Kinematics-2D

$v = \sqrt{v_{r}^{2} + v_{\theta}^{2}}$
Here $v_{r}=\frac{dr}{dt}=2\ ms^{-1}$
Since $v_{\theta} = r \cdot \frac{d\theta}{dt} = (2 \times 1)(4) = 8 \, \mathrm{ms}^{-1}$
$\Rightarrow v = \sqrt {6 4 + 4} = \sqrt {6 8} = 8. 2 5 \mathrm{ms} ^ {- 1}$
Hence, the correct answer is (C).

$a _ {T} = \frac {v ^ {2} - u ^ {2}}{2 \ell} = \frac {(1 5) ^ {2} - (5) ^ {2}}{2 (2 0 0)} = 0. 5 \mathrm{ms} ^ {- 2}$
$a _ {C} = \frac {v _ {\text {final}}}{r} = \frac {(1 5) ^ {2}}{\left(\frac {4 0 0}{\pi}\right)} \quad \left\{\because l = \frac {2 \pi r}{4} = 2 0 0 \right\}$
$\Rightarrow a _ {C} = 1. 8 \mathrm{ms} ^ {- 2}$
$\Rightarrow a = \sqrt {(1 . 8) ^ {2} + (0 . 5) ^ {2}}$
$\Rightarrow a = 1. 9 \mathrm{ms} ^ {- 2}$
Hence, the correct answer is (C).

For a particle moving in a circle with constant angular velocity $\omega$ , the figure shows the discussion at time t.

Initial position vector of the particle is
$\vec {r} _ {i} = a \hat {i}$
Final position vector of the particle at time t is
$\vec {r} _ {f} = (a \cos \theta) \hat {i} + (a \sin \theta) \hat {j}$
So, displacement of the particle is
$\begin{array}{r l} & {\overrightarrow {A B} = \Delta \vec {r} = \vec {r} _ {f} - \vec {r} _ {i}} \\ {\Rightarrow} & {\Delta \vec {r} = (a \cos \theta - a) \hat {i} + (a \sin \theta) \hat {j}} \\ {\Rightarrow} & {| \Delta \vec {r} | = a \sqrt {(\cos \theta - 1) ^ {2} + \sin^ {2} \theta}} \\ {\Rightarrow} & {| \Delta \vec {r} | = a \sqrt {\cos^ {2} \theta + 1 - 2 \cos \theta + \sin^ {2} \theta}} \end{array}$
Since $\cos^2\theta +\sin^2\theta = 1$
$\Rightarrow \quad | \Delta \vec {r} | = a \sqrt {1 + 1 - 2 \cos \theta} = \sqrt {2} a \sqrt {1 - \cos \theta}$
$\text { Further } 1 - \cos \theta = 2 \sin^ {2} \left(\frac {\theta}{2}\right)$
$\Rightarrow \quad | \Delta \vec {r} | = \sqrt {2} a \sqrt {2 \sin^ {2} \left(\frac {\theta}{2}\right)} = 2 \sin \left(\frac {\theta}{2}\right)$
Since $\theta = \omega t$
$\Rightarrow \quad | \Delta \vec {r} | = 2 a \sin \left(\frac {\omega t}{2}\right)$
Hence, the correct answer is (D).

Since both bodies are moving under influence of gravity hence relative acceleration is zero i.e. both move with constant velocity relative to each other

$\vec{v}_{12}=$ Relative velocity of 1 with respect to 2
$\vec {v} _ {1 2} = \vec {v} _ {1} - \vec {v} _ {2}$
$\left| \vec {v} _ {1 2} \right| = \sqrt {v _ {1} ^ {2} + v _ {2} ^ {2} - 2 v _ {1} v _ {2} \cos \left(\frac {\pi}{2} - \alpha\right)}$
$\left| \vec {v} _ {1 2} \right| = v _ {0} \sqrt {2 (1 - \sin \alpha)} = 2 v _ {0} \sin \left(\frac {\pi}{4} - \frac {\alpha}{2}\right)$
$\left| \vec {r} _ {1 2} \right| = \left| \vec {v} _ {1 2} \right| t$
$\left| \vec {r} _ {1 2} \right| = 2 v _ {0} t \sin \left(\frac {\pi}{4} - \frac {\alpha}{2}\right)$
Hence, the correct answer is (C).

