3 Marks Question

Question 13 Marks

Find the maximum revolution per minute of a body of 500 gram tied to a 1.5 m long string, if the string can bear a maximum tension of 40 N .

Answer

$\begin{array}{l}
m=\text { mass of body }=500 g=0.5 kg \\
R=\text { radius of circle }=1.5 m \\
F_c=\text { centripetal force }=40 N
\end{array}$
Now,
$\begin{aligned}
F_c & =\text { mass } \times \text { acceleration } \\
F_c & =m a_r \\
& =\frac{m v^2}{R}=\frac{m}{R}(R \omega)^2 \quad \because v=R \omega \\
& =\frac{m R^2 \omega^2}{R}=m R \omega^2 \\
& =m R(2 \pi f)^2=4 \pi^2 m R f^2 \\
f^2 & =\frac{F_c}{4 \pi^2 m R} \\
f & =\sqrt{\frac{F_c}{4 \pi^2 m R}}=\sqrt{\frac{40}{4 \times(3.14)^2 \times 0.5 \times 1.5}} \\
& =\sqrt{1.351}=1.162 revolution / sec \\
& =1.162 \times 60=69.74 rev / min
\end{aligned}$
Hence, the body can revolve at rate of 69 revolution/ minute without breakage of string.

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Question 23 Marks

The horizontal range of cannon ball is $R$. If the greatest heights of two paths, for which this is possible, are $h$ and $h ^{\prime}$ then prove that
$4 \sqrt{h h^{\prime}}=R$

Answer

Consider the cannon ball with velocity $u$ and at angles $\theta$ and $\left(\frac{\pi}{2}-\theta\right)$ with the horizontal then range
$R=\frac{u^2 \sin 2 \theta}{g}$
If projectile angle is $\theta$, then maximum height
$h=\frac{u^2 \sin ^2 \theta}{2 g} \ \ldots(1)$
If projectile angle is $\left(\frac{\pi}{2}-\theta\right)$, then maximum height is
$\begin{array}{l}
h^{\prime}=\frac{u^2 \sin ^2\left(\frac{\pi}{2}-\theta\right)}{2 g} \\
h^{\prime}=\frac{u^2 \cos ^2 \theta}{2 g} \ \ldots(2)
\end{array}$
From equations (1) and (2)
$\begin{aligned}
h h^{\prime} & =\frac{u^2 \sin ^2 \theta}{2 g} \times \frac{u^2 \cos ^2 \theta}{2 g} \\
h h^{\prime} & =\frac{u^4 \sin ^2 \theta \cos ^2 \theta}{4 g^2} \\
\therefore \quad 4 h h^{\prime} & =\frac{u^4 \sin ^2 \theta \cos ^2 \theta}{g^2} \\
\text { or } \quad 16 h h^{\prime} & =\frac{u^4 \times(2 \sin \theta \cos \theta)^2}{g^2} \\
16 h h^{\prime} & =\frac{u^4 \sin ^2 2 \theta}{g^2}
\end{aligned}$
We take square root on both side,
$\begin{aligned}
4 \sqrt{h h^{\prime}} & =\frac{u^2 \sin 2 \theta}{g}=R \\
R & =4 \sqrt{h h^{\prime}}
\end{aligned}$ Hence Proved.

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Question 33 Marks

A player can throw a stone to a maximum horizontal distance of 100 m . What is the maximum vertical height to which the player can throw the same stone?

Answer

Given that
But
$\begin{aligned}
R_{\max } & =100 m \\
R_{\max } & =\frac{u^2}{g}=100 m \\
u^2 & =100 g
\end{aligned}$
$\therefore$
For vertical position,
$\begin{aligned}
v^2 & =u^2+2 g h \\
0 & =100 g-2 g H \\
2 g H & =100 g \\
H & =\frac{100 g}{2 g}=50 m
\end{aligned}$

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Question 43 Marks

The position vector of a particle depends on time as follows (in meters).
$\vec{r}=\left(2+3 t-t^3\right) \hat{i}+\left(-t+t^2\right) \hat{j}+\left(7 t+t^3\right) \hat{k}$
Find (i) Displacement of particle from $t=0 sec$ to $t=1 sec$ (ii) At $t=1 sec$, find the magnitude of instantaneous velocity and instantaneous acceleration of particle.

Answer

(1) At $t=0$, Position vector
$\begin{array}{l}
\overrightarrow{r_1}=2 \hat{i} \quad \text { At } t=1, \text { Position vector } \\
\overrightarrow{r_2}=\left(2+3 \times 1-1^3\right) \hat{i}+\left(-1+1^2\right) \hat{j}+\left(7 \times 1+1^3\right) \hat{k} \\
\overrightarrow{r_2}=4 \hat{i}+0 . \hat{j}+8 \hat{k}
\end{array}$
Hence, displacement
$\begin{aligned}
\Delta \vec{r} & =\overrightarrow{r_2}-\overrightarrow{r_1} \\
& =4 \hat{i}+0 \cdot \hat{j}+8 \hat{k}-2 \hat{i} \\
& =2 \hat{i}+8 \hat{k} m
\end{aligned}$
(ii)
$\begin{aligned}
\vec{v} & =\frac{d \vec{r}}{d t} \\
& =\frac{d}{d t}\left\{\left(2-3 t-t^3\right) \hat{i}+\left(-t+t^2\right) \hat{j}+\left(7 t+t^3\right) \hat{k}\right\} \\
v & =\left(3-3 t^2\right) \hat{i}+(-1+2 t) \hat{j}+\left(7+3 t^2\right) \hat{k}
\end{aligned}$
and $\quad \vec{a}=\frac{d \overrightarrow{ v }}{d t}=(0-6 t) \hat{i}+(0+2) \hat{j}+(0+6 t) \hat{k}$
$\begin{aligned}
\vec{a} & =-6 \hat{i}+2 \hat{j}+6 \hat{k} \quad t=1 sec \\
\vec{v} & =0 . \hat{i}+\hat{j}+\left(7+3 \times 1^2\right) \hat{k} \\
& =\hat{j}+10 \hat{k} m / s
\end{aligned}$
$t=1 sec$
$\begin{aligned}
\therefore|\vec{v}| & =\sqrt{(1)^2+(10)^2}=\sqrt{1+100}=\sqrt{101} \\
& =10.05 m / s
\end{aligned}$
Similarly, put $t=1 sec$
$\begin{aligned}
\vec{a} & =-6 \hat{i}+2 \hat{j}+6 \hat{k} ms^{-2} \\
\therefore \quad|\vec{a}| & =\sqrt{(-6)^2+(2)^2+(6)^2}=\sqrt{36+4+36}=\sqrt{76} \\
& =8.72 m / s^2
\end{aligned}$

