Kinematics-2D

Given, $u = 98 \, ms^{-1}$ and $\theta = 30^{\circ}$
∵ Range of a projectile,
$\begin{array}{l} {R = \frac {u ^ {2} \sin (2 \theta)}{g} = \frac {9 8 \times 9 8 \times \sin 6 0 ^ {\circ}}{9 . 8}} \\ {R = 4 9 0 \sqrt {3} \mathrm{m}} \end{array}$

Horizontal range, $R = \frac{u^{2} \sin 2\theta}{g}$
When $\theta = 45^{\circ}$ , maximum horizontal range,
$R _ {\mathrm{max}} = \frac {u ^ {2}}{g} \sin 9 0 ^ {\circ} = \frac {u ^ {2}}{g}$
When $\theta = 135^{\circ}$ , maximum horizontal range,
$R = \frac {u ^ {2}}{g} \sin 2 7 0 ^ {\circ} = \frac {- u ^ {2}}{g}$
Negative sign implies opposite direction.

Given, $x = a \cos(pt)$ , $y = b \sin(pt)$
$\therefore \frac{x^2}{a^2} +\frac{y^2}{b^2} = 1$ , i.e. equation of ellipse
Now, $\mathbf{r} = x\hat{\mathrm{i}} +y\hat{\mathrm{j}} = a\cos (pt)\hat{\mathrm{i}} +b\sin (pt)\hat{\mathrm{j}}$
$\mathbf {v} = \frac {d \mathbf {r}}{d t} = - p a \sin (p t) \hat {\mathbf {i}} + p b \cos (p t) \hat {\mathbf {j}}$
$\mathbf {a} = \frac {d \mathbf {v}}{d t} = - p ^ {2} a \cos (p t) \hat {\mathbf {i}} - p ^ {2} b \sin (p t) \hat {\mathbf {j}}$
At
$\begin{array}{l} {t = \frac {\pi}{2 p},} \\ {{\bf v} = - p a \hat {\bf i} \mathrm{and} {\bf a} = - p ^ {2} b \hat {\bf j}} \end{array}$
Thus, velocity is perpendicular to acceleration.
Also, $a = -p^{2}b$ , i.e. directed towards a fixed point as p and b are positive constants.
$\mathrm{As}, \qquad v = \frac {d r}{d t} \Rightarrow \Delta r = \int_ {0} ^ {\frac {\pi}{2 p}} v d t$
$\Rightarrow \Delta r = [ a \cos (p t) \hat {\mathbf {i}} + b \sin (p t) \hat {\mathbf {j}} ] _ {0} ^ {\frac {\pi}{2 p}} = - a \hat {\mathbf {i}} + b \hat {\mathbf {j}}$
$\mathrm{So}, \qquad | \mathbf {r} | = \sqrt {a ^ {2} + b ^ {2}}$

$x = 5t - 2t^2$
$V _ {x} = \frac {d x}{d t} = 5 - 4 t$
$a _ {x} = - 4$
y = 1 0 t
$V _ {y} = 1 0$
$a _ {y} = 0$
$\because \vec {a} = a _ {x} \hat {i} + a _ {y} \hat {j}$
$\mathrm{so} \vec {a} = - 4 \hat {i}$

Position vector of the particle is given by $\mathbf{r} = \cos \omega t\hat{\mathbf{x}} +\sin \omega t\hat{\mathbf{y}}$
where, $\omega$ is a constant.
Velocity of the particle is
$\begin{array}{r l} & {\mathbf {v} = \frac {d \mathbf {r}}{d t} = \frac {d}{d t} (\cos \omega t \hat {\mathbf {x}} + \sin \omega t \hat {\mathbf {y}})} \\ & {\quad = (- \sin \omega t) \omega \hat {\mathbf {x}} + (\cos \omega t) \omega \hat {\mathbf {y}}} \\ & {\quad = - \omega (\sin \omega t \hat {\mathbf {x}} - \cos \omega t \hat {\mathbf {y}})} \end{array}$
Acceleration of the particle,
$\begin{array}{r l} & {\mathbf {a} = \frac {d \mathbf {v}}{d t} = \frac {d}{d t} (- \boldsymbol {\omega} \sin \boldsymbol {\omega} t \hat {\mathbf {x}} + \boldsymbol {\omega} \cos \boldsymbol {\omega} t \hat {\mathbf {y}})} \\ & {\quad = - \boldsymbol {\omega} ^ {2} \cos \boldsymbol {\omega} t \hat {\mathbf {x}} - \boldsymbol {\omega} ^ {2} \sin \boldsymbol {\omega} t \hat {\mathbf {y}}} \end{array}$
$\Rightarrow \mathbf {a} = - \omega^ {2} \mathbf {r} = \omega^ {2} (- \mathbf {r})$
Assuming the particle is at P, then its position vector is directed as shown in the diagram.

