Kinematics-2D

$\mathrm{A} \rightarrow (\mathrm{r})$
$\mathrm{B} \rightarrow (\mathrm{t})$
$\mathrm{C} \rightarrow (\mathrm{q})$
$\mathrm{D} \rightarrow (\mathrm{p})$
$\mathrm{E} \rightarrow (\mathrm{s})$
$\vec {u} = u \left(\cos (4 5 ^ {\circ}) \hat {i} + \sin (4 5 ^ {\circ}) \hat {j}\right)$
$\Rightarrow \vec {u} = 2 0 \hat {i} + 2 0 \hat {j} \mathrm{ms} ^ {- 1} \text {and} \vec {a} = - 1 0 \hat {j}$
Since
$\vec {r} = \vec {u} t + \frac {1}{2} \vec {a} t ^ {2}$
$\Rightarrow \vec {r} = (2 0 \hat {i} + 2 0 \hat {j}) t + \frac {1}{2} (- 1 0 \hat {j}) t ^ {2}$
$\Rightarrow \vec {r} = (2 0 t) \hat {i} + (2 0 - 5 t) \hat {j}$
So, at $t = 1$ s, we get
$\vec {r} = 2 0 \hat {i} + 1 5 \hat {j}$
$\Rightarrow$ Average Velocity at $t = 1$ s is
$\vec {v} _ {a v} = \frac {\vec {r}}{t} = 2 0 \hat {i} + 1 5 \hat {j}$
$\Rightarrow \quad | \vec {v} _ {a v} | = \sqrt {4 0 0 + 2 2 5} = 2 5 \mathrm{ms} ^ {- 1}$
Further $\vec{v} = \vec{u} +\vec{a} t$
$\Rightarrow \vec {v} = 2 0 \hat {i} + 2 0 \hat {j} - (1 0 \hat {j}) t$
$\Rightarrow \vec {v} = 2 0 \hat {i} + (2 0 - 1 0 t) \hat {j}$
$\Rightarrow \Delta \vec {v} = \vec {v} - \vec {u} = - (1 0 t) \hat {j}$
$\Rightarrow \Delta \vec {v} = - 2 0 \hat {j}$
$\left\{\mathrm{at} t = 2 \mathrm{s} \right\}$
So, average acceleration at t = 1 s, 2 s, 3 s and 4 s
$\vec {a} _ {a v} = \frac {\Delta \vec {v}}{\Delta t} = - 1 0 \hat {j}$
{as expected}
Radius of curvature at t = 0 is
$\begin{array}{l l} & r = \frac {u ^ {2}}{(g \cos \theta)} \\ \Rightarrow & r = \frac {(2 0 \sqrt {2}) ^ {2}}{\left(\frac {1 0}{\sqrt {2}}\right)} = 8 0 \sqrt {2} \mathrm{m} \end{array}$
Radius of curvature at t = 1 s is
$r = \frac {v ^ {2}}{g \cos \beta}$
where v is the velocity at t=1 s and $\beta$ is the angle which $v(\text{at } t=1\text{ s})$ makes with the x-axis
Since $v = 20\hat{i} + 10\hat{j}$ , so $\tan \beta = \frac{10}{20}$

$\Rightarrow \cos \beta = \frac {2 0}{\sqrt {2 0 ^ {2} + 1 0 ^ {2}}} = \frac {2 0}{1 0 \sqrt {5}} = \frac {2}{\sqrt {5}}$
$\mathrm{So}, r = \frac {\left(\sqrt {2 0 ^ {2} + 1 0 ^ {2}}\right) ^ {2}}{(1 0) \left(\frac {2}{\sqrt {5}}\right)}$
$\Rightarrow r = \frac {5 0 0 \sqrt {5}}{2 0}$
$\Rightarrow r = 2 5 \sqrt {5} \mathrm{m}$
Radius of curvature at t = 2 s is
$\begin{array}{r l} & r = \frac {(u \cos \theta) ^ {2}}{g} \\ \Rightarrow & r = \frac {\left[ (2 0 \sqrt {2}) \left(\frac {1}{\sqrt {2}}\right) \right] ^ {2}}{1 0} = \frac {(2 0) ^ {2}}{1 0} = 4 0 \mathrm{m} \end{array}$