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Question 53 Marks

The resultant of two vector $\vec{P}$ and $\vec{Q}$ is $\vec{R}$. If the direction of $\vec{Q}$ is reversed, then the resulting vector becomes $\vec{S}$. Then prove that,
$
R ^2+ S ^2=2\left( P ^2+ Q ^2\right)
$

Answer

It is given,
$\begin{array}{l}
\overrightarrow{R}=\overrightarrow{P}+\overrightarrow{Q} \\
\overrightarrow{S}=\overrightarrow{P}-\overrightarrow{Q}
\end{array}$
If angle between $\vec{P}$ and $\vec{Q}$ is $\theta$, then
$\begin{aligned}
R & =\sqrt{P^2+Q^2+2 PQ \cos \theta} \\
R^2 & =P^2+Q^2+2 PQ \cos \theta \ \ldots(1)
\end{aligned}$
If angle between $\vec{P}$ and $\vec{Q}$ is $\alpha$, then
$\begin{aligned}
\alpha & =(180-\theta) \\
\because \quad \overrightarrow{S} & =(\overrightarrow{P}-\overrightarrow{Q}) \\
\therefore \quad S & =\sqrt{P^2+Q^2+2 PQ \cos \alpha} \\
S & =\sqrt{P^2+Q^2+2 PQ \cos (180-\theta)} \\
& =\sqrt{P^2+Q^2-2 PQ \cos \theta} \\
S^2 & =P^2+Q^2-2 PQ \cos \theta \ \ldots(2)
\end{aligned}$
Add equations (1) and (2), we get
$R^2+S^2=2\left(P^2+Q^2\right)$

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Question 63 Marks

Prove that in projectile motion, when bodies are projected at angle $\left(45^{\circ}+\phi\right)$ and $\left(45^{\circ}-\phi\right)$ from the horizontal, their horizontal range will be same.

Answer

We know that by projecting an object at angle $\theta$,
$R=\frac{u^2 \sin 2 \theta}{g}$
But it is given that when object is projected at angle $(45+\phi)$ and $\left(45^{\circ}-\phi\right)$ from the horizontal, their ranges are same.
$

\begin{array}{l}

R_1=\frac{u^2 \sin 2\left(45^{\circ}+\phi\right)}{g} \\

R_1=\frac{u^2 \sin \left(90^{\circ}+2 \phi\right)}{g} \\

R_1=\frac{u^2 \cos 2 \phi}{g}.....(1)

\end{array}

$
similarly,
$

\begin{array}{l}

R_2=\frac{u^2 \sin 2\left(45^{\circ}-\phi\right)}{g} \\

R_2=\frac{u^2 \sin \left(90^{\circ}-2 \phi\right)}{g} \\

R_2=\frac{u^2 \cos 2 \phi}{g}.....(2)

\end{array}

From equations (1) and (2), it is clear
$R_1=R_2$

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Question 73 Marks

The time of flight of a projectile is $T$ and horizontal range is $R$. What will be the projectile angle?Or The time of flight is $T$ and horizontal range is $R$ of a projectile. Prove that the angle of projection of a projectile is $\theta=\tan ^{-1}\left(\frac{g T^2}{2 R}\right)$.

Answer

Time of flight, $\quad T =\frac{2 u \sin \theta}{g}$
Horizontal range, $\quad R =\frac{u^2 \sin 2 \theta}{g}$
On square of equation (1)
$T^2=\frac{4 u^2 \sin ^2 \theta}{g}$
By dividing equation (3) to equation (2)
$\begin{array}{rlrl}
\frac{T^2}{R} & =\frac{\frac{4 u^2 \sin ^2 \theta}{g}}{\frac{u^2 \sin 2 \theta}{g}} \\
\Rightarrow \quad & \frac{g T^2}{R} & =\frac{4 \sin ^2 \theta}{\sin 2 \theta}=\frac{4 \sin ^2 \theta}{2 \sin \theta \cos \theta} \\
\Rightarrow \quad & \frac{g T^2}{R} & =\frac{2 \sin \theta}{\cos \theta}=2 \tan \theta \\
\therefore \quad & \frac{g T^2}{R} & =\tan \theta \\
\therefore \quad & \theta & =\tan ^{-1}\left(\frac{g T^2}{2 R}\right)
\end{array}$

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Question 83 Marks

What is the meaning of the components of vectors along $X, Y, Z$ axes and prove that $\cos ^2 \alpha+\cos ^2$ $\beta+\cos ^2 r=1$.

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