Physics 1 Marks Question

1. According to the problem, unit vector

$\hat{\text{r}}=\cos\theta\hat{\text{i}}+\sin\theta\hat{\text{j}}\cdots(\text{i})$

$\hat{\theta}=-\sin\theta\hat{\text{i}}+\cos\theta\hat{\text{j}}\cdots(\text{ii})$

Multiplying Eq.(i) by $\sin\theta$ and Eq. (ii) with $\cos\theta$ and adding

$\Rightarrow\hat{\text{r}}\sin\theta+\hat{\theta}\cos\theta$$=\sin\theta\cdot\cos\theta\hat{\text{i}}+\sin^2\theta\hat{\text{j}}+\cos^2\theta\hat{\text{j}}-\sin\theta\cdot\cos\theta\hat{\text{i}}$

$=\hat{\text{j}}(\cos^2\theta+\sin^2\theta)=\hat{\text{j}}$

$\hat{\text{r}}\sin\theta+\hat{\theta}\cos\theta=\hat{\text{j}}$

By Eq.(i) $\text{x}\cos\theta-$Eq. (ii) $\text{x}\sin\theta$

$\text{n}(\hat{\text{r}}\cos\theta-\hat{\theta}\sin\theta)=\hat{\text{i}}$

2. $\hat{\text{r}}\cdot\hat{\theta}=(\cos\theta\hat{\text{i}}+\sin\theta\hat{\text{j}})\cdot(-\sin\theta\hat{\text{i}}+\cos\theta\hat{\text{j}})$

$=-\cos\theta\cdot\sin\theta+\sin\theta\cdot\cos\theta=0$

$\Rightarrow\theta=90^\circ,$ angle between $\hat{\text{r}}$ and $\hat{\theta}$.

3. Given, $\hat{\text{r}}=\cos\theta\hat{\text{i}}+\sin\theta\hat{\text{j}}$

$\frac{\text{d}\hat{\text{r}}}{\text{dt}}=\frac{\text{d}}{\text{dt}}(\cos\theta\hat{\text{i}}+\sin\theta\hat{\text{j}})$

$=-\sin\theta\cdot\frac{\text{d}\theta}{\text{dt}}\hat{\text{i}}+\cos\theta\cdot\frac{\text{d}\theta}{\text{dt}}\hat{\text{j}}$

$=\omega[-\sin\theta\hat{\text{i}}+\cos\theta\hat{\text{j}}$ $\Big[\because\theta=\frac{\text{d}\theta}{\text{dt}}\Big]$

4. Given,$\text{r}=\text{a}\theta\hat{\text{r}},$ here writing dimension.

$[\text{r}]=[\text{a}][\theta][\hat{\text{r}}]\Rightarrow\text{L}=\frac{[\text{a}]}{1}$

$\Rightarrow[\text{a}]=\text{L}=[\text{M}^0\text{L}^1\text{T}^0]$

5. Given, $\text{a}=1$ unit, $\text{r}=\theta\hat{\text{r}}=\theta[\cos\theta\hat{\text{i}}+\sin\theta\hat{\text{j}}]$

By differentiating this equation,we get

Velocity $\text{v}=\frac{\text{dr}}{\text{dt}}=\frac{\text{d}\theta}{\text{dt}}\hat{\text{r}}+\theta\frac{\text{d}}{\text{dt}}\hat{\text{r}}$

$=\frac{\text{d}\theta}{\text{dt}}\hat{\text{r}}+\theta\frac{\text{d}}{\text{dt}}[(\cos\theta\hat{ \text{i}}+\sin\theta\hat{\text{j}})]$

$=\frac{\text{d}\theta}{\text{dt}}\hat{\text{r}}+\theta\Big[(-\sin\theta\hat{\text{i}}+\cos\theta\hat{\text{j}})\frac{\text{d}\theta}{\text{dt}}\Big]$

$=\frac{\text{d}\theta}{\text{dt}}\hat{\text{r}}+\theta\hat{\theta}\omega=\omega\hat{\text{r}}+\omega\theta\hat{\theta}$