For shell to hit the boat, time taken by the boat to go from B to C with uniform velocity $v_{1}$ equals the time taken by shell to go from O to A to C i.e. $\frac{2v_{2}\sin\alpha}{g}=T$ . Further, we observe
$\begin{array}{r l} & x + v _ {1} T = R = \frac {2}{g} (v _ {2} \sin \alpha) (v _ {2} \cos \alpha) \\ \Rightarrow & x = \frac {2}{g} (v _ {2} \sin \alpha) (v _ {2} \cos \alpha - v _ {1}) \end{array}$
Hence, the correct answer is (C).

When range is maximum, the $h=\frac{R}{4}$ . So h=100 m.
Velocity of the projectile is minimum at the maximum height. So this point has coordinates (200, 100)
Hence, the correct answer is (B).

$\begin{array}{r l} y & = x \tan \theta - \frac {g x ^ {2}}{2 u ^ {2} \cos^ {2} \theta} \\ y & = m x - \frac {g x ^ {2}}{2 u ^ {2}} (1 + m ^ {2}), \text { where } m = \tan \theta \\ \Rightarrow & \left(\frac {g x ^ {2}}{2 u ^ {2}}\right) m ^ {2} - x m + \left(\frac {g x ^ {2}}{2 u ^ {2}} + y\right) = 0 \end{array}\tag{... (1}$
Equation (1) is a quadratic in $m$ with two roots $m_1$ and $m_2$ .
These two roots $m_{1}$ and $m_{2}$ of equation (1) give us the two firing angles for the two trajectories shown (in the statement) such that the projectiles pass through a common point A. This point A will approach the envelope E as the two roots approach equality i.e. the DISCRIMINANT for quadratic in m must equal zero.
$\begin{array}{r l} & x ^ {2} - 4 \left(\frac {g x ^ {2}}{2 u ^ {2}}\right) \left(\frac {g x ^ {2}}{2 u ^ {2}} + y\right) = 0 \\ \Rightarrow & 1 - \left(\frac {2 g}{u ^ {2}}\right) \left(\frac {g x ^ {2}}{2 u ^ {2}} + y\right) = 0 \\ \Rightarrow & \frac {u ^ {2}}{2 g} - \frac {g x ^ {2}}{2 u ^ {2}} - y = 0 \\ \Rightarrow & y = \frac {u ^ {2}}{2 g} - \frac {g x ^ {2}}{2 u ^ {2}} \end{array}$
is the equation of the envelope $E$ .
Hence, the correct answer is (A).

For both the particles launched simultaneously to hit the point $C$ , we have
$t _ {A} = t _ {B}$
$\begin{array}{r l} & {\frac {2 u \sin \theta}{g} = \sqrt {\frac {2 h}{g}}} \\ {\Rightarrow} & {\frac {2 (1 0) (\sin (6 0 ^ {\circ}))}{1 0} = \sqrt {\frac {2 h}{1 0}}} \\ {\Rightarrow} & {2 \frac {\sqrt {3}}{2} = \sqrt {\frac {2 h}{1 0}}} \\ {\Rightarrow} & {\sqrt {3} = \sqrt {\frac {2 h}{1 0}}} \\ {\Rightarrow} & {h = 1 5 \mathrm{m}} \end{array}$
Hence, the correct answer is (C).

$y = x \tan \alpha - \frac{gx^{2}}{2u^{2} \cos^{2} \alpha}$
For y to be maximum
$\begin{array}{r l} \Rightarrow & \frac {d y}{d \alpha} = 0 \\ \Rightarrow & \tan \alpha = \frac {u ^ {2}}{g x} \end{array}$
Hence, the correct answer is (C).

