Kinematics-2D

Apply formula of time of flight on inclined plane. Hence, the correct answer is (B).

$T = \frac{2u \sin \theta}{g}$ $\Rightarrow R = \frac{u^{2} \sin(2\theta)}{g} - \frac{2u^{2} \sin^{2} \theta}{g}$ R will be maximum when, $\frac{dR}{d\theta} = 0$ $\Rightarrow 2 \cos(2\theta) - 2 \sin(2\theta) = 0$ $\Rightarrow \tan(2\theta) = 1$ $\Rightarrow 2\theta = \frac{\pi}{4}$ $\Rightarrow \theta = \frac{\pi}{8}$ radian
Hence, the correct answer is (B).

$u = 20\mathrm{ms}^{-1},\theta = 45^{\circ}$ Given $h = Ax - Bx^{2}$ Comparing with, $y = x\tan \theta -\frac{gx^2}{2u^2\cos^2\theta}$
We get $A = \tan\theta = \tan45^{\circ} = 1$ and $B = \frac{b}{2u^{2}\cos^{2}\theta} = \frac{10}{2(20)^{2}\left(\frac{1}{2}\right)} = \frac{1}{40}$ $\Rightarrow \frac{A}{B}=\frac{40}{1}$
Hence, the correct answer is (D).

$\begin{array}{l} v _ {x} = \frac {d x}{d t} = 2 + 8 t \\ \Rightarrow a _ {x} = \frac {d v _ {x}}{d t} = 8 \\ v _ {y} = \frac {d y}{d t} = 5 \\ \Rightarrow a _ {y} = \frac {d v _ {y}}{d t} = 0 \end{array}$
Hence, the correct answer is (C).

$v_{x}=\frac{dx}{dt}=v_{0}$ $\Rightarrow x=v_{0}t$ Since, $v_{y}=\frac{dy}{dt}=a\omega\cos(\omega t)$ $\Rightarrow y=\int_{0}^{t}a\omega\cos(\omega t)dt=\frac{a\omega\sin(\omega t)}{\omega}$ $\Rightarrow y=a\sin(\omega t)=a\sin\left(\frac{\omega x}{v_{0}}\right)$ Hence, the correct answer is (C).

$\begin{array}{l} \frac {\frac {1}{2} m u _ {1} ^ {2} \cos^ {2} \theta_ {1}}{\frac {1}{2} m u _ {2} ^ {2} \cos^ {2} \theta_ {2}} = \frac {4}{1} \\ \Rightarrow \quad \frac {u _ {1} \cos \theta_ {1}}{u _ {2} \cos \theta_ {2}} = 2 \\ \text { and } \quad \frac {u _ {1} ^ {2} \sin^ {2} \theta_ {1}}{u _ {2} ^ {2} \sin^ {2} \theta_ {2}} = \frac {4}{1} \\ \Rightarrow \quad \frac {u _ {1} \sin \theta_ {1}}{u _ {2} \sin \theta_ {2}} = \frac {2}{1} \end{array}$
From equations (1) and (2), we get
$\frac {(u _ {1} \sin \theta_ {1}) (u _ {1} \cos \theta_ {1})}{(u _ {2} \sin \theta_ {2}) (u _ {2} \cos \theta_ {2})} = \frac {4}{1}$
$\Rightarrow \quad \frac {\frac {g R _ {1}}{2}}{\frac {g R _ {2}}{2}} = \frac {4}{1}$
$\Rightarrow \frac {R _ {1}}{R _ {2}} = \frac {4}{1}$
Hence, the correct answer is (B).

$\begin{array}{l} {2 = \sqrt {u ^ {2} \sin^ {2} \theta - 2 h g}} \\ {\Rightarrow \quad 4 = u ^ {2} \sin^ {2} \theta - 2 (0. 4) (1 0)} \end{array}$

$\begin{array}{l l} \Rightarrow & u ^ {2} \sin^ {2} \theta = 1 2 \\ \Rightarrow & u \sin \theta = 2 \sqrt {3} \text {and} u \cos \theta = 6 \\ \Rightarrow & \tan \theta = \frac {1}{\sqrt {3}} \\ \Rightarrow & \theta = 3 0 ^ {\circ} \end{array}$
Hence, the correct answer is (C).

