Kinematics-2D

Coordinates of the point P are $(R, -h)$ . Since the point P lies on the trajectory, so $(R, -h)$ must satisfy
$y = x \tan \theta - \frac {g x ^ {2}}{2 u ^ {2}} \big (1 + \tan^ {2} \theta \big)$
$\begin{array}{r l} \Rightarrow & - h = R \tan \theta - \frac {g R ^ {2}}{2 (2 a g)} (1 + \tan^ {2} \theta) \\ \Rightarrow & R ^ {2} \tan^ {2} \theta - 4 a R \tan \theta + (R ^ {2} - 4 a h) = 0 \end{array}$
For $\theta$ to be real, we must have
$\begin{array}{r l} & {\mathrm{DISCRIMINANT} \geq 0} \\ & {(4 a R) ^ {2} - 4 R ^ {2} (R ^ {2} - 4 a h) \geq 0} \\ {\Rightarrow} & {4 a ^ {2} \geq R ^ {2} - 4 a h} \\ {\Rightarrow} & {R ^ {2} \leq 4 a (a + h)} \\ {\Rightarrow} & {R \leq 2 \sqrt {a (a + h)}} \\ {\Rightarrow} & {R _ {M A X} = 2 \sqrt {a (a + h)}} \end{array}$
Hence, the correct answer is (D).

$\begin{array}{l} \vec {r} = (2 t) \hat {i} + (2 t ^ {2}) \hat {j} \\ \Rightarrow x = 2 t \text {and} y = 2 t ^ {2} \\ \Rightarrow \frac {d x}{d t} = v _ {x} = 2 \text {and} \frac {d y}{d t} = v _ {y} = 4 t \end{array}$
Since $\theta$ is the angle which $\vec{v}=v_{x}\hat{i}+v_{y}\hat{j}$ makes with the positive x-axis, so
... (1)

Differentiating both sides w.r.t. t we get
$\begin{array}{r l} & {\sec^ {2} \theta \left(\frac {d \theta}{d t}\right) = 2} \\ {\Rightarrow} & {\frac {d \theta}{d t} = \frac {2}{\sec^ {2} \theta} = \frac {2}{1 + \tan^ {2} \theta}} \end{array}$
From (1),
$\begin{array}{r l} & \tan \theta = 2 t \\ \Rightarrow & \frac {d \theta}{d t} = \frac {2}{1 + 4 t ^ {2}} \\ \text { So, } & \left. \frac {d \theta}{d t} \right| _ {t = 0. 5 \mathrm{s}} = \frac {2}{1 + 4 (0 . 2 5)} = 1 \mathrm{rads} ^ {- 1} \end{array}$
Hence, the correct answer is (D).

Done in theory.
Hence, the correct answer is (B).

For Q
Since acceleration due to gravity has no component along AB. So motion of the particle along AB is non-accelerated motion with uniform velocity v. If $t_{Q}$ is the time taken by particle Q to go from A to B, then
$t _ {Q} = \frac {A B}{v}\tag{... (1}$
For P
Since motion of particle P from A to C is accelerated and that from C to B is retarded. So from A to C the horizontal component of velocity gradually increases to attain a maximum value at the lowest point which further decreases gradually to attain the same value v at point B.
That is the motion from A to B to C experiences a horizontal velocity which has a value more than v. So $t_{Q} > t_{P}$ .
Hence, the correct answer is (A).

Tangential acceleration, for a simple pendulum is
$a _ {T} = g \sin \theta$
So, at $\theta = 30^{\circ}$ , we have
$a _ {T} = g \sin (3 0 ^ {\circ}) = \frac {g}{2} = 5 \mathrm{ms} ^ {- 2}$
Further we know that $a_{C} = \ell \omega^{2}$
$\begin{array}{l l} \Rightarrow & a _ {C} = \left(\frac {2 0 0}{1 0 0 0}\right) (9) ^ {2} \\ \Rightarrow & a _ {C} = \left(\frac {2}{1 0}\right) 8 1 = 1 6. 2 \mathrm{ms} ^ {- 2} \end{array}$
Since the total acceleration is
Hence, the correct answer is (D).

Velocity of train relative to bomber is $u_{r}$ , then
$\begin{array}{l} u _ {r} = 5 0 0 - 1 0 0 = 4 0 0 \mathrm{kmh} ^ {- 1} \\ \Rightarrow u _ {r} = 4 0 0 \times \frac {5}{1 8} \mathrm{ms} ^ {- 1} = \frac {1 0 0 0}{9} \mathrm{ms} ^ {- 1} \end{array}$

Since, $y = \frac{gx^{2}}{2u_{r}^{2}}$
$\Rightarrow \quad 5 0 0 0 = \frac {g x ^ {2}}{2 \left(\frac {1 0 0 0}{9}\right) ^ {2}}$
$\Rightarrow x = 3 5 1 3. 6 5 \mathrm{m}$
$\mathrm{Further} \tan \theta = \frac {y}{x}$
$\Rightarrow \tan \theta = \frac {5 0 0 0}{3 5 1 3 . 6 5}$
$\Rightarrow \quad \theta \simeq 5 5 ^ {o}$
Hence, the correct answer is (A).

Let AC = x and BE = y. Then
$B E ^ {2} + A E ^ {2} = \ell^ {2}$

$\Rightarrow y ^ {2} + \left(\frac {x}{2}\right) ^ {2} = \ell^ {2}$
$\Rightarrow 2 y \left(\frac {d y}{d t}\right) + \frac {x}{2} \left(\frac {d x}{d t}\right) = 0$
$\Rightarrow - \frac {d y}{d t} = \frac {1}{2} \left(\frac {x}{2 y}\right) \left(\frac {d x}{d t}\right)$
When the rhombus is a square, then x = 2y
$\Rightarrow v _ {B} = \frac {1}{2} v _ {c} = \frac {v}{2}$
Hence, the correct answer is (B).

$v = \sqrt{v_{r}^{2} + v_{\theta}^{2}}$
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).