$\begin{array}{l} y = x \tan \alpha - \frac {g x ^ {2}}{2 u ^ {2} \cos^ {2} \alpha} \\ \Rightarrow \quad y = x \tan \alpha - \frac {g x ^ {2}}{2 u ^ {2}} (1 + \tan^ {2} \alpha) \\ \Rightarrow \quad y _ {\max} = \frac {u ^ {2}}{2 g} - \frac {g x ^ {2}}{2 u ^ {2}} = h \end{array} \qquad \left\{\because \tan \alpha = \frac {u ^ {2}}{g x} \right\}$
Hence, the correct answer is (D).

Motion of one projectile as seen from other is a straight line inclined at acute angle with the horizontal. Hence, the correct answer is (D).

$\left(u _ {y}\right) _ {1} = 1 0 \sin (3 0 ^ {\circ}) = 5 \mathrm{ms} ^ {- 1}\tag{53.}$
$\left(u _ {y}\right) _ {2} = \frac {1 0}{\sqrt {3}} \sin (6 0 ^ {\circ}) = 5 \mathrm{ms} ^ {- 1}$
Since $\left(u_y\right)_1 = \left(u_y\right)_2$ and time of flight is $T = \frac{2u_y}{g}$
$\Rightarrow T _ {1} = T _ {2}$
i.e., both the particles strike the ground at the same instant.
So, the maximum separation between them is the difference of their horizontal ranges.
$\Rightarrow \Delta x = R _ {1} - R _ {2}$
$\begin{array}{r l} & {\Rightarrow \Delta x = \frac {(2 0) ^ {2} \sin (6 0 ^ {\circ})}{g} - \frac {\left(\frac {2 0}{\sqrt {3}}\right) ^ {2} \sin (1 2 0 ^ {\circ})}{g}} \\ & {\Rightarrow \Delta x = 2 0 \sqrt {3} - \frac {2 0 \sqrt {3}}{3}} \\ & {\Rightarrow \Delta x = 2 0 \sqrt {3} \bigg (1 - \frac {1}{3} \bigg)} \\ & {\Rightarrow \Delta x = \frac {4 0}{\sqrt {3}} = 2 3 \mathrm{m}} \end{array}$
Hence, the correct answer is (C).

Since horizontal components of both are the same, so relative horizontal velocity $(v_{12})_{x}$ is zero and hence $\vec{v}_{12}$ (velocity of 1 w.r.t. 2) has only a vertical component please note that the angles $\theta_{1}$ and $\theta_{2}$ are with the vertical and hence the horizontal component of velocities are $u_{1}\sin\theta_{1}$ and $u_{2}\sin\theta_{2}$ which are actually equal.
Hence, the correct answer is (D).

Since $y=\frac{gx^{2}}{2u^{2}}$
... (1)
When x = nb and y = nh, then
$n = \frac {2 h u ^ {2}}{g b ^ {2}}$
{Put values in (1)}
Hence, the correct answer is (B).

METHOD I
$y = a x - b x ^ {2}$
Compare with
$y = x \tan \theta - \left(\frac {g}{2 u ^ {2} \cos^ {2} \theta}\right) x ^ {2}$
we get
$\begin{array}{l} \tan \theta = a \text {and} \frac {g}{2 u ^ {2} \cos^ {2} \theta} = b \\ H _ {M A X} = \frac {u ^ {2} \sin^ {2} \theta}{2 g} = \left(\frac {g}{2 b \cos^ {2} \theta}\right) \frac {\sin^ {2} \theta}{2 g} \\ \Rightarrow H _ {M A X} = \frac {\tan^ {2} \theta}{4 b} = \frac {a ^ {2}}{4 b} \end{array}$
and $\theta = \tan^{-1}(a)$
METHOD II
At maximum height
${\frac {d y}{d x}} = 0$
$\begin{array}{r l} \Rightarrow & a - 2 b x = 0 \\ \Rightarrow & x = \frac {a}{2 b} \\ \Rightarrow & y _ {M A X} = y \Bigg | _ {x = \frac {a}{2 b}} = a \left(\frac {a}{2 b}\right) - b \left(\frac {a ^ {2}}{4 b ^ {2}}\right) \\ \Rightarrow & y _ {M A X} = \frac {a ^ {2}}{4 b} \end{array}$
$\mathrm{Now} \frac {d y}{d x} = a - 2 b x$
So, the slope at the point of launch $(x=0)$ is $\tan\theta$ . Hence
$\begin{array}{r l} & {\left. \frac {d y}{d x} \right| _ {x = 0} = a = \tan \theta} \\ {\Rightarrow} & {\theta = \tan^ {- 1} (a)} \end{array}$
Hence, the correct answer is (C).