$\begin{array}{l} u _ {x} = 1 4. 7 \mathrm{ms} ^ {- 1}, \\ u _ {y} = 9. 8 \times 2 = 1 9. 6 \mathrm{ms} ^ {- 1} \\ \Rightarrow u = \sqrt {(1 4 . 7) ^ {2} + (1 9 . 6) ^ {2}} = 2 4. 5 \mathrm{ms} ^ {- 1} \\ \Rightarrow \tan \theta = \frac {1 9 . 6}{1 4 . 7} = \frac {4}{3} \\ \Rightarrow \theta = 5 3 ^ {\circ} \\ \text { Hence,the correct answer is(D). } \end{array}$

Since, both projectiles have equal height, so
$H = \frac {u ^ {2} \sin^ {2} \theta}{2 g} = \frac {u _ {y} ^ {2}}{2 g}$
So, time of flight of both will be same.
$\Rightarrow T _ {1} = T _ {2} = \frac {2 u \sin \theta}{g} = \frac {2 u _ {y}}{g}$
Since Range is greater for second, i.e.
$R _ {2} > R _ {1}$
Thus $\left(u_x\right)_2 > \left(u_x\right)_1$
$\Rightarrow u _ {2} > u _ {1}$
Hence, the correct answer is (D).

Acceleration of the bead down the wire is $g \cos \theta$ .
Hence $v^{2}=0^{2}+2(g\cos\theta)(2R\cos\theta)$ Where $P_{1}P_{2}=2R\cos\theta$ $\Rightarrow v=2\left(\sqrt{gR}\right)\cos\theta$ Since, $t=\frac{v}{a}=\frac{2\sqrt{gR}\cos\theta}{g\cos\theta}=2\sqrt{\frac{R}{g}}$
Hence, (B), (C) and (D) are correct.

$y = x \tan \theta \left(1 - \frac {x}{R}\right)$
where $x = (u\cos \theta)t$ and $R = (u\cos \theta)T$
$\ldots(2)$
Also, we observe that
$\begin{array}{l l} & R = \frac {2 u ^ {2} \sin \theta \cos \theta}{g} \text {and} H = \frac {u ^ {2} \sin^ {2} \theta}{2 g} \\ \Rightarrow & \frac {H}{R} = \frac {\sin^ {2} \theta}{4 \sin \theta \cos \theta} = \frac {1}{4} \tan \theta \\ \Rightarrow & \tan \theta = \frac {4 H}{R} \end{array}\tag{... (3}$
So, we get
$y = \frac {4 H x}{R} \left(1 - \frac {x}{R}\right)$
Substitute x and R from (2), we get
$y = 4 H \left(\frac {t}{T}\right) \left(1 - \frac {t}{T}\right)$
Hence, (A) and (C) are correct.

$\vec {u} = (u \cos \theta) \hat {i} + (u \sin \theta) \hat {j}\tag{... (1}$
Let the velocity at point Q, at time t be $\vec{v}$ , then
$\vec {v} = (u \cos \theta) \hat {i} + (u \sin \theta - g t) \hat {j}$
Since $\vec{v} \perp \vec{u}$ , so
$\begin{array}{r l} & {\vec {v} \cdot \vec {u} = 0} \\ {\Rightarrow} & {u ^ {2} - u g t \sin \theta = 0} \end{array}$
$\Rightarrow t = \frac {u}{g \sin \theta} = \left(\frac {u}{g}\right) \mathrm{cosec} \theta$
Speed of the particle at Q is $|\vec{v}|$ , where
$| \vec {v} | = \sqrt {u ^ {2} + g ^ {2} t ^ {2} - 2 u g t \sin \theta}$
Substituting the value of t, we get
$| \vec {v} | = u \cot \theta$
Hence, (C) and (D) are correct.