By differentiating this equation, we get

Acceleration,

$\text{a}=\frac{\text{d}\theta}{\text{dt}}[\omega\hat{\text{r}}+\omega\theta\hat{\theta}]$

$=\frac{\text{d}}{\text{dt}}\Big[\frac{\text{d}\theta}{\text{dt}}\hat{\text{r}}+\frac{\text{d}\theta}{\text{dt}}(\theta\hat{\theta})\Big]$

$=\frac{\text{d}^2\theta}{\text{dt}^2}\hat{\text{r}}+\frac{\text{d}\theta}{\text{dt}}\cdot\frac{\text{d}\hat{\text{r}}}{\text{dt}}+\frac{\text{d}^2\theta}{\text{dt}^2}\theta\hat{\theta}+\frac{\text{d}\theta}{\text{dt}}\frac{\text{d}}{\text{dt}}(\theta\hat{\theta})\frac{\text{d}^2\theta}{\text{dt}^2}\hat{\text{r}}$

$+\omega[-\sin\theta\hat{\text{i}}+\sin\theta\hat{\text{j}}]+\frac{\text{d}^2\theta}{\text{dt}^2}\theta\hat{\theta}+\frac{\omega\text{d}}{\text{dt}}(\theta\hat{\theta})$

$=\frac{\text{d}^2\theta}{\text{dt}^2}\hat{\text{r}}+\omega^2\hat{\theta}+\frac{\text{d}^2\theta}{\text{dt}^2}\times\theta\hat{\theta}+\omega^2\hat{\theta}+\omega^2\theta(-\hat{\text{r}})$

$\Big(\frac{\text{d}^2\theta}{\text{dtv}^2}-\omega^2\Big)\hat{\text{r}}+\Big(2\omega^2+\frac{\text{d}^2\theta}{\text{dt}^2}\theta\Big)\theta$

$\text{u}_\text{x}=\text{u}+\text{v}_0\cos\theta$

Initial velocity in y-direction,

$\text{u}_\text{y}=\text{v}_0\sin\theta$

Where angle of projection is $\theta$.

1. Now, the angle of projection with horizontal seen by spectator will be

$\tan\theta=\frac{\text{u}_\text{y}}{\text{u}_\text{x}}=\frac{\text{u}_0\sin\theta}{\text{u}+\text{u}_0\cos\theta}$

$\Rightarrow\theta=\tan^{-1}\Big(\frac{\text{v}_0\sin\theta}{\text{u}+\text{v}_0\cos\theta}\Big)$

2. As net displacement along y-axis is zero over time period T(time of flight).

$\text{y}=0,\text{u}_\text{y}=\text{v}_0\sin\theta,\text{a}_\text{y}=-\text{g},\text{t}=\text{T}$

We know that $\text{y}=\text{u}_\text{y}\text{t}+\frac{1}{2}\text{a}_\text{y}\text{t}^2$

$\Rightarrow0=\text{v}_0\sin\theta\text{T}+\frac{1}{2}(-\text{g})\text{T}^2$

$\Rightarrow\text{T}\Big[\text{v}_0\sin\theta-\frac{\text{g}}{2}\text{T}\Big]=0$

$\Rightarrow\text{T}=0,\frac{2\text{v}_0\sin\theta}{\text{g}}$

$\text{T}=0,$ corresponds to point O.

Hence, $\text{T}=\frac{2\text{u}_0\sin\theta}{\text{g}}$

3. Horizontal range,

$\text{R}=(\text{u+v}_0\cos\theta)\text{T}$

$=(\text{u}+\text{v}_0\cos\theta)\frac{2\text{v}_0\sin\theta}{\text{g}}$

$=\frac{\text{v}_0}{\text{g}}[2\text{u}\sin\theta+\text{v}_0\sin2\theta]$

4. For horizontal range to be maximum,$\frac{\text{dR}}{\text{d}\theta}=0$

$\Rightarrow\frac{\text{v}_0}{\text{g}}[2\text{u}\cos\theta+\text{v}_0\cos2\theta\times2]=0$

