Kinematics-2D

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

For shell to hit the boat the time lapsed equals the time of flight of the shell = $\frac{2v_{2}\sin\alpha}{g}$ $\Rightarrow$ Distance travelled by boat = $v_{1}\left(\frac{2v_{2}\sin\alpha}{g}\right)$ Distance of boat from the cannon at the instant the shell is fired is $x = R - \left(\frac{2v_{1}v_{2}\sin\alpha}{g}\right)$
$\begin{array}{l l} \Rightarrow & x = \frac {2}{g} (v _ {2} \sin \alpha) (v _ {2} \cos \alpha) - \frac {2 v _ {1} v _ {2} \sin \alpha}{g} \\ \Rightarrow & x = \frac {2}{g} (v _ {2} \sin \alpha) (v _ {2} \cos \alpha - v _ {1}) \end{array}$
and the distance of boat from the cannon when the shell hits the boat is equal to the range of the shell
$R = \frac {2}{g} (v _ {2} \sin \alpha) (v _ {2} \cos \alpha)$
Hence, (A), (B), (C) and (D) are correct.

$\begin{array}{r l} & u _ {x} = u \cos (4 5 ^ {\circ}) = 4 0 \mathrm{ms} ^ {- 1} \text {and} \\ & u _ {y} = u \sin (4 5 ^ {\circ}) = 4 0 \mathrm{ms} ^ {- 1} \\ & \text {At} t = 2 \mathrm{s}, \\ & v _ {x} = u _ {x} = 4 0 \mathrm{ms} ^ {- 1} \\ & v _ {y} = u _ {y} - g t = 4 0 - (1 0) (2) = 2 0 \mathrm{ms} ^ {- 1} \end{array}\tag{... (1}$
So, (B) is correct
$\mathrm{Further} x = u _ {x} t = 8 0 \mathrm{m}$
$\mathrm{and} y = u _ {y} t + \frac {1}{2} a _ {y} t ^ {2}$
$\Rightarrow y = (4 0) (2) + \frac {1}{2} (- 1 0) (4)$
$\Rightarrow y = 6 0 \mathrm{m}$
$\mathrm{Hence(D)iscorrect}$
So, displacement of the particle is $r = \sqrt{x^{2} + y^{2}}$
$\Rightarrow r = \sqrt {(8 0) ^ {2} + (6 0) ^ {2}} = 1 0 0 \mathrm{m}$
Hence (A) is also correct.
Also, we observe that if $\alpha$ is the angle made by v with the horizontal, then
$\tan \alpha = \frac {v _ {y}}{v _ {x}} = \frac {2 0}{4 0} = \frac {1}{2}$
Hence (C) is not correct.
Hence, (A), (B) and (D) are correct.

$x_0 + ut = vt$
$\begin{array}{l} x _ {0} + u t = v t \\ \Rightarrow \quad \left(6 \hat {i} + 8 \hat {j}\right) + (2 t) \hat {i} = \left(a \hat {i} + b \hat {j}\right) t \\ \text { At } t = 2 \text { s, we get } \\ \Rightarrow \quad 1 0 \hat {i} + 8 \hat {j} = 2 a \hat {i} + 2 b \hat {j} \\ \Rightarrow \quad 2 a = 1 0 \text { and } 2 b = 8 \\ \Rightarrow \quad a = 5 \text { and } b = 4 \\ \text { Hence, (B) and (C) are correct. } \end{array}$

$y = h - x\tan \alpha$
$\mathrm{Further} y = \big (v _ {0} \sin \theta \big) t - \frac {1}{2} g t ^ {2}\tag{... (2}$
and $x = (v_{0} \cos \theta)t$
Equating (1) and (2) and put value of x from (3) to get
$\Rightarrow t ^ {2} - \frac {2 v _ {0}}{g} (\cos \theta \tan \alpha + \sin \theta) t + \frac {2 h}{g} = 0$
$\Rightarrow t ^ {2} - \left[ \frac {2 v _ {0} \sin (\theta + \alpha)}{g \cos \alpha} \right] t + \frac {2 h}{g} = 0\tag{... (4}$
For $t$ to be MINIMUM
$\begin{array}{l l} & \frac {d t}{d \theta} = 0 \\ \Rightarrow & 2 t \frac {d t}{d \theta} - \frac {2 v _ {0}}{g \cos \alpha} \frac {d}{d \theta} [ t \sin (\alpha + \theta) ] + 0 = 0 \\ \Rightarrow & 0 - \frac {2 v _ {0}}{g \cos \alpha} \left[ \sin (\alpha + \theta) \frac {d t}{d \theta} + t \cos (\alpha + \theta) \right] = 0 \\ \Rightarrow & 0 - \frac {2 v _ {0} t \cos (\alpha + \theta)}{g \cos \alpha} = 0 \quad \left\{\because \frac {d t}{d \theta} = 0 \right\} \\ \Rightarrow & \cos (\theta + \alpha) = 0 \quad \{\because t \neq 0 \} \\ \Rightarrow & \theta + \alpha = \frac {\pi}{2} \quad \{\text {OPTION (A)} \} \end{array}$
Substituting $\theta +\alpha = \frac{\pi}{2}$ in (4), we get
$t _ {\mathrm{min}} ^ {2} - \frac {2 v _ {0}}{g \cos \alpha} t _ {\mathrm{min}} + \frac {2 h}{g} = 0$
$\Rightarrow t _ {\mathrm{min}} = \frac {v _ {0} - \sqrt {v _ {0} ^ {2} - 2 g h \cos^ {2} \alpha}}{g \cos \alpha}$
{OPTION (C)}
For $t_{\mathrm{min}}$ to be real $v_0^2 > 2gh\cos^2\alpha$
$\Rightarrow v _ {0} > \sqrt {2 g h} \cos \alpha$
So if $v_{0} < \sqrt{2gh}\cos\alpha$ , then projectile never reaches the roof.
{OPTION (D)}
Hence, (A), (C) and (D) are correct.