$\begin{array}{l l} \frac {d x}{d t} = 8 \pi \sin (2 \pi t) \\ \Rightarrow & d x = 8 \pi \sin (2 \pi t) d t \\ \Rightarrow & \int_ {8} ^ {x} d x = 8 \pi \int_ {0} ^ {t} \sin (2 \pi t) d t \\ \Rightarrow & x - 8 = - 8 \pi \frac {\cos (2 \pi t)}{2 \pi} \Big | _ {0} ^ {t} \\ \Rightarrow & x - 8 = 4 \cos (2 \pi t - 1) \\ \Rightarrow & x - 8 = 4 \cos (2 \pi t) - 4 \\ \Rightarrow & x - 4 = 4 \cos (2 \pi t) \\ \text {Further since,} & \frac {d y}{d t} = 8 \pi \cos (2 \pi t) \\ \Rightarrow & d y = 8 \pi \cos (2 \pi t) d t \\ \Rightarrow & \int_ {0} ^ {y} d y = 8 \pi \int_ {0} ^ {t} \cos (2 \pi t) d t \\ \Rightarrow & y = \frac {8 \pi \sin (2 \pi t)}{2 \pi} \Big | _ {0} ^ {t} \\ \Rightarrow & y = 4 \sin (2 \pi t) \end{array}\tag{... (1}$
From (1) and (2), we have
$(x - 4) ^ {2} + y ^ {2} = 4\tag{... (2}$
This is the equation of a circle with centre at $(4, 0)$ and radius 2.
Hence, the correct answer is (B).

Done Earlier.
Hence, the correct answer is (D).

$H = \frac {u ^ {2} \sin^ {2} \theta}{2 g} = \frac {u _ {y} ^ {2}}{2 g}$
Since $u_{y}$ and $a_{y}$ remain the same as previous, so H remains same.
Also, $T = \frac{2u \sin \theta}{g} = \frac{2u_{y}}{g}$
So, $T$ also remains the same
Now, $R_{new} = u_{x}T + \frac{1}{2}a_{x}T^{2}$
$\Rightarrow R _ {\text { new }} = R + \frac {1}{2} \left(\frac {g}{2}\right) \left(\frac {4 u _ {y} ^ {2}}{g ^ {2}}\right) = R + \frac {H}{2}$
Hence, the correct answer is (B).

$\begin{array}{l} \text {Since} \frac {d x}{d t} = \frac {d y}{d t} = c \\ \Rightarrow \frac {d ^ {2} x}{d t ^ {2}} = \frac {d ^ {2} y}{d t ^ {2}} = 0 \\ \text {Further} z = a x ^ {3} + b y ^ {2} \\ \Rightarrow \frac {d z}{d t} = 3 a x ^ {2} \frac {d x}{d t} + 2 b y \frac {d y}{d t} \\ \Rightarrow \frac {d z}{d t} = 3 a c x ^ {2} + 2 b c y \qquad \left\{\because \frac {d x}{d t} = c = \frac {d y}{d t} \right\} \\ \Rightarrow \frac {d ^ {2} z}{d t ^ {2}} = 6 a c x \bigg (\frac {d x}{d t} \bigg) + 2 b c \bigg (\frac {d y}{d t} \bigg) \\ \Rightarrow \frac {d ^ {2} z}{d t ^ {2}} = 6 a c ^ {2} x + 2 b c ^ {2} \end{array}$
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).