$\Rightarrow2\text{u}\cos\theta+2\text{v}_0[2\cos^2\theta-1]=0$

$\Rightarrow4\text{v}_0\cos^2\theta+2\text{u}\cos\theta-2\text{v}_0=0$

$\Rightarrow2\text{v}_0\cos^2\theta+\text{u}\cos\theta-\text{v}_0=0$

$\Rightarrow\cos\theta=\frac{-\text{u}\pm\sqrt{\text{u}^2+{8\text{v}_\text{0}}^2}}{4\text{v}_0}$

$\Rightarrow\theta_\text{max}=\cos^{-1}\Bigg[\frac{-\text{u}\pm\sqrt{\text{u}^2+{8\text{v}_0}^2}}{4\text{v}_0}\Bigg]$

$=\cos^{-1}\Bigg[\frac{-\text{u}\pm\sqrt{\text{u}^2+{8\text{v}_0}^2}}{4\text{v}_0}\Bigg]$

5. If $\text{u}=\text{v}_0,$

$\cos\theta=\frac{-\text{v}_0\pm\sqrt{\text{v}^2_0+8\text{v}^2_0}}{4\text{v}^2_0}=\frac{-1+3}{4}=\frac{1}{2}$

$\Rightarrow\theta=60^\circ$

If $\text{u}<<\text{v}_0,$ then $8\text{v}^2_0+\text{u}^2\approx8\text{v}^2_0$

$\theta_\text{max}=\cos^{-1}\Bigg[\frac{-\text{u}\pm2\sqrt{2}\text{v}_0}{4\text{v}_0}\Bigg]$$=\cos^{-1}\Big[\frac{1}{\sqrt{2}}-\frac{\text{u}}{4\text{v}_0}\Big]$

If $\text{u}<<\text{v}_0$ then $\theta_\text{max}=\cos^{-1}\Big(\frac{1}{\sqrt{2}}\Big)=\frac{\pi}{4}$

If $\text{u}>\text{u}_0$ and $\text{u}>>\text{v}_0$

$\theta_\text{max}=\cos^{-1}\Big[\frac{-\text{u}\pm\text{u}}{4\text{v}_0}\Big]=0$

$\Rightarrow\theta_\text{max}=\frac{\pi}{2}$

6. If $\text{u}=0,\theta_\text{max}=\cos^{-1}\Bigg[\frac{0\pm\sqrt{8\text{v}^2_0}}{4\text{v}_0}\Bigg]$

$=\cos^{-1}\Big(\frac{1}{\sqrt{2}}\Big)=45^\circ$

Hence, when $\text{u}=0,\theta\geq45^\circ$

$\text{a}_\text{y}=-\text{g}\cos\alpha$

$\text{a}_\text{x}=\text{g}\sin\alpha$

$\text{x}=\text{L}, \text{u}_{\text{x}}=\text{v}_{0}\cos\beta, \text{a}_\text{x}=-\text{g}\sin\alpha$

$\text{t} = \text{T}=\frac{2\text{v}_0\sin\beta}{\text{g}\cos\alpha}$

$\text{s} = \text{u}_\text{x}\text{t}+\frac{1}{2}\text{a}_\text{x}\text{t}^{2}$

$\text{L}=\text{v}_0\cos\beta(\text{T})+\frac{1}{2}(-\text{g}\sin\alpha)\text{T}^2=\text{T}\big[\text{v}_0\cos\beta-\frac{1}{2}\text{g}\sin\alpha.\text{T}\big]$

$=\frac{2\text{v}_0\sin\beta}{\text{g}\cos\alpha}\Big[\text{v}_0\cos\beta-\frac{1}{2}\text{g}\sin\alpha.\frac{2\text{v}_0\sin\beta}{\text{g}\cos\alpha}\Big]$

$=\frac{2\text{v}^{2}_{0}\sin\beta}{\text{g}\cos^2\alpha}[\cos\beta.\cos\alpha-\sin\beta\sin\alpha]$

$\text{L}=\frac{2\text{v}^{2}_{0}\sin\beta}{\text{g}\cos^2\alpha}\cos(\alpha+\beta)$ [Range on the plane surface]

The main concept used: This problem can be solved in two different ways:

1. The target is at horizontal distance $(\text{R}+\Delta\text{x})$ and below the point of projection h metre below i.e., y = -h.
2. The motion of projectile after point P to T: projection speed is at an angle$(-\theta)$ i.e., $\theta^\circ$ below horizontal and vertical height covered it is (-h) and horizontal range $\Delta\text{x}.$

Solution:

$\text{R}=\frac{\text{v}^2_0}{\text{g}}\ \dots(\text{i})$ that it is maximum range of projectile

$\therefore$ The angle of projection $\theta=45^\circ$

Let the gun is raised to a heighth from the horizontal level of target T. So that the projectile can hit the target T.