$A\rightarrow r$
$\mathrm{B} \rightarrow (\mathrm{s})$
$\mathrm{C} \rightarrow (\mathrm{p})$
$\mathrm{D} \rightarrow (\mathrm{q})$
Since $v = 2t$
So, distance travelled is
$s = \int_ {0} ^ {2} v d t = \int_ {0} ^ {2} 2 t d t = 2 \left(\frac {t ^ {2}}{2} \Bigg | _ {0} ^ {2}\right)$
$\Rightarrow \quad s = 4 \mathrm{m}$
Hence average speed is
$v _ {a v} = \frac {s}{t} = \frac {4}{2} = 2 \mathrm{ms} ^ {- 1}$
$\mathrm{So}, (\mathrm{A}) \rightarrow (\mathrm{r}), (\mathrm{C}) \rightarrow (\mathrm{p})$
Further since the particle is moving in a circular path, so we have
$\Rightarrow \quad \theta = 2 \int_ {0} ^ {2} t d t$
$\Rightarrow \theta = 4 \mathrm{rad}$
$\Delta \vec {R} = \mathrm{Displacement} = \vec {R} _ {f} - \vec {R} _ {i}$
$\Rightarrow \quad | \text {Displacement} | = \sqrt {R _ {f} ^ {2} + R _ {i} ^ {2} + 2 R _ {f} R _ {i} \cos (1 8 0 - \theta)}$
Since $\left|\vec{R}_f\right| = \left|\vec{R}_i\right| = R$
$\Rightarrow \quad | \mathrm{Displacement} | = \sqrt {R ^ {2} + R ^ {2} - 2 R ^ {2} \cos \theta}$
$R = 1\mathrm{m}$
$\Rightarrow \quad | \text { Displacement } | = 2 R \sin \left(\frac {\theta}{2}\right) = 2 \sin (2)$
$\omega = \frac {v}{r} = 2 t \mathrm{rads} ^ {- 1}$

$A \rightarrow r$
$B \rightarrow s$
$C\rightarrow p$
$D \rightarrow q$
Since $\vec{a} = \frac{d\vec{v}}{dt}$
$\Rightarrow \vec{a} = 2\hat{i} + (2t)\hat{j}$
So, at $t=1,s$, we get
$\vec{v} = 2\hat{i} + \hat{j}$ and $\vec{a} = 2\hat{i} + 2\hat{j}$
$\Rightarrow |\vec{a}| = \sqrt{2^2 + 2^2} = 2\sqrt{2},ms^{-2}$
Hence C $\rightarrow$ p
Now, since the tangential acceleration is
$a_T = \frac{\vec{a}\cdot\vec{v}}{|\vec{v}|}$
$\Rightarrow a_T = \frac{(2\hat{i}+2\hat{j})\cdot(2\hat{i}+\hat{j})}{\sqrt{2^2+1^2}}$
$\Rightarrow a_T = \frac{4+2}{\sqrt{5}} = \frac{6}{\sqrt{5}},ms^{-2}$
So, A $\rightarrow$ r
Now since we know that
$a^2 = a_C^2 + a_T^2$, where $a_C$ is the radial acceleration
$\Rightarrow a_C = \sqrt{a^2-a_T^2}$
$\Rightarrow a_C = \sqrt{8-\frac{36}{5}}$
$\Rightarrow a_C = \frac{2}{\sqrt{5}},ms^{-2}$
So, B $\rightarrow$ s
Finally, the radius of curvature is given by
$r = \frac{v^2}{a_C} = \frac{5}{\frac{2}{\sqrt{5}}} = \frac{5\sqrt{5}}{2},m$
So, D $\rightarrow$ q

