Kinematics-2D

11. (7.50)

Range $R = \frac{2u_xu_y}{g} = \frac{2\times 5\times 5}{10} = 5m$
Time of flight $T = \frac{2u_y}{g} = \frac{2 \times 5}{10} = 1 \sec$


∵ Time of motion of one part falling vertically
downwards is = 0.5 sec = $\frac{T}{2}$
$\Rightarrow$ Time of motion of another part, $t = \frac{T}{2} = 0.5\mathrm{sec}$
From momentum conservation $\Rightarrow P_{i} = P_{f}$
$2m \times 5 = m \times v \Rightarrow v = 10m / s$
Displacement of other part in 0.5 sec in horizontal direction = v $\frac{T}{2}$ = 10 × 0.5 = 5 m = R
Total distance of second part from point 'O' is,
$x = \frac {3 R}{2} = 3 \times \frac {5}{2}$
x = 7. 5 m
$\Rightarrow t = 0. 5 \sec .$

Average velocity $\langle v\rangle$ is given by
$\langle v \rangle = \frac {\Sigma R}{\Sigma T}$
$\Sigma R = \left(\frac {2 u _ {0} ^ {2} \sin \theta \cos \theta}{g}\right) \left(1 + \frac {1}{\alpha^ {2}} + \frac {1}{\alpha^ {4}} + \dots .\right)$
$\Sigma R = \left(\frac {2 u _ {0} ^ {2} \sin \theta \cos \theta}{g}\right) \left(\frac {1}{1 - \frac {1}{\alpha^ {2}}}\right)$
$\Rightarrow \Sigma T = \left(\frac {2 u _ {0} \sin \theta}{g}\right) \left(\frac {1}{1 - \frac {1}{\alpha}}\right)$
Since $\langle v\rangle = 0.8V_{1}$ , where $V_{1} = u_{0}\cos \theta$
$\Rightarrow 0. 8 V _ {1} = \frac {u _ {0} \cos \theta \left(\frac {1}{1 - \frac {1}{\alpha^ {2}}}\right)}{\frac {1}{1 - \frac {1}{\alpha}}}$
$\begin{array}{r l} \Rightarrow & \left(\frac {\alpha^ {2}}{\alpha^ {2} - 1}\right) \left(\frac {\alpha - 1}{\alpha}\right) = 0. 8 \\ \Rightarrow & \frac {\alpha}{\alpha + 1} = 0. 8 \\ \Rightarrow & 5 \alpha = 4 \alpha + 4 \\ \Rightarrow & \alpha = 4 \end{array}$
Net acceleration,
$a = \sqrt {\left(a _ {c}\right) ^ {2} + \left(a _ {t}\right) ^ {2}} = \sqrt {(7 . 2) ^ {2} + (1 2) ^ {2}} = 1 4 \mathrm{ms} ^ {- 2}$
Hence, the correct answer is (A).

Average velocity $< v > = \frac{\text{Total displacement}}{\text{Total time}}$
Total time taken = $t_{1} + t_{2} + t_{3} + \ldots\ldots$
$= t _ {1} + \frac {t _ {1}}{\alpha} + \frac {t _ {1}}{\alpha^ {2}} +$
$\left[ T = \frac {2 u _ {y _ {1}}}{g} \right]$
Total displacement = $v_{1}t_{1} + v_{2}t_{2} + \dots\dots$ .
$= v _ {1} t _ {1} + \frac {v _ {1}}{\alpha}. \frac {t _ {1}}{\alpha} + \dots \dots = \frac {v _ {1} t _ {1}}{1 - \frac {1}{\alpha^ {2}}}$
On solving
$< v > = \frac {v _ {1} \alpha}{\alpha + 1} = 0. 8 v _ {1}$
$\Rightarrow \alpha = 4. 0 0.$

