Kinematics-2D

Since the projectile is at same height at times $t_{1}$ and $t_{2}$ . As studied, the sum of these times equals the time of flight of the projectile. So,
$T = t _ {1} + t _ {2} = 3 + 5 = 8 \mathrm{s}$
Hence, the correct answer is (D).

Taking upward direction as positive, we get
$\begin{array}{r l} & {- 8 0 = 1 2 t + \frac {1}{2} (- 1 0) t ^ {2}} \\ {\Rightarrow} & {- 8 0 = 1 2 t - 5 t ^ {2}} \\ {\Rightarrow} & {5 t ^ {2} - 1 2 t - 8 0 = 0} \end{array}$
$\begin{array}{r l} \Rightarrow & t = \frac {1 2 \pm \sqrt {1 4 4 + 1 6 0 0}}{2 (5)} \\ \Rightarrow & t \cong 5. 5 \mathrm{s} \end{array}$
Hence, the correct answer is (B).

$4 = (u\cos \alpha)t$
... (1)
$2 = \frac {u ^ {2} \sin^ {2} \alpha}{2 g}\tag{... (2}$
$- 3 = (u \sin \alpha) t - \frac {1}{2} g t ^ {2}\tag{... (3}$
From (2)
$u \sin \alpha = 2 \sqrt {g}\tag{... (4}$
Substituting in (3), we get
$\Rightarrow 4. 9 t ^ {2} - 6. 3 t - 3 = 0$
So, from (1), $4 = (u\cos \alpha)(8.1)$
$\Rightarrow u \cos \alpha = \frac {4}{8} \cong \frac {1}{2}\tag{... (5}$
From (4) and (5), we get
$\begin{array}{r l} & u ^ {2} \sin^ {2} \alpha + u ^ {2} \cos^ {2} \alpha = 4 0 + \frac {1}{4} \\ \Rightarrow & u = \sqrt {4 0 . 2 5} = 6. 3 \mathrm{ms} ^ {- 1} \end{array}$
Hence, the correct answer is (A).

$\tan \theta = \tan \alpha + \tan \beta$
$\Rightarrow \quad \tan \alpha + \tan \beta = \tan 6 0$
$\Rightarrow \quad \tan \alpha + \tan \beta = \sqrt {3}$
Hence, the correct answer is (D).

The radius of curvature at the instant the projectile has a velocity v making an angle $\beta$ with the horizontal is
$R = \frac {v ^ {2}}{g \cos \beta}$
where $\beta = 30^{\circ}$ and v has to be found. Since horizontal motion is non-accelerated motion, so
$\begin{array}{r l} & v _ {x} = u _ {x} \\ \Rightarrow & v \cos \beta = u \cos \theta \end{array}$
$\Rightarrow v \left(\frac {\sqrt {3}}{2}\right) = 2 0 \left(\frac {1}{2}\right)$
$\Rightarrow v = \frac {2 0}{\sqrt {3}} \mathrm{ms} ^ {- 1}$
$\Rightarrow R = \frac {\left(\frac {2 0}{\sqrt {3}}\right) ^ {2}}{g \cos (3 0 ^ {\circ})} = \frac {\frac {4 0 0}{3}}{1 0 \left(\frac {\sqrt {3}}{2}\right)} = \frac {8 0}{3 \sqrt {3}} = 1 5. 4 \mathrm{m}$
Hence, the correct answer is (C).

If $T$ be the time of flight, then
$\begin{array}{r l} & T = \frac {2 u \sin (\alpha - 3 0 ^ {\circ})}{g \cos (3 0 ^ {\circ})} \\ & \text {For the projectile to hit the plane normally, v_{x} = 0} \\ & \Rightarrow u _ {x} + a _ {x} T = 0 \\ & \Rightarrow u \cos (\alpha - 3 0 ^ {\circ}) - g \sin (3 0 ^ {\circ}) T = 0 \\ & \Rightarrow u \cos (\alpha - 3 0 ^ {\circ}) = g \sin (3 0 ^ {\circ}) \left(\frac {2 u \sin (\alpha - 3 0 ^ {\circ})}{g \cos (3 0 ^ {\circ})}\right) \\ & \Rightarrow \tan (\alpha - 3 0 ^ {\circ}) = \frac {1}{2} \cot (3 0 ^ {\circ}) \\ & \Rightarrow \tan (\alpha - 3 0 ^ {\circ}) = \frac {\sqrt {3}}{2} \\ & \Rightarrow (\alpha - 3 0 ^ {\circ}) = \tan^ {- 1} \left(\frac {\sqrt {3}}{2}\right) \\ & \Rightarrow \alpha = 3 0 ^ {\circ} + \tan^ {- 1} \left(\frac {\sqrt {3}}{2}\right) \end{array}$
In general, for the particle to hit the incline normally, we have
$\begin{array}{l} 2 \tan (\alpha - \beta) = \cot \beta \\ \text {OR} \\ \cot (\alpha - \beta) = 2 \tan \beta \end{array}$
Hence, the correct answer is (D).