$A\rightarrow q$
$B\rightarrow s$
$C\rightarrow p$
$D\rightarrow r$
Let us choose the $x$ and $y$ directions along $OB$ and $OA$ respectively. Then
$u_x = u = 10\sqrt{3},ms^{-1}, \quad u_y = 0$
$a_x = -g\sin(60^\circ) = -5\sqrt{3},ms^{-2}$
and $a_y = -g\cos(60^\circ) = -5,ms^{-2}$
At point $Q$, $x$-component of velocity is zero. Hence, substituting in
$v_x = u_x + a_xt$
$\Rightarrow 0 = 10\sqrt{3} - 5\sqrt{3}t$
$\Rightarrow t = \frac{10\sqrt{3}}{5\sqrt{3}} = 2s$
So, A $\rightarrow$ q
At point $Q$, $\vec{v} = v_y = u_y + a_yt$
$\Rightarrow v = 0 - (5)(2) = -10,ms^{-1}$
Here, negative sign implies that velocity of particle at $Q$ is along negative $y$ direction. So, $v = 10,ms^{-1}$, along $-y$ direction.
Hence B $\rightarrow$ s
Distance $PO = \left|\text{Displacement of particle along }y\text{-direction}\right| = |s_y|$
Since, $s_y = u_yt + \frac{1}{2}a_yt^2 = 0 - \frac{1}{2}(5)(2)^2 = -10,m$
$\Rightarrow PO = 10,m$
Since, $h = PO\sin(30^\circ) = (10)\left(\frac{1}{2}\right)$
$\Rightarrow h = 5,m$
So, C $\rightarrow$ p
Distance $OQ = \left|\text{Displacement of particle along }x\text{-direction}\right| = |s_x|$
Since, $s_x = u_xt + \frac{1}{2}a_xt^2$
$\Rightarrow s_x = (10\sqrt{3})(2) - \frac{1}{2}(5\sqrt{3})(2)^2 = 10\sqrt{3},m$
$\Rightarrow OQ = 10\sqrt{3},m$
Since, $PQ = \sqrt{(PO)^2 + (OQ)^2}$
$\Rightarrow PQ = \sqrt{(10)^2 + (10\sqrt{3})^2} = \sqrt{400}$
$\Rightarrow \boxed{PQ = 20,m}$

$A\to q)$
$B\to r)$
$C\to s)$
$D\to t)$
$E\to p)$
$t=\frac{x}{\alpha}\qquad ...(1)$
$y=\alpha\left(\frac{x}{\alpha}\right)\left[1-\beta\left(\frac{x}{\alpha}\right)\right]$
$\Rightarrow y=x-\left(\frac{\beta}{\alpha}\right)x^2\qquad ...(2)$
$y=0$ at $x=R$
$\Rightarrow 0=R-\frac{\beta}{\alpha}R^2$
$\Rightarrow R\left(1-\frac{\beta}{\alpha}R\right)=0$
$\Rightarrow R=\frac{\alpha}{\beta}=2\left(\frac{\alpha}{2\beta}\right)$
So, $A\to q$
At maximum height,
$\frac{dy}{dx}=0$
$\Rightarrow 1-\frac{\beta}{\alpha}(2x)=0$
$\Rightarrow x=\frac{\alpha}{2\beta}$
$H_{\max}=\frac{\alpha}{2\beta}-\frac{\beta}{\alpha}\left(\frac{\alpha^2}{4\beta^2}\right)$
$\Rightarrow H_{\max}=\frac{\alpha}{2\beta}-\frac{\alpha}{4\beta}=\frac{\alpha}{4\beta}=4\left(\frac{\alpha}{16\beta}\right)$
So, $B\to r$
$T=\frac{R}{u_x}$
${\text{or from (1), we have }t =\frac{x}{\alpha}}$
Since $x=\alpha t$, so $v_x=\frac{dx}{dt}=\alpha=\text{constant}$
$\Rightarrow v_x=u_x=\alpha$
$\Rightarrow T=\frac{\left(\frac{\alpha}{\beta}\right)}{\alpha}=\frac{1}{\beta}=\frac{1}{8\beta}$
So, $C\to s$
Also, we observe that $y=\alpha t-(\alpha\beta)t^2$
$\Rightarrow \frac{dy}{dt}=\alpha-2\alpha\beta t=\alpha(1-2\beta t)$
$\Rightarrow v_y=\alpha(1-2\beta t)$
$\Rightarrow \vec v=v_x\hat i+v_y\hat j$
$\Rightarrow \vec v=\alpha\hat i+\alpha(1-2\beta t)\hat j$
Now, $a_x=\frac{dv_x}{dt}=0$ and $a_y=-2(\alpha\beta)\hat j=\text{constant}$
$\Rightarrow \vec a=a_x\hat i+a_y\hat j=-(2\alpha\beta)\hat j$
$\Rightarrow |\vec a|=2\alpha\beta=16\left(\frac{\alpha\beta}{8}\right)$
So, $D\to t$
Now at $t=\frac{1}{2\beta}$, we have
$\vec v=\alpha\hat i+\alpha\left[1-(2\beta)\left(\frac{1}{2\beta}\right)\right]\hat j=\alpha\hat i$
$\Rightarrow |\vec v|=\alpha=1(\alpha)$
So, $E\to p$

