Kinematics-2D

$\begin{array}{l} \text {Since} \frac {d x}{d t} = \frac {d y}{d t} = c \\ \Rightarrow \frac {d ^ {2} x}{d t ^ {2}} = \frac {d ^ {2} y}{d t ^ {2}} = 0 \\ \text {Further} z = a x ^ {3} + b y ^ {2} \\ \Rightarrow \frac {d z}{d t} = 3 a x ^ {2} \frac {d x}{d t} + 2 b y \frac {d y}{d t} \\ \Rightarrow \frac {d z}{d t} = 3 a c x ^ {2} + 2 b c y \qquad \left\{\because \frac {d x}{d t} = c = \frac {d y}{d t} \right\} \\ \Rightarrow \frac {d ^ {2} z}{d t ^ {2}} = 6 a c x \bigg (\frac {d x}{d t} \bigg) + 2 b c \bigg (\frac {d y}{d t} \bigg) \\ \Rightarrow \frac {d ^ {2} z}{d t ^ {2}} = 6 a c ^ {2} x + 2 b c ^ {2} \end{array}$
Now acceleration of particle is
$\begin{array}{r l} & {\vec {a} = \frac {d ^ {2} x}{d t ^ {2}} \hat {i} + \frac {d ^ {2} y}{d t ^ {2}} \hat {j} + \frac {d ^ {2} z}{d t ^ {2}} \hat {k}} \\ {\Rightarrow} & {\vec {a} = (6 a c ^ {2} x + 2 b c ^ {2}) \hat {k}} \end{array}$
Hence, the correct answer is (B).

$\begin{array}{r l} & {\mathrm{Since,} y = x \tan \theta - \frac {g x ^ {2}}{2 u ^ {2} \cos^ {2} \theta}} \\ & {\Rightarrow b = a \tan \alpha - \frac {g a ^ {2}}{2 u ^ {2} \cos^ {2} \alpha}} \\ & {\Rightarrow a \tan \alpha - b = \frac {g a ^ {2}}{2 u ^ {2} \cos^ {2} \alpha}} \\ & {\Rightarrow \frac {u ^ {2}}{g} = \frac {a ^ {2} \sec^ {2} \alpha}{2 (a \tan \alpha - b)}} \end{array}\tag{... (1}$
Further, maximum height $h$ is
$h = \frac {u ^ {2} \sin^ {2} \alpha}{2 g}$
$\Rightarrow h = \frac {a ^ {2} \sec^ {2} \alpha \sin^ {2} \alpha}{4 (a \tan \alpha - b)}$
$\Rightarrow \quad h = \frac {a ^ {2} \tan^ {2} \alpha}{4 (a \tan \alpha - b)}$
Hence, the correct answer is (A).

In all the cases, we observe that the horizontal component of the velocity of the food packet is same as the horizontal component of the velocity of the aeroplane and due to this, at all the instants, both have the same horizontal displacements.
Hence, the correct answer is (D).

$x + a\cos \theta = a$

and $y = a\sin \theta = a\sin (\omega t)$
Also we observe that at t=0, $|\vec{r}|=a$ Hence, the correct answer is (C).

Kinetic energy is minimum at the maximum height. So,
$\begin{array}{l} K _ {\mathrm{MIN}} = \frac {1}{2} m u _ {x} ^ {2} \\ \frac {(K _ {M I N}) _ {1}}{(K _ {M I N}) _ {2}} = \frac {(u _ {x}) _ {1} ^ {2}}{(u _ {x}) _ {2} ^ {2}} = \frac {4}{1} \\ \Rightarrow \frac {(u _ {x}) _ {1}}{(u _ {x}) _ {2}} = 2 \end{array}\tag{... (1}$
Further, $H = \frac{u_{y}^{2}}{2g}$
$\mathrm{Since,} \frac {H _ {1}}{H _ {2}} = \frac {9}{1}$
$\Rightarrow \frac {H _ {1}}{H _ {2}} = \frac {(u _ {y}) _ {1} ^ {2}}{(u _ {y}) _ {2} ^ {2}} = \frac {9}{1}$
$\Rightarrow \frac {(u _ {y}) _ {1}}{(u _ {y}) _ {2}} = \frac {3}{1}$
Now $R = \frac{2}{g}\big(u_{x}\big)\big(u_{y}\big)$
$\Rightarrow \frac {R _ {1}}{R _ {2}} = \frac {\left(u _ {x}\right) _ {1}}{\left(u _ {x}\right) _ {2}} \frac {\left(u _ {y}\right) _ {1}}{\left(u _ {y}\right) _ {2}} = (2) (3)$
$\Rightarrow \frac {R _ {1}}{R _ {2}} = \frac {6}{1}$
Hence, the correct answer is (C).