$H = \frac{u^2\sin^2(45^\circ)}{2g} = 120\mathrm{m}$
$\Rightarrow \frac {u ^ {2}}{4 g} = 1 2 0 \mathrm{m}$
If speed is v after the first collision, then speed should remain $\frac{1}{\sqrt{2}}$ times, because kinetic energy has reduced to half.
$\begin{array}{r l} \Rightarrow & v = \frac {u}{\sqrt {2}} \\ \Rightarrow & h _ {\max} = \frac {v ^ {2} \sin^ {2} (3 0 ^ {\circ})}{2 g} \\ \Rightarrow & h _ {\max} = \frac {\left(\frac {u}{\sqrt {2}}\right) ^ {2} \sin^ {2} 3 0 ^ {\circ}}{2 g} \\ \Rightarrow & h _ {\max} = \left(\frac {u ^ {2} / 4 g}{4}\right) = \frac {1 2 0}{4} \\ \Rightarrow & h _ {\max} = 3 0 \mathrm{m} \end{array}$

$H _ {1} = \frac {u ^ {2} \sin^ {2} 4 5}{2 g} = 1 2 0$
$\Rightarrow \frac {u ^ {2}}{4 g} = 1 2 0\tag{... (i}$
When half of kinetic energy is lost $v = \frac{u}{\sqrt{2}}$
$H _ {2} = \frac {\left(\frac {u}{\sqrt {2}}\right) ^ {2} \sin^ {2} 3 0}{2 g} = \frac {u ^ {2}}{1 6 g}\tag{... (ii}$
From (i) and (ii)
$H _ {2} = \frac {H _ {1}}{4} = 3 0 m \text { or } 3 0. 0 0.$

$T = \frac{2u_y}{g} = \frac{2\times 10\sin(60^\circ)}{g} = \sqrt{3}s$
Since $R = 1.15 \mathrm{~m}$ and
$\begin{array}{r l} & R = u _ {x} T - \frac {1}{2} a _ {x} T ^ {2} \\ \Rightarrow & 1. 1 5 = 1 0 \cos (6 0 ^ {\circ}) \sqrt {3} - \frac {1}{2} a (\sqrt {3}) ^ {2} \\ \Rightarrow & a = 5 \mathrm{ms} ^ {- 2} \end{array}$

$T = \frac{2u\sin\theta}{g} \Rightarrow T = \frac{2\times 10\times\sqrt{3}}{10\times 2} = \sqrt{3}\sec$
$R = u \cos \theta . T - \frac {1}{2} a T ^ {2}$

Since $R=\frac{2u^{2}\sin(\alpha-\beta)\cos\alpha}{g\cos^{2}\beta}$ $\Rightarrow R=\frac{2(21)^{2}\sin(60^{\circ}-30^{\circ})\cos(60^{\circ})}{9.8\cos^{2}(30^{\circ})}$ $\Rightarrow R=\frac{(2)(441)\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)}{(9.8)\left(\frac{3}{4}\right)}=30\ m$

At P, at height h,
$\begin{array}{r l} & v ^ {2} = u ^ {2} - 2 g h \\ \Rightarrow & v ^ {2} = 4 g h - 2 g h \\ \Rightarrow & v = \sqrt {2 g h} \end{array}$
$\left\{\because u = 2 \sqrt {g h} \right\}\tag{... (1}$

Assuming PQ to be horizontal, then Range = 2h = PQ
$\Rightarrow \frac {v ^ {2} \sin (2 \theta)}{g} = 2 h$
$\Rightarrow \frac {2 (g h) \sin (2 \theta)}{g} = 2 h$
$\Rightarrow \sin (2 \theta) = 1$
$\Rightarrow \quad \theta = 4 5 ^ {\circ}$
So, to fly from $P$ to $Q$ , the projectile will take a time $t$ given by
$t = \frac {2 v \sin (4 5 ^ {\circ})}{g} = \left(\frac {2 \sqrt {2 g h}}{g}\right) \left(\frac {1}{\sqrt {2}}\right) = 2 \sqrt {\frac {h}{g}}$
$\mathrm{So}, = 2$