$y = 9 x ^ {2}$
$\Rightarrow \frac {d y}{d t} = v _ {y} = 1 8 x \frac {d x}{d t} = 1 8 x \left(\frac {1}{3}\right)$
$\Rightarrow \frac {d y}{d t} = v _ {y} = 6 x$
$\Rightarrow a _ {y} = \frac {d ^ {2} y}{d t ^ {2}} = \frac {d v _ {y}}{d t} = 6 \left(\frac {d x}{d t}\right) = 6 \left(\frac {1}{3}\right)$
$\Rightarrow a _ {y} = 2 m s ^ {- 2} \mathrm{alongy-axis}$
Hence $\vec{a}_y = (2ms^{-2})\hat{j}$
Hence, the correct answer is (D).

$\omega_{centre}=\frac{d\theta}{dt}$
$\omega_ {\text { circumference }} = \frac {d}{d t} \left(\frac {\theta}{2}\right) = \frac {1}{2} \omega_ {\text { centre }}$
$\Rightarrow \omega_ {\mathrm{centre}} = 2 \omega_ {\mathrm{circumference}}$
$\Rightarrow \omega_ {\mathrm{centre}} = 4 \mathrm{rads} ^ {- 1}$
Now $v = R\omega_{\mathrm{centre}} = (2)(4) = 8\mathrm{ms}^{-1}$
Hence, the correct answer is (D).

Since we know that ranges are equal for complimentary angles i.e., $\phi$ and $90 - \phi$ . Also, we know that
$\begin{array}{r l} & R _ {\phi} = R _ {9 0 - \phi} = 4 \sqrt {H _ {\phi} H _ {9 0 - \phi}} \\ \Rightarrow & R _ {\phi} = R _ {9 0 - \phi} = 4 \sqrt {(4) (1 6)} \\ \Rightarrow & R _ {\phi} = R _ {9 0 - \phi} = 3 2 \mathrm{m} \end{array}$
Hence, the correct answer is (D).

When the particle is moving in a circular path with uniform speed, then the tangential acceleration is zero. However, radial acceleration is non-zero.
Hence, the correct answer is (C).

$y = \alpha x - \beta x^{2}$
Comparing with standard equation of projectile
$y = x \tan \theta - \frac {a x ^ {2}}{2 u ^ {2} \cos^ {2} \theta} \mathrm{weget}$
$\alpha = \tan \theta \mathrm{and} \beta = \frac {a}{2 u ^ {2}} \sec^ {2} \theta$
Since $\sec^2\theta = 1 + \tan^2\theta$
$\Rightarrow \beta = \frac {a}{2 u ^ {2}} (1 + \alpha^ {2})$
$\Rightarrow u = \sqrt {\frac {a}{2 \beta} (1 + \alpha^ {2})}$
Hence, the correct answer is (A).

For a horizontal projectile, we have
$y = \frac {g x ^ {2}}{2 u ^ {2}}. \mathrm{So},$
$h = \frac {g (2 5 0) ^ {2}}{2 v ^ {2}}\tag{... (1}$
Similarly,
$4 h = \frac {g x ^ {2}}{2 (v / 2) ^ {2}}\tag{... (2}$
Dividing (1) & (2), we get
x = 2 5 0 m
Hence, the correct answer is (A).