$\delta = \frac{1}{2} gt^2 = \frac{gx^2}{2u^2\cos^2\alpha}$
$\left\{\because t = \frac {x}{u \cos \alpha} \right\}$
Hence, the correct answer is (D).
$\left\{\because x = \frac {t ^ {2}}{2} \right\}$

$y=\frac{x^{2}}{2}=\frac{\left(\frac{t^{2}}{2}\right)^{2}}{2}=\frac{t^{4}}{8}$ $\Rightarrow \frac{dy}{dt}=\frac{t^{3}}{2}$ $\Rightarrow \left.\frac{dy}{dt}\right|_{t=2s}=4\ ms^{-1}$ Similarly, for $x=\frac{t^{2}}{2}$ $\frac{dx}{dt}=t$ $\Rightarrow \left.\frac{dx}{dt}\right|_{t=2s}=2\ ms^{-1}$ $\Rightarrow \left.\vec{v}\right|_{t=2s}=\left(4\hat{i}+2\hat{j}\right)\mathrm{ms}^{-1}$
Hence, the correct answer is (C).

$v_{x} = \frac{dx}{dt} = 8t + 4$
$v _ {y} = \frac {d y}{d t} = - 3 t ^ {2} + 1 2$
$a _ {x} = \frac {d v _ {x}}{d t} = 8$
$a _ {y} = \frac {d v _ {y}}{d t} = - 6 t$
$\Rightarrow \vec {a} = 8 \hat {i} - (6 t) \hat {j}$
$\Rightarrow \quad \vec {a} \bigg | _ {t = 1 \mathrm{s}} = 8 \hat {i} - 6 \hat {j}$
$\Rightarrow | \vec {a} | = 1 0 \mathrm{ms} ^ {- 2}$
Hence, the correct answer is (B).

Since $v^{2}=v_{x}^{2}+v_{y}^{2}$ $\Rightarrow$ $2v\frac{dv}{dt}=2v_{x}\frac{dv_{x}}{dt}+2v_{y}\frac{dv_{y}}{dt}$ $\Rightarrow$ $\frac{dv}{dt}=\frac{v_{x}a_{x}+v_{y}a_{y}}{v}=\frac{v_{x}a_{x}+v_{y}a_{y}}{\sqrt{v_{x}^{2}+v_{y}^{2}}}$ $\Rightarrow$ $\frac{dv}{dt}=\frac{(3)(2)+(4)(1)}{5}=2\ ms^{-2}$
Hence, the correct answer is (C).

Since horizontal motion is non accelerated, so $u_{x} = v_{x}$
$\Rightarrow u \cos 6 0 = v \cos 3 0$
$\Rightarrow 1 0 \left(\frac {1}{2}\right) = v \left(\frac {\sqrt {3}}{2}\right)$
$\Rightarrow v = \frac {1 0}{\sqrt {3}} \mathrm{ms} ^ {- 1}$
Hence, the correct answer is (D).

$u_{x} = v_{x}$ $u\sin \theta = v\sin \phi$ $\{\theta$ and $\phi$ are with the vertical} $\Rightarrow v = u\sin \theta \csc \phi = u\left(\frac{\csc \phi}{\csc \theta}\right)$ Hence, the correct answer is (D).

At $h = 0.4 \, \text{m}$ , $\vec{v} = 6\hat{i} + 2\hat{j}$ $\Rightarrow v_x = 6 \, \text{ms}^{-1}$ and $v_y = 2 \, \text{ms}^{-1}$ Since $v_x = u_x$ $\Rightarrow u_x = 6 \, \text{ms}^{-1}$ Also, $v_y^2 - u_y^2 = 2a_yy$ $\Rightarrow 2^2 - u_y^2 = 2(-g)(0.4)$ $\Rightarrow u_y^2 = 4 + 8$ $\Rightarrow u_y = 2\sqrt{3} \, \text{ms}^{-1}$
If $\theta$ is the angle made by the initial velocity with the vertical, then
$u _ {x} = u \sin \theta = 6 \text { and } u _ {y} = u \cos \theta = 2 \sqrt {3}$
$\Rightarrow \frac {u \sin \theta}{u \cos \theta} = \frac {6}{2 \sqrt {3}} = \sqrt {3}$
$\Rightarrow \tan \theta = \sqrt {3}$
$\Rightarrow \quad \theta = 6 0 ^ {\circ}$
Hence, the correct answer is (B).