$\begin{array}{l l} & h = 2 \sqrt {h g} (\sin \theta) t - \frac {1}{2} g t ^ {2} \\ \Rightarrow & t ^ {2} - 4 \sqrt {\frac {h}{g}} (\sin \theta) t + \frac {2 h}{g} = 0 \\ \Rightarrow & t = \frac {4 \sqrt {\frac {h}{g}} \pm \sqrt {1 6 \left(\frac {h}{g}\right) \sin^ {2} \theta - \frac {8 h}{g}}}{2} \\ \Rightarrow & t = \frac {4 \sqrt {\frac {h}{g}} \pm \sqrt {\frac {8 h}{g}} \sqrt {2 \sin^ {2} \theta - 1}}{2} \\ \Rightarrow & t _ {1} = \frac {4 \sqrt {\frac {h}{g}} - \sqrt {\frac {8 h}{g}} \sqrt {2 \sin^ {2} \theta - 1}}{2} \\ \Rightarrow & t _ {2} = \frac {4 \sqrt {\frac {h}{g}} + \sqrt {\frac {8 h}{g}} \sqrt {2 \sin^ {2} \theta - 1}}{2} \\ \Rightarrow & (t _ {2} - t _ {1}) = \sqrt {\frac {8 h}{g}} \sqrt {2 \sin^ {2} \theta - 1} \end{array}$
Further horizontal distance travelled during this time is 2h
$\begin{array}{l l} \Rightarrow & 2 h = 2 \sqrt {h g} \cos \theta (t _ {2} - t _ {1}) \\ \Rightarrow & 2 h = 2 \sqrt {h g} \cos \theta \sqrt {\frac {8 h}{g}} \sqrt {2 \sin^ {2} \theta - 1} \\ & \text {Solving we get} \cos \theta = \frac {1}{2} \end{array}$
$\theta = 60^{\circ}$ with horizontal or $30^{\circ}$ with vertical
$\Rightarrow \quad (t _ {2} - t _ {1}) = 2 \sqrt {\frac {h}{g}}$
Hence, (A) and (D) are correct.

$\begin{aligned} &\vec{u} = 3\hat{i}\text{ m/s} \quad \text{and} \quad \vec{a} = -\hat{i} - 0.5\hat{j}\text{ m/s}^2 \\ &\Rightarrow u_x = 3\text{ ms}^{-1}, \ a_x = -1\text{ ms}^{-2} \quad \text{and} \quad a_y = -0.5\text{ ms}^{-2} \end{aligned}$
Since $u_x$ and $a_x$ have opposite signs, the particle must first reverse its direction of motion along the x-axis. Let this happen when the x-coordinate of the particle is $x$. Then:
$\begin{aligned} & 0^2 - u_x^2 = 2 a_x x \\ \Rightarrow & -(3)^2 = 2 (-1) x \\ \Rightarrow & x = 4.5\text{ m} \end{aligned}$
Also, if this happens at time $t$, then:
$\begin{aligned} & v_x = u_x + a_x t \quad \text{gives} \quad 0 = 3 + (-1) t \\ \Rightarrow & t = 3\text{ s} \end{aligned}$
At this time $t$, the y-coordinate of the particle is given by:
$\begin{aligned} & y = u_y t + \frac{1}{2} a_y t^2 \\ \Rightarrow & y = 0 + \frac{1}{2} (-0.5) (3)^2 \\ \Rightarrow & y = -\frac{9}{4}\text{ m} = -2.25\text{ m} \\ \text{Also}, \quad & v_y = 0 + a_y t = -(0.5) (3) = -1.5\text{ ms}^{-1} \\ \Rightarrow & \vec{v} = v_x \hat{i} + v_y \hat{j} = 0 \hat{i} + (-1.5) \hat{j} = -1.5\hat{j}\text{ ms}^{-1} \\ \text{and} \quad & \vec{r} = x \hat{i} + y \hat{j} = (4.5\hat{i} - 2.25\hat{j})\text{ m} \end{aligned}$
Hence, (A) and (D) are correct.