From the diagram, we observe that
$\vec {r} _ {P} + \vec {r} _ {\mathrm{Fields}} = \vec {r} _ {\mathrm{Ball}}\tag{... (1}$
where $\vec{r}_P = (46\hat{i} + 28\hat{j})$ m
$\ldots(2)$
Let fielder run for the ball making an angle $\theta$ with the horizontal.
Then $\vec{r}_{\text{Fielder}} = 5t\left(\cos \theta \hat{i} + \sin \theta \hat{j}\right)$
... (3)

where $t$ is the time taken by the fielder to go from point $P$ to the interception point, which equals the time taken by the ball to go from the origin to the interception point with a velocity of $(7.5\hat{i} + 10\hat{j})\text{ ms}^{-1}$.
$\vec{r}_{\text{Ball}} = (7.5t\hat{i} + 10t\hat{j})$
Substituting (2), (3), and (4) in (1), we get:
$(46 + 5t\cos\theta)\hat{i} + (28 + 5t\sin\theta)\hat{j} = (7.5t)\hat{i} + (10t)\hat{j}$
$\Rightarrow 46 + 5t\cos\theta = 7.5t \tag{5}$
$\Rightarrow 28 + 5t\sin\theta = 10t \tag{6}$
From (5) and (6), we get:
$\cos\theta = \frac{7.5t - 46}{5t} \quad \text{and} \quad \sin\theta = \frac{10t - 28}{5t}$
Since $\cos^2\theta + \sin^2\theta = 1$:
$\left(\frac{7.5t - 46}{5t}\right)^2 + \left(\frac{10t - 28}{5t}\right)^2 = 1$
Solving to get $t = 4\text{ s}$ or $t = \frac{116}{21}\text{ s}$.
So, the shortest time for interception is $t = 4\text{ s}$

Let the particle be projected from O with velocity u and strike the plane at a point P horizontally. Then
P Q = O Q.

$\Rightarrow \quad \text { Maximum height } = \frac {\text { Horizontal range }}{2}$
$\Rightarrow \frac {u ^ {2} \sin^ {2} \phi}{2 g} = \frac {u ^ {2} \sin 2 \phi}{2 g} = \frac {u ^ {2} \sin \phi \cos \phi}{g}$
$\Rightarrow \tan \phi = 2$

The angular speed is
$\omega = \frac {v}{r} = \frac {2}{0 . 2 5} = 8 \mathrm{rads} ^ {- 1}$

Since $y = u_{y}t + \frac{1}{2}a_{y}t^{2}$ , where $u_{y} = 0$ and $a_{y} = g$ , y = h.
So $h = \frac{1}{2}gt^{2}$ $\Rightarrow t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2(20)}{10}} = 2s$ Further since $x = u_{x}t + \frac{1}{2}a_{x}t^{2}$ where $u_{x} = v$ (say), $a_{x} = 0$ , $x = 10 \, m$ , so $10 = vt = v(2)$ $\Rightarrow v = 5 \, ms^{-1}$ Since $a = \frac{v^{2}}{R}$ $\Rightarrow a = \frac{(5)^{2}}{0.5} = \frac{25}{0.5} = 50 \, ms^{-2}$ $\Rightarrow a = 50 \, ms^{-2}$

Since the car is travelling with a constant speed, its tangential component of acceleration is zero, i.e., $a_{T}=0$ . Thus,
$a = a _ {N} = \frac {v ^ {2}}{r}$
$\Rightarrow 3 = \frac {2 5 ^ {2}}{r}$
$\Rightarrow r = 2 0 8 \mathrm{m}$