Since we know that $\ell = r\theta$
$\begin{array}{r l} & {\Rightarrow 1 0 0 = r \left(\frac {\pi}{4}\right)} \\ & {\Rightarrow r = \frac {4 0 0}{\pi} \mathrm{m}} \\ & {\mathrm{Since} a = \frac {v ^ {2}}{r}} \\ & {\Rightarrow a = \frac {(2 0) ^ {2}}{\left(\frac {4 0 0}{\pi}\right)} = \pi \mathrm{ms} ^ {- 2} \left\{\because v = 7 2 \mathrm{kmh} ^ {- 1} = 2 0 \mathrm{ms} ^ {- 1} \right\}} \\ & {\Rightarrow a = 3. 1 4 \mathrm{ms} ^ {- 2}} \end{array}$
Hence, the correct answer is (B).

$\tan\beta=\frac{u\sin\theta-gt}{u\cos\theta}$
$\begin{array}{l l} \Rightarrow & \tan (4 5 ^ {\circ}) = \frac {u \sin \theta - g t}{u \cos \theta} \\ \Rightarrow & u \sin \theta - 1 0 = u \cos \theta \end{array}\tag{75.}$
... (1)
After three more second it is travelling horizontally i.e., at t = 4 s $\beta = 0$
$\begin{array}{r l} \Rightarrow & u \sin \theta - g (4) = 0 \\ \Rightarrow & u \sin \theta = 4 0 \end{array}\tag{... (2}$
$\begin{array}{c} \text {Substituting in (1), we get} \\ u \cos \theta = 3 0 \end{array}\tag{... (3}$
Squaring and adding, we get
$u = 5 0 \mathrm{ms} ^ {- 1}$
$\tan \theta = \frac {4}{3}$
$\Rightarrow \quad \theta = \tan^ {- 1} \left(\frac {4}{3}\right)$
Hence, the correct answer is (C).

Since we know that
$\begin{array}{r l} & y = x \tan \theta - \frac {g x ^ {2}}{2 u ^ {2} \cos^ {2} \theta} \\ \Rightarrow & y = x \tan \theta - \frac {g x ^ {2}}{2 u ^ {2}} (1 + \tan^ {2} \theta) \end{array}$
For the particle to pass through the point (30 m, 40 m), we have
$4 0 = 3 0 \tan \theta - \frac {g (3 0) ^ {2}}{2 u ^ {2}} (1 + \tan^ {2} \theta)$
$\Rightarrow 9 0 0 \tan^ {2} \theta - 6 u ^ {2} \tan \theta + (9 0 0 + 8 u ^ {2}) = 0$
For real values of $\theta$ , we must have
$\begin{array}{r l} & {\big (6 u ^ {2} \big) ^ {2} - 4 (9 0 0) \big (9 0 0 + 8 u ^ {2} \big) \geq 0} \\ & {3 6 u ^ {4} - 3 6 0 0 \big (9 0 0 + 8 u ^ {2} \big) \geq 0} \\ {\Rightarrow} & {u ^ {4} - 1 0 0 \big (9 0 0 + 8 u ^ {2} \big) \geq 0} \\ {\Rightarrow} & {u ^ {4} - 8 0 0 u ^ {2} - 9 0 0 0 0 \geq 0} \\ {\Rightarrow} & {u ^ {4} - 8 0 0 u ^ {2} + (4 0 0) ^ {2} - (4 0 0) ^ {2} - 9 0 0 0 0 \geq 0} \\ {\Rightarrow} & {\big (u ^ {2} - 4 0 0 \big) ^ {2} - 2 5 0 0 0 0 \geq 0} \\ {\Rightarrow} & {\big (u ^ {2} - 4 0 0 \big) ^ {2} \geq 2 5 0 0 0 0} \\ {\Rightarrow} & {\big (u ^ {2} - 4 0 0 \big) \geq 5 0 0} \\ {\Rightarrow} & {u ^ {2} \geq 9 0 0} \\ {\Rightarrow} & {u \geq 3 0 \mathrm{ms} ^ {- 1}} \\ {\Rightarrow} & {u _ {M I N} = 3 0 \mathrm{ms} ^ {- 1}} \end{array}$
Hence, the correct answer is (D).