Therefore, acceleration is directed towards -r, i.e. towards O (origin).
$\mathbf{v}\cdot \mathbf{r} = -\boldsymbol {\omega}(\sin \omega t\hat{\mathbf{x}} -\cos \omega t\hat{\mathbf{y}})\cdot (\cos \omega t\hat{\mathbf{x}} +\sin \omega t\hat{\mathbf{y}})\\ = -\boldsymbol {\omega}\left[\sin \omega t\cdot \cos \omega t + 0 + 0 - \sin \omega t\cdot \cos \omega t\right]\\ = -\boldsymbol {\omega}(0) = 0$
$\Rightarrow \mathbf{v} \perp \mathbf{r}$
Thus, velocity is perpendicular to r.

From C to D, $\vec{L}_{0}=mv\left[\frac{R}{\sqrt{2}}+a\right]\hat{k}$ from B to C, $\vec{L}_{0}=mv\left[\frac{R}{\sqrt{2}}+a\right]\hat{k}$ from D to A, $\vec{L}_{0}=\frac{mv}{\sqrt{2}}R(-\hat{k})$ from A to B, $\vec{L}_{0}=\frac{mv}{\sqrt{2}}R(-\hat{k})$ .

When minimum speed of body is $\sqrt{5gR}$ , then no matter from where it enters the loop, it will complete full vertical loop.

$a_{c} = \frac{V^{2}}{r}$ $15\cos 30^{0} = \frac{V^{2}}{2.5}$ $V^{2} = 32.44$ $V = 5.7m / \sec$

$\frac{1}{2} mv^2 = E \Rightarrow \frac{1}{2}\left(\frac{10}{1000}\right)v^2 = 8 \times 10^{-4}$ $\Rightarrow v^2 = 16 \times 10^{-2} \Rightarrow v = 4 \times 10^{-1} = 0.4m/s$ Now, $v^2 = u^2 + 2a_t s$ ( $s = 4\pi R$ ) $\Rightarrow \frac{16}{100} = 0^2 + 2a_t \left(4 \times \frac{22}{7} \times \frac{6.4}{100}\right)$ $\Rightarrow a_t = \frac{16}{100} \times \frac{7 \times 100}{8 \times 22 \times 6.4} = 0.1m/s^2$

Particle at periphery will have both radial and tangential acceleration
$a _ {t} = R \alpha = 0. 5 \times 2 = 1 m / s ^ {2}$
$\omega = \omega_ {0} + \alpha t = 0 + 2 \times 2 = 4 r a d / s e c$
$a _ {c} = \omega^ {2} R = (4) ^ {2} \times 0. 5 = 1 6 \times 0. 5 = 8 m / s ^ {2}$
$a _ {t o t a l} = \sqrt {a _ {t} ^ {2} + a _ {c} ^ {2}} = \sqrt {1 ^ {2} + 8 ^ {2}} \approx 8 m / s ^ {2}.$
[Note -: In this question we have assumed the point to be located at the periphery of the disc.]

The equation of trajectory,
$y = x \tan \alpha \left[ 1 - \frac {x}{R} \right]\tag{Gives}$
$Q = P \tan \theta \left[ 1 - \frac {P}{R} \right]\tag{... (i}$
and
$P = Q \tan \theta \left[ 1 - \frac {Q}{R} \right]\tag{... (ii}$
On dividing Eq. (i) by Eq. (ii), we get
$\frac {Q ^ {2}}{P ^ {2}} = \frac {[ 1 - P / R ]}{[ 1 - Q / R ]}$
$\frac {1}{R} [ P ^ {3} - Q ^ {3} ] = P ^ {2} - Q ^ {2}$
⇒
$R = \frac {P ^ {3} - Q ^ {3}}{P ^ {2} - Q ^ {2}} = \frac {P ^ {2} + P Q + Q ^ {2}}{P + Q}$
Now,
$\begin{array}{r} \frac {Q}{P} = \tan \theta \left[ 1 - \frac {P (P + Q)}{P ^ {2} + P Q + Q ^ {2}} \right] \\ = \tan \theta \left[ \frac {P ^ {2} + P Q + Q ^ {2} - P ^ {2} - P Q}{P ^ {2} + P Q + Q ^ {2}} \right] \end{array}$
⇒
$\tan \theta = \frac {P ^ {2} + Q ^ {2} + P Q}{P Q}$
⇒
$\theta = \tan^ {- 1} \left[ \frac {P ^ {2} + P Q + Q ^ {2}}{P Q} \right]$