As studied and discussed in theory, we observe that all (A), (B), (C) and (D) are correct.
Hence, (A), (B), (C) and (D) are correct.

The vertical velocity at P is zero.
Hence $v \sin (\theta + \alpha) - gt = 0$
$\Rightarrow t = \frac {v \sin (\theta + \alpha)}{g}\tag{... (1}$

Since displacement perpendicular to the plane is zero hence
$0 = (v \sin \theta) t + \frac {1}{2} (- g \cos \alpha) t ^ {2}$
$\Rightarrow t = \frac {2 v \sin \theta}{g \cos \alpha}\tag{... (2}$
Equating (1) & (2) and rearranging, we get
$\frac {v \sin (\theta + \alpha)}{g} = \frac {2 v \sin \theta}{g \cos \alpha}$
$\Rightarrow 2 \sin \theta = \cos \alpha (\sin \theta \cos \alpha + \cos \theta \sin \alpha)$
$\Rightarrow 2\sin \theta = \cos^2\alpha \sin \theta +\cos \theta \sin \alpha \cos \alpha$
$\Rightarrow \quad (2 - \cos^ {2} \alpha) \sin \theta = \cos \theta \sin \alpha \cos \alpha$
$\Rightarrow \tan \theta = \frac {\sin \alpha \cos \alpha}{2 - \cos^ {2} \alpha}$
$\Rightarrow \tan \theta = \frac {\sin \alpha \cos \alpha}{(1 - \cos^ {2} \alpha) + 1} = \frac {\sin \alpha \cos \alpha}{1 + \sin^ {2} \alpha}$
$\Rightarrow \quad \tan \theta = \frac {\tan \alpha}{1 + 2 \tan^ {2} \alpha}$
Hence, (A) and (D) are correct.

Time of ascent $= 2+1=3,s$
If $\alpha$ is the angle of projection, then
$\frac{u\sin\alpha}{g}=3$
$\Rightarrow u\sin\alpha=30 \tag{1}$
If $\beta=30^\circ$ is the angle which the projectile makes with the horizontal at $t=2,s$ and $v$ is the velocity at this instant of time, then
$v\cos\beta=u\cos\alpha \tag{2}$
and
$v\sin\beta=u\sin\alpha-g(2)$
$\Rightarrow v\sin\beta=30-20=10$
$\Rightarrow v\sin30^\circ=10$
$\Rightarrow v=20,ms^{-1}$
Substituting $\beta=30^\circ$ and $v=20,ms^{-1}$ in (2), we get
$v\frac{\sqrt{3}}{2}=u\cos\alpha$
$\Rightarrow u\cos\alpha=10\sqrt{3} \tag{3}$
Squaring (1) and (3) and adding,
$u^2=30^2+(10\sqrt{3})^2$
$\Rightarrow u^2=1200$
$\Rightarrow u=20\sqrt{3},ms^{-1}$
and
$\tan\alpha=\frac{u\sin\alpha}{u\cos\alpha}$
$=\frac{30}{10\sqrt{3}}$
$=\sqrt{3}$
$\Rightarrow \alpha=60^\circ$ with horizontal or $30^\circ$ with vertical.
Hence, (B) and (C) are correct.

At the instant shown, we have

$v \cos \theta = u \cos (6 0 ^ {\circ}) = \frac {u}{2}$
$\Rightarrow v = \frac {u}{2 \cos \theta}\tag{... (1}$
The radius of curvature is given by
$R = \frac {v ^ {2}}{g \cos \theta} = \frac {u ^ {2}}{4 g \cos^ {3} \theta}$
$\Rightarrow R _ {\min} = \frac {u ^ {2}}{4 g}, \text { when } \theta = 0 ^ {\circ}$
$\mathrm{Since}, R = \frac {8}{3 \sqrt {3}} R _ {\mathrm{min}}$
$\Rightarrow \frac {u ^ {2}}{4 g \cos^ {3} \theta} = \frac {8}{3 \sqrt {3}} \left(\frac {u ^ {2}}{4 g}\right)$
$\Rightarrow \cos^ {3} \theta = \frac {3 \sqrt {3}}{8} = \left(\frac {\sqrt {3}}{2}\right) ^ {3}$
$\Rightarrow \cos \theta = \frac {\sqrt {3}}{2}$
$\Rightarrow \theta = 3 0 ^ {\circ}$
So, from equation (1), we get
$v = \frac {u}{\sqrt {3}}$
Hence, (B) and (C) are correct.