$\begin{array}{l} x ^ {2} + y ^ {2} = l ^ {2} \\ \Rightarrow 2 x \frac {d x}{d t} + 2 y \frac {d y}{d t} = 0 \\ \Rightarrow v _ {b} = \frac {x}{y} \frac {d x}{d t} = 1 0 \mathrm{ms} ^ {- 1} \end{array}$
$\left(x = y \text { at } 4 5 ^ {\circ}\right)$
Hence, the correct answer is (D).

$b = a\tan \alpha -\frac{ga^2}{2u^2\cos^2\alpha}$
$\mathrm{and} a = b \tan \alpha - \frac {g b ^ {2}}{2 u ^ {2} \cos^ {2} \alpha}$
$\Rightarrow \frac {a \tan \alpha - b}{b \tan \alpha - a} = \frac {a ^ {2}}{b ^ {2}}$
$\Rightarrow a b ^ {2} \tan \alpha - b ^ {3} = b a ^ {2} \tan \alpha - a ^ {3}$
$\Rightarrow a b ^ {2} \tan \alpha - b a ^ {2} \tan \alpha = b ^ {3} - a ^ {3}$
$\Rightarrow a b (b - a) \tan \alpha = (b - a) (b ^ {2} + a ^ {2} + a b)$
$\Rightarrow \tan \alpha = \frac {b ^ {2} + a ^ {2} + a b}{a b}$
$\Rightarrow \quad \tan \alpha = \frac {b ^ {2} + a ^ {2} + 2 a b + a b - 2 a b}{a b}$
$\Rightarrow \tan \alpha = \frac {(b - a) ^ {2} + 3 a b}{a b}$
$\Rightarrow \tan \alpha = 3 + \frac {(b - a) ^ {2}}{a b}$
$\mathrm{Since}, (b - a) ^ {2} > 0$
$\Rightarrow \tan \alpha > 3$
$\Rightarrow \quad (\tan \alpha) _ {\mathrm{min}} = 3$
Hence, the correct answer is (B).

$H_{1}=\frac{u^{2}\sin^{2}\alpha}{2g}$
$H _ {2} = \frac {u ^ {2} \sin^ {2} (9 0 - \alpha)}{2 g} = \frac {u ^ {2} \cos^ {2} \alpha}{2 g}$
{Because range is same for complimentary angles}
$\begin{array}{l l} \text {Further} & H _ {1} - H _ {2} = \frac {H _ {1} + H _ {2}}{2} \\ \Rightarrow & H _ {1} = 3 H _ {2} \\ \Rightarrow & \tan \alpha = \sqrt {3} \\ \Rightarrow & \alpha = 6 0 ^ {o} \end{array}$
Hence, the correct answer is (C).

Let the particles pass through the common point $P(x,y)$ at times $t_1$ and $t_2$.
$u_1 t_1 = u_2 t_2 \tag{1}$
$v_1 t_1 - \frac{1}{2} g t_1^2 = v_2 t_2 - \frac{1}{2} g t_2^2 \tag{2}$
Rearranging and solving equations (1) and (2) to get value of $(t_1 - t_2)$.
$\Rightarrow (t_1 - t_2) = \frac{2}{g} \left(\frac{v_1 u_2 - v_2 u_1}{u_1 + u_2}\right)$
Hence, the correct answer is (D).

Rate of change of speed is
$\frac {d v}{d t} = \mathrm{tangentialacceleration}$

$\begin{array}{l l} \Rightarrow & \frac {d v}{d t} = g \sin (3 0 ^ {\circ}) \\ \Rightarrow & \frac {d v}{d t} = 1 0 \left(\frac {1}{2}\right) \mathrm{ms} ^ {- 2} = 5 \mathrm{ms} ^ {- 2} \end{array}$
Hence, the correct answer is (B).

$T_{0} = \frac{2u\sin\theta}{g}$ (is time of flight when no drag exists)
$T = \frac {u \sin \theta}{g + \frac {g}{1 0}} + \frac {u \sin \theta}{g - \frac {g}{1 0}}$
$T = \frac {u \sin \theta}{g} \left(\frac {1 0}{1 1} + \frac {1 0}{9}\right)$
$T = \frac {2 u \sin \theta}{g} \left[ 5 \left(\frac {2 0}{9 9}\right) \right]$
$T = T _ {0} \frac {1 0 0}{9 9}$
$\Rightarrow$ Percentage increase $\simeq$ 1%
Hence, the correct answer is (A).

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).