Consider the diagram

For horizontal motion, $50 = 10\cos 30^{\circ}\times t$
⇒
$t = \frac {5}{\cos 3 0 ^ {\circ}} = \frac {5}{\frac {\sqrt {3}}{2}} = \frac {1 0}{\sqrt {3}}$
For vertical motion,
$\begin{array}{r l} & h = 1 0 \sin 3 0 ^ {\circ} \times \frac {1 0}{\sqrt {3}} - \frac {1}{2} \times 1 0 \times \left(\frac {1 0}{\sqrt {3}}\right) ^ {2} \\ & \quad = 1 0 \times \frac {1}{2} \times \frac {1 0}{\sqrt {3}} - \frac {1 0 0}{\sqrt {3}} \times \frac {5}{\sqrt {3}} = \frac {1 0 0}{\sqrt {3}} \left(\frac {1}{2} - \frac {5}{\sqrt {3}}\right) \\ & \quad = \frac {5 0}{\sqrt {3}} \left[ 1 - \frac {1 0}{\sqrt {3}} \right] \mathrm{m} \end{array}$

Range of projectile is same for $\theta$ and $90^{\circ}-\theta$ .
Given, $\theta = 40^{\circ}$
Another angle $= 90^{\circ} - \theta = 90^{\circ} - 40^{\circ} = 50^{\circ}$

Given, a particle having horizontal range is twice the greatest height attained by it, i.e. R = 2H
$\frac {v ^ {2} \sin 2 \theta}{g} = \frac {2 v ^ {2} \sin^ {2} \theta}{2 g}$

$\Rightarrow 2\sin \theta \cos \theta = \sin^{2}\theta \Rightarrow \tan \theta = 2$
$\Rightarrow \sin \theta = \frac {2}{\sqrt {5}} \Rightarrow \cos \theta = \frac {1}{\sqrt {5}}$
$\therefore R = \frac{v^2\sin 2\theta}{g} = \frac{v^2\times 2\sin\theta\cos\theta}{g}$
$\Rightarrow R = v^{2}\times \frac{2}{g}\times \frac{2}{\sqrt{5}}\times \frac{1}{\sqrt{5}}$
i.e. $R = \frac{4v^2}{5g}$

$F_{c} = \frac{mv_{1}^{2}}{r} = \frac{2mv_{2}^{2}}{(r / 2)} = \frac{4mv_{2}^{2}}{r}$ So $v_{1} = 2v_{2}$

$x = 4\sin (2\pi t),y = 4\cos (2\pi t)$ Squiring and adding $x^{2} + y^{2} = 4^{2}\Rightarrow R = 4$ $\Rightarrow$ Circular motion $V = \omega R = (2\pi)(4) = 8\pi .$

B

From the figure $T = mg\cos \theta + mv^2 / L$ .

D

$= 2 0 - 8 + 8 \cos \theta$
$\cos \theta = - \frac {7}{8}$

\frac {3 \pi}{4} < \theta < \pi .

For the second stone time required to reach the ground is given by
$y = u t - \frac {1}{2} g t ^ {2} \Rightarrow - 2 4 0 = 4 0 t - \frac {1}{2} \times 1 0 \times t ^ {2}$
$\therefore 5 t ^ {2} - 4 0 t - 2 4 0 = 0$
$(t - 1 2) (t + 8) = 0 \Rightarrow \therefore t = 1 2 s$
For the first stone: $-240 = 10t - \frac{1}{2} \times 10 \times t^2$

Given, $\theta_{1} = \theta_{2} = 45^{\circ}$
$\theta_ {1} + \theta_ {2} = (4 5 ^ {\circ} + \alpha) + (4 5 ^ {\circ} - \alpha) = 9 0 ^ {\circ}$
$u _ {1} = u _ {2} = u \Rightarrow R _ {1} = R _ {2} = ?$
$R _ {1} = \frac {u _ {1} ^ {2} \sin (\theta_ {1} + \theta_ {2})}{g}$
⇒
$R _ {1} = \frac {u ^ {2} \sin 9 0 ^ {\circ}}{g} = \frac {u ^ {2}}{g}\tag{... (i}$
Similarly, $R_{2} = \frac{u_{2}^{2}\sin(\theta_{1} + \theta_{2})}{g}$
⇒
$R _ {2} = \frac {u ^ {2} \sin 9 0 ^ {\circ}}{g} = \frac {u ^ {2}}{g}\tag{... (ii}$
The ratio of horizontal ranges
$\frac {R _ {1}}{R _ {2}} = \frac {u ^ {2} / g}{u ^ {2} / g} = 1: 1 \Rightarrow R _ {1}: R _ {2} = 1: 1$