Total range of projectile must be $=(\text{R}+\Delta\text{x})$

Horizontal component of velocity at $\text{A}=\text{v}_0\cos\theta$

The motion of the projectile from P to T: As A and P are on the same level. So the magnitude of velocity will be same at A and P.

$\therefore$ But the direction of velocity will be below horizontal,

so horizontal velocity at $\text{P}=\text{v}_\text{x}=-\text{v}_0\cos\theta$

So vertical velocity at $\text{P}=\text{v}_\text{y}=\text{v}_0\sin\theta$

$\text{h}=\text{ut}+\frac{1}2\text{at}^2$

$\text{h}=-\text{v}_0\sin\theta(\text{t})+\frac{1}2\text{gt}^2\ \dots(\text{ii})$

Consider horizontal motion from A to T = distance $(\text{R}+\Delta\text{x})=\text{v}_0\cos\theta.\text{t}$

$\text{t}=\frac{\text{R}+\Delta\text{x}}{\text{v}_0\cos\theta}$

Substitue t in (ii) $\text{h}=-\text{v}_02\sin\theta\Big[\frac{\text{R}+\Delta\text{x}}{\text{v}_0\cos\theta}\Big]+\frac{1}2\text{g}\frac{(\text{R}+\Delta\text{x})^2}{\text{v}^2_0\cos^2\theta}$

$\text{h}=-\tan\theta(\text{R}+\Delta\text{x})+\frac{1}2\Big(\frac{\text{g}}{\text{v}^2_0}\Big)\frac{(\text{R}+\Delta\text{x})^2}{\frac{1}2}\theta=45^\circ$

$\text{h}=-(\text{R}+\Delta\text{x})+\frac{1}{\text{R}}(\text{R}^2+\Delta\text{x}^2+2\text{R}\Delta\text{x})$ $\Big[\because\frac{\text{g}}{\text{v}^2_0}=\frac{1}{\text{R}}\Big]$

$=-\text{R}-\Delta\text{x}+\text{R}+\frac{\Delta\text{x}^2}{\text{R}}+2\Delta\text{x}=\Delta\text{x}+\frac{\Delta\text{x}^2}{\text{R}}$

$\text{h}=\Delta\text{x}\Big[1+\frac{\Delta\text{x}}{\text{R}}\Big]$ Hence proved.

$\vec{\text{V}}_\text{MR}=$ Velocity of man relative to river flow

$\vec{\text{V}}_\text{RE}=$ Velocity of river flow relative to earth

$\vec{\text{V}}_\text{ME}=$ Velocity of man relative to earth

Denoting $|\vec{\text{V}}_\text{MR}|=\text{v},|\vec{\text{V}}_\text{RE}|=\text{u},|\vec{\text{V}}_\text{ME}|=\text{v}_\text{R}$

Case (i): Man swims towards flow

$\vec{\text{V}}_\text{MR}=\text{v}\sin\theta\hat{\text{i}}+\text{v}\cos\theta\hat{\text{j}};\vec{\text{V}}_\text{ME}=\text{u}\hat{\text{i}}$ as $\vec{\text{V}}_\text{ME}=\vec{\text{V}}_\text{MR}+\vec{\text{V}}_\text{RE}$

$\vec{\text{V}}_\text{R}=(\text{v}\sin\theta\hat{\text{i}}+\cos\theta\hat{\text{j}})+\text{u}\hat{\text{i}}$

$=(\text{v}\sin\theta+\text{u})\hat{\text{i}}+\text{v}\cos\theta\hat{\text{j}}$

Suppose the man reaches the point B(x, y)and time of motion is t. We have

$\text{x}=\text{v}_{\text{RX}}\text{t}=(\text{v}\sin\theta+\text{u})\text{t}\ \cdots(\text{i})$

$\text{y}=\text{v}_{\text{RY}}\text{t}=\text{v}\cos\theta\text{t}\ \cdots(\text{ii})$

We have used the fact that the motion along the x- and y axes takes place with the respective to x- and y- components of resultant velocity.

