = \frac{\vec u_2 - \vec u_1}{|\vec u_2 - \vec u_1|}$

Range $= \frac{u^2\sin2\theta}{g}$; for identical trajectories at the same angle, $g \propto u^2$.
$g_{planet} = \left(\frac{3}{5}\right)^2\times 9.8 = 3.5$ $ms^{-2}$

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