Here, the car's tangential component of acceleration of $a_{T} = -3 \, \mathrm{ms}^{-2}$ . Thus,
$a = \sqrt {a _ {T} ^ {2} + a _ {N} ^ {2}}$
$\Rightarrow 5 = \sqrt {(- 3) ^ {2} + a _ {N} ^ {2}}$
$\Rightarrow a _ {N} = 4 \mathrm{ms} ^ {- 2}$
Since $a_{N} = \frac{v^{2}}{r}$
$\Rightarrow 4 = \frac {2 0 ^ {2}}{r}$
$\Rightarrow r = 1 0 0 \mathrm{m}$

$t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2\times 0.8}{10}} = 0.4\mathrm{s}$
$u _ {\mathrm{MIN}} = \frac {(1 2 0 - 3 0 + 1 0) \times 1 0 ^ {- 3}}{0 . 4} = 0. 2 5 \mathrm{ms} ^ {- 1}$
and $u_{\mathrm{MAX}}=\frac{(120+30-10)\times10^{-3}}{0.4}=0.35\ \mathrm{ms}^{-1}$
$\Rightarrow u _ {M I N} = 2 5 \mathrm{cms} ^ {- 1} \mathrm{and} u _ {M A X} = 3 5 \mathrm{cms} ^ {- 1}$

Let the ball clear the point C at time $t_{1}$ . Then
$\begin{array}{r l r} & y = u _ {y} t + \frac {1}{2} a _ {y} t ^ {2} \\ \Rightarrow & (1 + 3. 9) = 0 + \frac {1}{2} (9. 8) t _ {1} ^ {2} & \left\{\because u _ {y} = 0 \right\} \\ \Rightarrow & 4. 9 = \frac {1}{2} (9. 8) t _ {1} ^ {2} \\ \Rightarrow & t _ {1} = 1 \mathrm{s} \\ \text {Since} & B C = u _ {x} t \\ \Rightarrow & 6 = u (1) \\ \Rightarrow & u = 6 \mathrm{ms} ^ {- 1} \end{array}$
Let the ball hit the ground at D in time t. Then
$6 + x = 6 t\tag{... (1}$
$\text { Further } (1 + 3. 9 + 1 4. 7) = \frac {1}{2} (9. 8) t ^ {2}$
$\Rightarrow t = 2 \mathrm{s}$
$\Rightarrow 6 + x = 1 2$
$\Rightarrow x = 6 \mathrm{m}$

At time $t = T = \frac{2u \sin(\phi - 45^{\circ})}{g \cos 45^{\circ}}$
component of velocity along the plane is zero.
$\Rightarrow 0 = u \cos (\phi - 4 5 ^ {\circ}) - (g \sin 4 5 ^ {\circ}) t$

$\Rightarrow u \cos (\phi - 4 5 ^ {\circ}) = (g \sin 4 5 ^ {\circ}) \left(\frac {2 u \sin (\phi - 4 5 ^ {\circ})}{g \cos 4 5 ^ {\circ}}\right)$
$\Rightarrow 2\tan (\phi -45^{\circ}) = \cot 45^{\circ} = 1$
$\Rightarrow 2 \left(\frac {\tan \phi - \tan 4 5 ^ {\circ}}{1 + \tan \phi \tan 4 5 ^ {\circ}}\right) = 1$
$\Rightarrow \tan \phi = 3$

As $y = x \tan \theta - \frac{gx^{2}}{2u^{2}}(1 + \tan^{2} \theta)$
for $(a, b)$ , i.e. when $x = a$ and $y = b$ , we have
$g a ^ {2} \tan^ {2} \theta - 2 a u ^ {2} \tan \theta + (g a ^ {2} + 2 b u ^ {2}) = 0$
This is a quadratic in $\tan\theta$ and since discriminant of a quadratic must be positive, so we have
$4 a ^ {2} u ^ {2} - 4 g a ^ {2} \left(g a ^ {2} + 2 b u ^ {2}\right) \geq 0$
Solving, we get
$u \geq \sqrt {b g + g \sqrt {a ^ {2} + b ^ {2}}}$
On substituting the values, we get
$u _ {\mathrm{min}} = 3 0 \mathrm{ms} ^ {- 1}$