Case (ii): Man swims against flow

$\vec{\text{v}}_\text{R}=(-\text{v}\sin\theta+\text{u})\hat{\text{i}}+\text{v}\cos\theta\hat{\text{j}}$

$\text{x}=\text{v}_\text{RX}\text{t}=(\text{u}-\text{v}\sin\theta)\text{t}\ \ \cdots(\text{iii})$

$\text{y}=\text{v}_\text{RX}\text{t}=\text{v}\cos\theta\text{ t}\cdots(\text{iv})$

• If width of the river is d, then time taken to cross the river is

$\text{t}=\frac{\text{y}}{\text{v}_{\text{RY}}}=\frac{\text{d}}{\text{v}\cos\theta}$

• Time of crossing is minimum if $\cos\theta=1$, i.e. the swimmer begins to swim at right angle to flow $\text{t}_{\text{min}}=\frac{\text{d}}{\text{v}}\cdot$
• Actual velocity of the swimmer is $|\vec{\text{v}}_\text{R}|=\sqrt{\text{v}^2+\text{u}^2}$ at an angle

$\tan\alpha=\frac{|\vec{\text{u}|}}{|\vec{\text{v}}|}$.

• If the man wishes to swim straight (perpendicular to flow), his resultant velocity must be zero. $\text{v}_{\text{RX}}=\text{u - v}\sin\theta=0$

$|\vec{\text{v}}_{\text{R}}|=\sqrt{\text{v}^2+\text{u}^2}$

$\tan\alpha=\frac{|\vec{\text{u}}|}{|\vec{\text{v}}|}$

$|\vec{\text{v}}_{\text{R}}|=\sqrt{\text{v}^2-\text{u}^2}$

$\sin\alpha=\frac{|\vec{\text{u}}|}{|\vec{\text{v}}|}$

1. As the swimmer starts swimming due north, then y-component of his resultant velocity is 4 m/ s (north) and x-component (produced by river flow) is 3 m/ s (east). Then his resultant velocity will be

$\text{v}=\sqrt{\text{v}^2_\text{r}+\text{v}^2_\text{s}}=\sqrt{(3)^2+(4)^2}$

$=\sqrt{9+16}=\sqrt{25}=5\text{m/ s}$

$\tan\theta=\frac{\text{v}_\text{r}}{\text{v}_\text{s}}=\frac{3}{4}=0.75=\tan36^\circ54'$

Hence, $\theta=36^\circ54'\text{ N or }\theta=37^\circ\text{N}$

2. If he wants to reach on the point directly opposite to him (points B), the swimmer should swim at an angle θ of north.

Resultant speed of the swimmer

$\text{v}=\sqrt{\text{v}^2_\text{s}-\text{v}^2_\text{r}}=\sqrt{(4)^2-(3)^2}$

$=\sqrt{16-9}=\sqrt{7}\text{m/ s}$

$\tan\theta=\frac{\text{v}_\text{r}}{\text{v}}=\frac{3}{\sqrt{7}}$

$\theta=\tan^{-1}\Big(\frac{3}{\sqrt{7}}\Big)\text{of north}$

3. In case (a)

Time taken by the swimmer to cross the river, $\text{t}_1=\frac{\text{d}}{\text{v}_\text{s}}=\frac{\text{d}}{4}\text{s}$

In case (b),

Time taken by the swimmer to cross the river,

$\text{t}_1=\frac{\text{d}}{\text{v}}=\frac{\text{d}}{\sqrt{7}}$

As $\frac{\text{d}}{\text{v}}<\frac{\text{d}}{\sqrt{7}},$ therefore $\text{t}_1<\text{t}_2.$

Hence, the swimmer will cross the river in shorter time in case (a).