Average velocity, $v_{av} = \frac{Net displacement}{Time taken}$ $= \frac{(13 - 2)\hat{i} + (14 - 3)\hat{j}}{5}$ $= \frac{11\hat{i} + 11\hat{j}}{5} = \frac{11}{5} (\hat{i} + \hat{j})$

For vertical motion, $h_{1}=u\sin\theta t_{1}-\frac{1}{2}gt_{1}^{2}$
(For $h_{1}$ )
$t _ {1} = \frac {h _ {1} + \frac {1}{2} g t _ {1} ^ {2}}{u \sin \theta}\tag{... (i}$
⇒
$h _ {2} = u \sin \theta t _ {2} - \frac {1}{2} g t _ {2} ^ {2}\tag{For \(h_2\)}$
$t _ {2} = \frac {h _ {2} + \frac {1}{2} g t _ {2} ^ {2}}{u \sin \theta}$
⇒
... (ii)
On dividing Eq. (i) by Eq. (ii), we get
$\frac {t _ {1}}{t _ {2}} = \frac {\left(h _ {1} + \frac {1}{2} g t _ {1} ^ {2}\right) / u \sin \theta}{\left(h _ {2} + \frac {1}{2} g t _ {2} ^ {2}\right) / u \sin \theta}$
$\Rightarrow h _ {1} t _ {2} - h _ {2} t _ {1} = \frac {g}{2} (t _ {1} t _ {2} ^ {2} - t _ {1} ^ {2} t _ {2})$
The time of flight of the ball,
$\begin{array}{r l r} & {T = \frac {2 u \sin \theta}{g} = \frac {2}{g} (u \sin \theta)} \\ & {\quad = \frac {2}{g} \left(\frac {h _ {1} + 1 / 2 g t _ {1} ^ {2}}{t _ {1}}\right)} & {[ \mathrm{FromEq.(i)} ]} \\ & {\quad = \frac {2}{t _ {1}} \left(\frac {h _ {1}}{g} + \frac {t _ {1} ^ {2}}{2}\right)} \\ & {\quad = \frac {h _ {1}}{t _ {1}} \times \frac {2}{g} + t _ {1} = \frac {h _ {1}}{t _ {1}} \times \left(\frac {t _ {1} t _ {2} ^ {2} - t _ {1} ^ {2} t _ {2}}{h _ {1} t _ {2} - h _ {2} t _ {1}}\right) + t _ {1}} \\ & {\quad = \frac {h _ {1} t _ {1} t _ {2} ^ {2} - h _ {1} t _ {1} ^ {2} t _ {2} + h _ {1} t _ {1} ^ {2} t _ {2} - h _ {2} t _ {1} ^ {3}}{t _ {1} (h _ {1} t _ {2} - h _ {2} t _ {1})} = \left(\frac {h _ {1} t _ {2} ^ {2} - h _ {2} t _ {1} ^ {2}}{h _ {1} t _ {2} - h _ {2} t _ {1}}\right)} \end{array}$

As we know that, maximum height of a projectile is given by
$H _ {\mathrm{max}} = \frac {u ^ {2} \sin^ {2} \theta}{2 g}$
where, u = initial velocity of projectile
g = acceleration due to gravity
and $\theta = \text{angle of projection.}$
As per question,
$H _ {\mathrm{max}} = \frac {v ^ {2} \sin^ {2} 4 5 ^ {\circ}}{2 g}$
$\dots (\mathrm{i}) (\mathrm{As}, u = v, \theta = 4 5 ^ {\circ})$
Now, range of a projectile is given by
$R = \frac {u ^ {2} \sin 2 \theta}{g}$
⇒
$R = \frac {v ^ {2} \sin (2 \times 4 5 ^ {\circ})}{g}$
⇒
$R = \frac {v ^ {2} \sin 9 0 ^ {\circ}}{g}\tag{... (ii}$
On dividing Eq. (i) by Eq. (ii), we get
$\frac {H _ {\mathrm{max}}}{R} = \frac {v ^ {2} \sin^ {2} 4 5 ^ {\circ} \times g}{2 g \times v ^ {2} \sin 9 0 ^ {\circ}} = \frac {1}{4 \times 1}$