$\Rightarrow \frac {d y}{d t} = \frac {3 d x}{d t}$
$\Rightarrow v _ {y} = 3 v _ {x} = 3$
$\left\{\because v _ {2} = 1 \mathrm{ms} ^ {- 1} \right\}$
Accelerated curvilinear
$\begin{array}{r l} \Rightarrow & \frac {d y _ {y}}{a t} = \frac {3 d v _ {x}}{d t} \\ \Rightarrow & a _ {y} = 3 a _ {x} = 3 \end{array}$
$\left\{\because a _ {x} = 1 \mathrm{ms} ^ {- 1} \right\}$
nothing can be said about velocity and acceleration as the given values are at particular instant.

$x = 2 + 2t + 4t^{2}$ and $y = 4t + 8t^{2}$
$\Rightarrow \quad y = 2 + 2 x$
So, the particle moves along the straight line.
$\begin{array}{r l} & {\mathrm{Since,} \frac {d x}{d t} = 2 + 8 t \mathrm{and} \frac {d y}{d t} = 4 + 1 6 t} \\ & {\Rightarrow \frac {d ^ {2} x}{d t ^ {2}} = 8 \mathrm{and} \frac {d ^ {2} y}{d t ^ {2}} = 1 6} \\ & {\Rightarrow \vec {v} = \frac {d x}{d t} \hat {i} + \frac {d y}{d t} \hat {j} = (2 + 8) \hat {i} + (4 + 1 6 t) \hat {j}} \\ & {\Rightarrow \vec {a} = 8 \hat {i} + 1 6 \hat {j} = (\mathrm{constant})} \end{array}$
Hence, (A) and (B) are correct.

$\vec{v}=2t\hat{i}+t^{2}\hat{j}$
$\begin{array}{r l} \Rightarrow & \frac {d \vec {v}}{d t} = 2 \hat {i} + 2 t \hat {j} \\ \Rightarrow & | \vec {a} _ {n e t} | = \sqrt {2 ^ {2} + 2 ^ {2}} = 2 \sqrt {2} \end{array}$
Since, tangential acceleration is defined as the rate of change of speed, which is
$\begin{array}{r l r} & & {| \vec {v} | = \sqrt {(2 t) ^ {2} + (t ^ {2}) ^ {2}} = \sqrt {4 t ^ {2} + t ^ {4}}} \\ & {\mathrm{So},} & {a _ {T} = \frac {d | \vec {v} |}{d t} = \frac {1}{2 \sqrt {4 t ^ {2} + t ^ {4}}} (8 t + 4 t ^ {3})} \\ & {\Rightarrow} & {a _ {T} = \frac {1}{2 \times \sqrt {5}} \times 1 2 = \frac {6}{\sqrt {5}} \mathrm{ms} ^ {- 2}} \end{array}$
Since, we know that, for a particle moving in a curvilinear path with variable speed, the particle possesses both tangential acceleration $(a_{T})$ and centripetal acceleration $(a_{C})$ , so we have
$\begin{array}{r l} & a _ {\mathrm{net}} ^ {2} = a _ {C} ^ {2} + a _ {T} ^ {2} \\ \Rightarrow & a _ {C} ^ {2} = a _ {\mathrm{net}} ^ {2} - a _ {T} ^ {2} \\ \Rightarrow & a _ {C} ^ {2} = (2 \sqrt {2}) ^ {2} - (\frac {6}{\sqrt {5}}) ^ {2} = 8 - \frac {3 6}{5} = \frac {4}{5} \\ \Rightarrow & a _ {C} = \frac {2}{\sqrt {5}} \mathrm{ms} ^ {- 2} \end{array}$
Hence, (B) and (C) are correct.

$R = H$
$\begin{array}{l l} \Rightarrow & \frac {2 u ^ {2} \sin \theta \cos \theta}{g} = \frac {u ^ {2} \sin^ {2} \theta}{2 g} \\ \Rightarrow & \tan \theta = 4 \end{array}$
$R _ {\mathrm{max}} \big | _ {\theta = 4 5 ^ {\circ}} = \frac {u ^ {2}}{g}$
Hence, the correct answer is (B).

In case of non-uniform circular motion, net acceleration will not be directed towards centre.
Hence, the correct answer is (D).

Direction of acceleration changes. Hence, the correct answer is (D).

The particle will move in a parabolic path till acceleration (due to gravity) is constant. For this the particle should be near the surface of the Earth and air resistance should be negligible.
Hence, the correct answer is (D).

In uniform circular motion direction of acceleration is always along the radial direction. As particle is rotating so radial vector keeps on changing.
Hence, the correct answer is (D).

Hence, the correct answer is (A).

In uniform circular motion, the magnitude of velocity and acceleration remains same, but due to change in direction of motion, the direction of velocity and acceleration changes. Furthermore, the centripetal acceleration is given by $a = \omega^{2} r$ .
Hence, the correct answer is (B).