A → (q, r) B → (p, r) C → (q, r, s) D → (r)

$A\to q$
$B\to p$
$C\to r$
$D\to s$
Since $R_1=R_2$
$\Rightarrow \frac{R_1}{R_2}=1$
$\Rightarrow \frac{H_1}{H_2}=\frac{u^2\sin^2(30^\circ)}{u^2\sin^2(60^\circ)}=\frac{1}{3}$
$\frac{T_2}{T_1}=\frac{u\sin(60^\circ)}{u\sin(30^\circ)}=\sqrt3$
$\frac{T_1H_1R_1}{T_2H_2R_2}=\frac{T_1H_1}{T_2H_2}=\frac{1}{3}\left(\frac{1}{\sqrt3}\right)=\frac{1}{3\sqrt3}$

$A\to q$
$B\to r$
$C\to p$
$D\to s$
$\vec u=20\hat i+20\hat j,\quad \vec a=-10\hat j\quad\text{and}\quad t=1,\mathrm{s}$
Since, $\Delta\vec r=\vec u t+\frac12\vec a t^2=20\hat i+15\hat j\qquad ...(1)$
and $\vec v=\vec u+\vec at=20\hat i+10\hat j\qquad ...(2)$
$\Rightarrow |\vec v_{\mathrm{av}}|=\left|\frac{\Delta\vec r}{t}\right|=\sqrt{(20)^2+(15)^2}=25,\mathrm{ms^{-1}}\quad\text{(from (1))}$
$\Rightarrow |\Delta\vec v|=|\vec v-\vec u|=10,\mathrm{ms^{-1}}\quad\text{(from (2))}$
$\Rightarrow |\vec v_{\mathrm{inst}}|=|\vec v|=\sqrt{(20)^2+(10)^2}=10\sqrt5,\mathrm{ms^{-1}}$
$\Rightarrow \Delta|\vec v|=|\vec v|-|\vec u|=20\sqrt2-10\sqrt5=6,\mathrm{ms^{-1}}$

$A\to r$
$B\to r$
$C\to p$
$D\to q$
Since, $y=x-\frac{x^2}{80}$
Comparing with the standard equation of projectile, i.e.
$y=x\tan\theta-\frac{gx^2}{2u^2\cos^2\theta}$, we get
$\tan\theta=1$
$\Rightarrow \theta=45^\circ$
and
$\frac{1}{80}=\frac{g}{2u^2\cos^2\theta}$
$\Rightarrow u=20\sqrt2,\mathrm{ms^{-1}}$
Since, $\vec v=u\cos\theta\hat i+(u\sin\theta-gt)\hat j=20\hat i-20\hat j$
When $x=R,\ y=0$
$\Rightarrow 0=R-\frac{R^2}{80}$
$\Rightarrow R=80,\mathrm{m}$

$\quad A\to(p,r)$
$B\to(q,s)$
$C\to(p,q,r,s)$
$D\to(p,q,r,s)$
For a particle to move in circular path, $\vec a\perp\vec v$
For a particle to undergo projectile motion, angle between $\vec a$ and $\vec v$ must be acute.

Change in momentum is $\Delta p = (mg)t$ in time $t$.
A) Change in momentum $\rightarrow$ q) Independent of projected velocity
B) Angle at the highest point $\rightarrow$ r) At highest point is zero
C) Kinetic energy of body $\rightarrow$ s) Minimum at highest point
D) Horizontal component of velocity $\rightarrow$ p) Remains unchanged.

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.
$\Rightarrow 0 = u \cos (\phi - 4 5 ^ {\circ}) - (g \sin 4 5 ^ {\circ}) t$

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