Kinematics-2D

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

$\begin{array}{r l} x & = \frac {2 u ^ {2} \sin (\alpha - \beta) \cos \alpha}{g \cos^ {2} \beta} \text { where } \alpha = 6 0 ^ {\circ} \text { and } \beta = 3 0 ^ {\circ} \\ & \Rightarrow x = \frac {2 (3 0) ^ {2} \sin (3 0 ^ {\circ}) \cos (6 0 ^ {\circ})}{g \cos^ {2} (3 0 ^ {\circ})} \\ & \Rightarrow x = \frac {(2) (9 0 0) (0 . 5) (0 . 5)}{(1 0) \left(\frac {3}{4}\right)} = 6 0 \mathrm{m} \end{array} \tag {85.}$
Hence, the correct answer is (C).

Let us first calculate the components of the velocity of ball relative to the lift.
$\begin{array}{l} u _ {x} = 4 \cos (3 0 ^ {\circ}) = 2 \sqrt {3} \mathrm{ms} ^ {- 1} \text {and} \\ u _ {y} = 4 \sin (3 0 ^ {\circ}) = 2 \mathrm{ms} ^ {- 1} \end{array}$

The acceleration of the ball with respect to the lift is $(10+2)=12\ \text{ms}^{-2}$ in the negative y-direction (i.e., vertically downwards)
So, the time of flight $T$ is
$T = \frac {2 u _ {y}}{a _ {y}} = \frac {2 (2)}{1 2} = \frac {1}{3} \mathrm{s}$
Hence, the correct answer is (C).

Since $R = u_{x}T$
$\Rightarrow R = (2 \sqrt {3}) \left(\frac {1}{3}\right) = \frac {2}{\sqrt {3}} \mathrm{m}$
Hence, the correct answer is (B).

$\begin{array}{r l} x _ {\text { AIRCRAFT }} & = x _ {\text { BULLET }} \\ \Rightarrow (2 7 7) t & = (5 4 0 \cos \theta) t \\ & \left\{\because 1 0 0 0 \mathrm{kmh} ^ {- 1} = 2 7 7 \mathrm{ms} ^ {- 1} \right\} \\ \Rightarrow \cos \theta & = \frac {2 7 7}{5 4 0} \\ \Rightarrow \theta & \cong 5 9 ^ {\circ} \end{array} \tag {88.}$
Just think that $\frac{277}{540} \cong \frac{270}{540} = \frac{1}{2}$ , so $\theta$ must be an angle close to $60^{\circ}$ .
Hence, the correct answer is (C).

$v_{x} = 8t - 2$
$\begin{array}{l l} \Rightarrow & \frac {d x}{d t} = 8 t - 2 \\ \Rightarrow & \int_ {1 4} ^ {x} d x = 8 \int_ {2} ^ {t} t d t - 2 \int_ {2} ^ {t} d t \\ \Rightarrow & x - 1 4 = 4 (t ^ {2} - 4) - 2 (t - 2) \\ \Rightarrow & x - 1 4 = 4 t ^ {2} - 1 6 - 2 t + 4 \\ \Rightarrow & x = 4 t ^ {2} - 2 t + 2 \\ \text {Since} v _ {y} = 2 \\ \Rightarrow & \frac {d y}{d t} = 2 \\ \Rightarrow & \int_ {4} ^ {y} d y = 2 \int_ {2} ^ {t} d t \\ \Rightarrow & y - 4 = 2 (t - 2) \\ \Rightarrow & y = 2 t \\ \Rightarrow & t = \frac {y}{2} \end{array}$
Substituting in (1), we get
$x = 4 \left(\frac {y}{2}\right) ^ {2} - 2 \left(\frac {y}{2}\right) + 2$
$\Rightarrow x = y ^ {2} - y + 2$
Hence, the correct answer is (A).

The situation discussed is shown at time t. Then we observe that at any instant
$\ell^ {2} = x ^ {2} + y ^ {2}\tag{... (1}$

$\Rightarrow 0 = 2 x \frac {d x}{d t} + 2 y \frac {d y}{d t}$
$\Rightarrow \frac {d x}{d t} = - \frac {y}{x} \left(\frac {d y}{d t}\right)$
$\mathrm{From(1)}, x = \sqrt {\ell^ {2} - y ^ {2}}$
$\Rightarrow \frac {d x}{d t} = \frac {y v _ {0}}{\sqrt {\ell^ {2} - y ^ {2}}}$
$\left\{\because \frac {d y}{d t} = - v _ {0} \right\}$
So, the speed of the lower end decreases and vanishes when $y \rightarrow 0$ .
Hence, the correct answer is (B).

Initial velocity, $\vec{u}=(u\cos\theta)\hat{i}+(u\sin\theta)\hat{j}$
$\begin{array}{l} \text {Final velocity,} \vec {v} = (u \cos \theta) \hat {i} + (u \sin \theta - g t) \hat {j} \\ \text {So,} \Delta \vec {v} = \vec {v} - \vec {u} = - (g t) \hat {j} \\ \Rightarrow \quad \Delta \vec {v} = - (2 0) \hat {j} \\ \Rightarrow | \Delta \vec {v} | = 2 0 \mathrm{ms} ^ {- 1} \text {along - y direction} \\ \text {Hence, the correct answer is (C).} \end{array}$