C,D

Since the particles have same ranges.
$\begin{array}{r l} & {\mathrm{Hence} \alpha + \beta = \frac {\pi}{2}} \\ & {\qquad R = \frac {2 u ^ {2} \sin \alpha \cos \alpha}{g} = \frac {2 u ^ {2} \sin \beta \cos \beta}{g}} \\ & {\qquad h _ {1} = \frac {u ^ {2} \sin^ {2} \alpha}{2 g}} \\ & {\qquad h _ {2} = \frac {u ^ {2} \sin^ {2} \beta}{2 g} = \frac {u ^ {2} \cos^ {2} \alpha}{2 g}} \\ {\Rightarrow} & {\quad R = 4 \sqrt {h _ {1} h _ {2}}} \\ & {\qquad t _ {1} = \frac {2 u \sin \alpha}{g}} \\ & {\qquad t _ {2} = \frac {2 u \sin \beta}{g} = \frac {2 u \cos \alpha}{g}} \\ {\Rightarrow} & {\quad \frac {t _ {1}}{t _ {2}} = \tan \alpha \text {and} \frac {h _ {1}}{h _ {2}} = \tan^ {2} \alpha} \end{array}$
$\Rightarrow \tan \alpha = \sqrt {\frac {h _ {1}}{h _ {2}}}$
Hence, (A), (B), (C) and (D) are correct.

$\Rightarrow \frac {x ^ {2}}{a ^ {2}} + \frac {y ^ {2}}{b ^ {2}} = 1$
$\ddot {y} = - p ^ {2} y$
so acceleration is directed towards the focus. Hence, (A), (B) and (C) are correct.

$0 = \sqrt {2 g h} (\cos \alpha) t - \frac {1}{2} (g \cos \alpha) t ^ {2}$
$\Rightarrow \quad t = \sqrt {\frac {8 h}{g}} \text { after it strikes the incline at A }$
$s = \sqrt {2 g h} (\sin \alpha) t + \frac {1}{2} (g \sin \alpha) t ^ {2}$
$\text { Substituting value of t,we get } s = 8 h \sin \alpha$
Hence,(A)and(D)are correct.

$\vec {u} = 4 \hat {i} + 3 \hat {j} + 3 \sqrt {3} \hat {k}$
$\Rightarrow \quad | \vec {u} | = 2 \sqrt {1 3} \mathrm{ms} ^ {- 1}$
$\text {Velocity at highest point is} \vec {v} = 4 \hat {i} + 3 \hat {j}$
$\{\text {because at the highest point v_{Z}} = 0 \}$
$\Rightarrow \quad | \vec {v} | = 5 \mathrm{ms} ^ {- 1}$
Hence, (B) and (D) are correct.

Let shells collide in mid-air at point $P(x, y)$. For the 1st shell:
$x = (u \cos \alpha) T \tag{1}$
$y = (u \sin \alpha) T - \frac{1}{2} g T^2 \tag{2}$
For the 2nd shell:
$x = u \cos \beta (T - t) \tag{3}$
$y = (u \sin \beta) (T - t) - \frac{1}{2} g (T - t)^2 \tag{4}$
Equating (1) and (3), we get:
$T \cos \alpha = (T - t) \cos \beta \tag{\text{OPTION (A)}}$
$\begin{aligned} \Rightarrow \frac{\cos \alpha}{\cos \beta} &= 1 - \frac{t}{T} \\ \Rightarrow \frac{\cos \alpha}{\cos \beta} &< 1 \\ \Rightarrow \cos \alpha &< \cos \beta \\ \Rightarrow \alpha &> \beta \end{aligned} \tag{\text{OPTION (B)}}$
Equating (2) and (4), we get:
$(u \sin \alpha) T - \frac{1}{2} g T^2 = (u \sin \beta) (T - t) - \frac{1}{2} g (T - t)^2 \tag{\text{OPTION (D)}}$
Hence, (A), (B) and (D) are correct.

A,B,C,D

Let the projectile reach the point of impact in time t. Then
$\begin{array}{r l r} & & {l \sin \theta = \frac {1}{2} g t ^ {2}} \\ & {\Rightarrow} & {t = \sqrt {\frac {2 \ell \sin \theta}{g}}} \end{array}\tag{... (1}$

Further
$\begin{array}{r l} & l \cos \theta + D = v t \\ \Rightarrow & D = v t - l \cos \theta \\ \Rightarrow & D = v \sqrt {\frac {2 l \sin \theta}{g}} - l \cos \theta \end{array}\tag{... (2}$
For projectile to pass over the observer's head
$\begin{array}{r l} & D > 0 \\ \Rightarrow & v \sqrt {\frac {2 l \sin \theta}{g}} > l \cos \theta \end{array}$
$\Rightarrow v ^ {2} \left(\frac {2 l \sin \theta}{g}\right) > l ^ {2} \cos^ {2} \theta$
\Rightarrow l < \frac {2 v ^ {2} \tan \theta \sec \theta}{g}
Hence, (B) and (C) are correct.