$- h = x \tan \alpha - \frac {g x ^ {2}}{2 u ^ {2} \cos^ {2} \alpha}\tag{... (1}$
and
$- h = x \tan \beta - \frac {g x ^ {2}}{2 u ^ {2} \cos^ {2} \beta}\tag{... (2}$
Equating (1) and (2), we get
$x = \frac {2 u ^ {2}}{g} \left(\frac {1}{\tan \alpha + \tan \beta}\right)\tag{... (3}$
Substituting (3) in (1), we get
$\begin{array}{r l} & h = \frac {2 u ^ {2}}{g} \left(\frac {1 - \tan \alpha \tan \beta}{\tan \alpha + \tan \beta}\right) \\ \Rightarrow & h = \frac {2 u ^ {2}}{g} \cot (\alpha + \beta) \end{array}$
Hence, the correct answer is (D).

$y = \frac {1}{2} g t ^ {2} \text { and } x = v _ {0} t$
$\Rightarrow y = \frac {1}{2} (1 0) (3) ^ {2}$
$\Rightarrow y = 4 5 \mathrm{m}$
$\text { Since } \tan (3 0 ^ {\circ}) = \frac {y}{x}$
$\Rightarrow x = \frac {y}{\tan (3 0 ^ {\circ})} = 4 5 \sqrt {3} \mathrm{m}$
$\text { Since } x = v _ {0} t$
$\Rightarrow 4 5 \sqrt {3} = v _ {0} (3)$
$\Rightarrow v _ {0} = \frac {4 5 \sqrt {3}}{3}$
$\Rightarrow v _ {0} = 1 5 \sqrt {3} \mathrm{ms} ^ {- 1}$
$\Rightarrow v _ {0} = 2 5. 9 8 \mathrm{ms} ^ {- 1}$
$\Rightarrow v _ {0} \cong 2 6 \mathrm{ms} ^ {- 1}$
Hence, the correct answer is (C).

Given $v_T = 10\text{ ms}^{-1}$ and $a = 100\text{ ms}^{-2}$
Also, $a_T = 60\text{ ms}^{-2}$
Since $a^2 = a_C^2 + a_T^2$
$\begin{aligned} &\Rightarrow (100)^2 = a_C^2 + (60)^2 \\ &\Rightarrow a_C = 80\text{ ms}^{-2} \\ &\Rightarrow \frac{v_T^2}{r} = 80 \\ &\Rightarrow r = \frac{(10)^2}{80} = \frac{100}{80} \\ &\Rightarrow r = 1.25\text{ m} \end{aligned}$
Hence, the correct answer is (D).

$y = 2\mathrm{m}, x = 3\mathrm{m}$
$\mathrm{Since} y = \frac {g x ^ {2}}{2 u ^ {2}}$
$\Rightarrow 2 = \frac {(1 0) 9}{2 u ^ {2}}$
$\Rightarrow u ^ {2} = \frac {9 0}{4}$
$\Rightarrow u \cong 4. 7 \mathrm{ms} ^ {- 1}$
Hence, the correct answer is (A).

where $H=\frac{u^{2}\sin^{2}\theta}{2g}$
$\begin{array}{l} T = \frac {2 u \sin \theta}{g} \\ R = \frac {2 u ^ {2} \sin \theta \cos \theta}{g} \\ \Rightarrow v _ {a v} = \frac {\sqrt {\frac {u ^ {4} \sin^ {4} \theta}{4 g ^ {2}} + \frac {u ^ {4} \sin^ {2} \theta \cos^ {2} \theta}{g ^ {2}}}}{\frac {u \sin \theta}{g}} \\ \Rightarrow v _ {a v} = \frac {u}{2} \sqrt {\frac {\sin^ {4} \theta}{\sin^ {2} \theta} + \frac {4 \sin^ {2} \theta \cos^ {2} \theta}{\sin^ {2} \theta}} \\ \Rightarrow v _ {a v} = \frac {u}{2} \sqrt {\sin^ {2} \theta + 4 \cos^ {2} \theta} \\ \Rightarrow v _ {a v} = \frac {u}{2} \sqrt {\sin^ {2} \theta + \cos^ {2} \theta + 3 \cos^ {2} \theta} \\ \Rightarrow v _ {a v} = \frac {u}{2} \sqrt {1 + 3 \cos^ {2} \theta} \end{array}$
Hence, the correct answer is (C).

