Kinematics-2D

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).

For the observer sitting in train, the initial horizontal velocity of coin and train are same, thus observer will find its path as a straight line. Whereas the observer on the ground will see the coin to follow a parabolic path (as of a horizontal projectile).
Hence, the correct answer is (D).

$u _ {x} = 0, a _ {x} = 0, u _ {y} = u \sin \theta \text {and} a _ {y} = - g$
$\Rightarrow v _ {x} = u _ {x} + a _ {x} T = u _ {x} \quad \left\{\text {where} T = \frac {2 u \sin \theta}{g} \right\}$
$\text {Also} v _ {y} = u \sin \theta - g t$
$\Rightarrow v _ {y} = u \sin \theta - g \left(\frac {2 u \sin \theta}{g}\right)$
$\Rightarrow v _ {y} = - u \sin \theta$
Hence, the correct answer is (A).

$H = \frac{u^{2} \sin^{2} \theta}{2g}$ i.e., it is independent of mass of projectile.
Hence, the correct answer is (A).

As $T=\frac{2u\sin\theta}{g}$
Hence, the correct answer is (C).

Since acceleration due to gravity acts vertically downwards, so it has no component along the horizontal and hence does not affect the horizontal velocity. Hence, the correct answer is (A).
$R _ {\mathrm{max}} = \frac {u ^ {2}}{g}$

Since,
$\{\mathrm{For} \theta = 4 5 ^ {\circ} \}$
$\Rightarrow H = \frac {u ^ {2} \sin^ {2} (4 5 ^ {\circ})}{2 g} = \frac {u ^ {2}}{4 g}$
So, H is 25% of range.
Hence, the correct answer is (A).

$x = kt, y = kt(1 - \alpha t)$
Equation of trajectory,
$\begin{array}{r l}&y = k t - k \alpha t ^ {2} = k \left(\frac {x}{k}\right) - k \alpha \left(\frac {x ^ {2}}{k ^ {2}}\right)\\\Rightarrow&y = x - \alpha \frac {x ^ {2}}{k} \rightarrow \mathrm{parabola}\end{array}$
Hence, the correct answer is (C).