Since, we know that the range is same for complimentary angles i.e., $R_{\theta} = R_{90-\theta} = \frac{2u^{2}\sin\theta\cos\theta}{g}$
Then, $T_{1} = \frac{2u\sin\theta}{g}$ and $T_{2} = \frac{2u\cos\theta}{g}$
$\Rightarrow T_{1}T_{2}\propto R$ and
$\frac {T _ {1}}{T _ {2}} = \tan \theta$
Hence, (C) and (D) are correct.

Let the second shell reach the point $P(x, y)$ at time $t_{2}$ and the first shell be fired at a time
$\left(t _ {2} - \Delta t\right) = t _ {1}.$

For shell 1
$x = \left(5 \sqrt {3}\right) t_1 \tag{1}$
$-(10 - y) = - \frac{1}{2} g t_1^2 \tag{2}$
For shell 2
$x = \left(5 \sqrt{3} \cos 60^\circ\right) t_2 \tag{3}$
$-(10 - y) = (5 \sqrt{3} \sin 60^\circ) t_2 - \frac{1}{2} g t_2^2 \tag{4}$
Equating (1) and (3), we get:
$t_2 = 2 t_1 \tag{5}$
Equating (2) and (4), we get:
$\begin{aligned} - \frac{1}{2} g t_1^2 &= (5 \sqrt{3} \sin 60^\circ) t_2 - \frac{1}{2} g t_2^2 \\ \Rightarrow - \frac{1}{2} g t_1^2 &= \frac{15}{2} (2 t_1) - \frac{1}{2} g (2 t_1)^2 \\ \Rightarrow - \frac{1}{2} g t_1^2 &= 15 t_1 - 2 g t_1^2 \\ \Rightarrow - 5 t_1^2 &= 15 t_1 - 20 t_1^2 \\ \Rightarrow 15 t_1^2 &= 15 t_1 \\ \Rightarrow 15 t_1^2 - 15 t_1 &= 0 \end{aligned}$
$\Rightarrow 15 t_1 (t_1 - 1) = 0$
$\Rightarrow t_1 = 1\text{ s}$
$\Rightarrow t_2 = 2\text{ s} \quad \text{i.e., } \Delta t = 1\text{ s}$
$\Rightarrow x = 5 \sqrt{3}\text{ m} \quad \text{and} \quad y = 5\text{ m}$
Hence, (A), (B), (C) and (D) are correct.

Since $R=\frac{u^2\sin(2\theta)}{g}$,
$\Rightarrow R=kg^{-1} \tag{1}$
$\Rightarrow dR=-kg^{-2},dg \tag{2}$
$\Rightarrow \frac{dR}{R}=\frac{-kg^{-2},dg}{kg^{-1}}$
$\Rightarrow \frac{dR}{R}=-\frac{dg}{g}
=-\left(\frac{g_p-g_e}{g}\right)$
$\Rightarrow \text{Fractional Decrement}=\frac{dR}{R}=\frac{1}{291}$
Range at equator is
$R_e=\frac{u^2\sin(2\theta)}{g_e}$
Range at pole is
$R_p=\frac{u^2\sin(2\theta)}{g_p}$
$R = \left(\begin{array}{c} \text{Original range with} \\ \text{no variation of }g \\ \text{taken into account} \end{array}\right) = \frac{u^2\sin(2\theta)}{g}$
$\frac{dR}{R} = \left(\begin{array}{c} \text{Fractional} \\ \text{Decrement} \end{array}\right) = \frac{R_e-R_p}{R}$
$\Rightarrow \frac{dR}{R} = \frac{u^2\sin(2\theta)\left(\frac{1}{g_e}-\frac{1}{g_p}\right)} {u^2\sin(2\theta)\frac{1}{g}}$
$\Rightarrow \frac{dR}{R} = -g\left(\frac{1}{g_p}-\frac{1}{g_e}\right)$
Hence, (B) and (D) are correct.