Let the origin be located at the point $O$ and $O'$ . For the particles to strike the object placed at $P$ simultaneously
$u \cos \alpha = v \cos \beta\tag{... (1}$

Also,
$y = l \tan \beta - \frac {g l ^ {2}}{2 v ^ {2} \cos^ {2} \beta}\tag{... (2}$
and
$- (h - y) = l \tan \alpha - \frac {g l ^ {2}}{2 u ^ {2} \cos^ {2} \alpha}\tag{... (3}$
Subtract (3) from (2) and use (1) we get
$h = l \left(\tan \beta - \tan \alpha\right)$
Hence, the correct answer is (D).

$y = x \tan \theta - \frac {g x ^ {2}}{2 u ^ {2} \cos^ {2} \theta}\tag{81.}$
Here $y = 5 \mathrm{~m}$ , $x = 20 \mathrm{~m}$ , $u = ?$ , $\theta = 30^{\circ}$
$\begin{array}{r l} \Rightarrow & 5 = 2 0 \tan (3 0 ^ {\circ}) - \frac {(1 0) (2 0) ^ {2}}{2 u ^ {2} \left(\frac {3}{4}\right)} \\ \Rightarrow & 5 = \frac {2 0}{\sqrt {3}} - \frac {5 \times 4 0 0 \times 4}{3 u ^ {2}} \\ \Rightarrow & \frac {5 \times 4 0 0 \times 4}{3 u ^ {2}} = \frac {2 0}{\sqrt {3}} - 5 \\ \Rightarrow & u ^ {2} = \frac {5 \times 1 6 0 0}{3 \left(\frac {2 0}{\sqrt {3}} - 5\right)} \\ \Rightarrow & u ^ {2} = \frac {5 \times 1 6 0 0}{2 0 \sqrt {3} - 1 5} \\ \Rightarrow & u ^ {2} = \frac {8 0 0 0}{1 9 . 6 4} \\ \Rightarrow & u ^ {2} \cong 4 0 0 \\ \Rightarrow & u = 2 0 \mathrm{ms} ^ {- 1} \end{array}\tag{83.}$
Hence, the correct answer is (A).

Relative acceleration between the particles is zero, because both are under influence of gravity.
If s be the relative separation between them at time t, then
$s ^ {2} = \left[ h - (v - v \sin \theta) t \right] ^ {2} + \left[ (v \cos \theta) t \right] ^ {2}\tag{84.}$
For s to be MINIMUM or $s^{2}$ to be MINIMUM, we have
$\frac {d}{d t} \big (s ^ {2} \big) = 0$

Hence, the correct answer is (D).

$|\vec{u}|=\sqrt{5}$ at an angle $\theta=\tan^{-1}(2)$ with the horizontal

$\begin{array}{l l} \text {Since} & y = x \tan \theta - \frac {g x ^ {2}}{2 u ^ {2} \cos^ {2} \theta} \\ \Rightarrow & y = x \tan \theta - \frac {g x}{2 u ^ {2}} (1 + \tan^ {2} \theta) \\ \Rightarrow & y = 2 x - \frac {(1 0) x ^ {2}}{2 (5)} (1 + 4) \\ \Rightarrow & y = 2 x - 5 x ^ {2} \end{array}$
Hence, the correct answer is (D).

$H = \frac {u ^ {2} \sin^ {2} \theta}{2 g} = \frac {(2 0 \sqrt {2}) ^ {2} \sin^ {2} (4 5 ^ {\circ})}{2 (1 0)} = 2 0 \mathrm{m}$

Since at maximum height $x = \frac{R}{2}$ , so we get
$r ^ {2} = \frac {R ^ {2}}{4} + H ^ {2}$
where $R = \frac{u^{2} \sin(2\theta)}{g} = 80 \, \text{m}$ $\Rightarrow r=20\sqrt{5}\ m$ Now $\omega=\frac{v_{\perp}}{r}=\frac{v\sin\theta}{r}$ where $\sin\theta=\frac{H}{r}$ $\Rightarrow \omega=\frac{vH}{r^{2}}=\frac{(20)(20)}{400(5)}=\frac{1}{5}=0.2\ rads^{-1}$
Hence, the correct answer is (B).