$v_{x}=8\sqrt{2}$ ms $^{-1}$ is the relative velocity along x-axis
$\Rightarrow x = 2 2 - (8 \sqrt {2}) t$
$v_{y}=6\sqrt{2}$ ms $^{-1}$ is the relative velocity along y-axis
$\Rightarrow y = 9 - (6 \sqrt {2}) t$
$\Rightarrow \quad r ^ {2} = x ^ {2} + y ^ {2}$
For minimum r, we have $r^{2}$ to be minimum, so
$\frac {d}{d t} \big (r ^ {2} \big) = 0$
$\Rightarrow t = \frac {2 3}{1 0 \sqrt {2}} \mathrm{s}$
Substituting the value of t in equation (1), we get
$r _ {\mathrm{min}} = 6 \mathrm{m}$

$v=\frac{u\cos(60^{\circ})}{\cos(45^{\circ})}$

$\Rightarrow K _ {f} = K _ {i} \left(\frac {\cos^ {2} (6 0 ^ {\circ})}{\cos^ {2} (4 5 ^ {\circ})}\right)$
$\Rightarrow \frac {K _ {i}}{K _ {f}} = 2$

For Range to be maximum, the projectile must be launched at an angle
$\alpha = \frac {\pi}{4} + \frac {\beta}{2}$
$\Rightarrow \alpha = 4 5 ^ {\circ} + 1 5 ^ {\circ} = 6 0 ^ {\circ}$

Since, $H_{1}=(2)^{2}=4\ m$ and
$H _ {2} = (1) ^ {2} = 1 \mathrm{m}$
By Conservation of Energy, we have
$\binom{\text {Loss in gravitational}}{\text {potential energy}} = \binom{\text {Gain in}}{\text {kinetic energy}}$

$\Rightarrow m g \left(H _ {1} - H _ {2}\right) = \frac {1}{2} m v ^ {2}$
$\begin{array}{r l} \Rightarrow & v = \sqrt {2 g (H _ {1} - H _ {2})} \\ \Rightarrow & v = \sqrt {2 \times g \times (H _ {2} - H _ {1})} \\ \Rightarrow & v = \sqrt {2 0 \times 3} \\ \Rightarrow & v = \sqrt {6 0} \end{array}$
Now $\tan \theta = \frac{dy}{dx}\bigg|_{x = 1} = 2x\big|_{x = 1} = 2$
$\Rightarrow \tan \theta = 2$
$\begin{array}{r l} & {\mathrm{Since,} y = x \tan \theta - \frac {g x ^ {2}}{2 v ^ {2} \cos^ {2} \theta}} \\ & {\Rightarrow - 1 = x (2) - \frac {1 0 x ^ {2}}{2 v ^ {2}} (1 + 4)} \\ & {\Rightarrow - 1 = 2 x - \frac {1 0 x ^ {2}}{2 (6 0)} (5)} \\ & {\Rightarrow - 1 = 2 x - \frac {5}{1 2} x ^ {2}} \\ & {\Rightarrow - 1 2 = 2 4 x - 5 x ^ {2}} \\ & {\Rightarrow 5 x ^ {2} - 2 4 x - 1 2 = 0} \end{array}$

Time of flight is
$T = \frac {2 u \sin \theta}{g} = 2 \sqrt {2} \mathrm{s}$
Range (along north) is
$R _ {1} = \frac {u ^ {2} \sin 2 \theta}{g} = 4 0 \mathrm{m}$
Range (along east) is
$R _ {2} = \frac {1}{2} a T ^ {2} = 3 0 \mathrm{m}$
So, range $R = \sqrt{R_{1}^{2} + R_{2}^{2}} = \sqrt{30^{2} + 40^{2}} = 50 \, m$