2 Marks Questions

Question 12 Marks

Pick out the only vector quantity in the following list:
Temperature, pressure, impulse, time, power, total path length, energy, gravitational potential, coefficient of friction, charge.

Answer

Since, Impulse = change in momentum = force × time. As momentum and force are vector quantities, hence impulse is a vector quantity.

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Question 22 Marks

An insect trapped in a circular groove of radius $12 cm$ moves along the groove steadily and completes 7 revolutions in $100 s$. (a) What is the angular speed, and the linear speed of the motion? (b) Is the acceleration vector a constant vector? What is its magnitude?

Answer

This is an example of uniform circular motion. Here $R=12 cm$. The angular speed $\omega$ is given by
$
\omega=2 \pi / T=2 \pi \quad 7 / 100=0.44 rad / s
$
The linear speed $v$ is :
$
v=\omega R=0.44 s ^{-1} \quad 12 cm =5.3 cm s ^{-1}
$
The direction of velocity $v$ is along the tangent to the circle at every point. The acceleration is directed towards the centre of the circle. Since this direction changes continuously, acceleration here is not a constant vector. However, the magnitude of acceleration is constant:
$
\begin{array}{l}
a=\omega^2 R=\left(0.44 s ^{-1}\right)^2(12 cm ) \\
=2.3 cm s ^{-2}
\end{array}
$

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Question 32 Marks

A cricket ball is thrown at a speed of $28 m s ^{-1}$ in a direction 30 above the horizontal. Calculate (a) the maximum height, (b) the time taken by the ball to return to the same level, and (c) the distance from the thrower to the point where the ball returns to the same level.

Answer

(a) The maximum height is given by
$
\begin{aligned}
h_m & =\frac{\left(v_o \sin \theta_o\right)^2}{2 g}=\frac{\left(28 \sin 30^{\circ}\right)^2}{2(9.8)} m \\
& =\frac{14 \times 14}{2 \times 9.8}=10.0 m
\end{aligned}
$
(b) The time taken to return to the same level is
$
\begin{array}{l}
T_f=\left(2 v_0 \sin \theta_0\right) / g=\left(\begin{array}{lll}
2 & 28 & \sin 30
\end{array}\right) / 9.8 \\
=28 / 9.8 s =2.9 s \\
\end{array}
$
(c) The distance from the thrower to the point where the ball returns to the same level is
$
R=\frac{\left(v_{ o }^2 \sin 2 \theta_{ o }\right)}{a}=\frac{28 \times 28 \times \sin 60^{\circ}}{98}=69 m
$

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Question 42 Marks

The position of a particle is given by
$
r =3.0 t \hat{ i }+2.0 t^2 \hat{ j }+5.0 \hat{ k }
$
where $t$ is in seconds and the coefficients have the proper units for $r$ to be in metres. (a) Find $v (t)$ and $a(t)$ of the particle. (b) Find the magnitude and direction of $v (t)$ at $t=1.0 s$.

Answer

$
\begin{aligned}
v (t) & =\frac{ d r }{ d t}=\frac{ d }{ d t}\left(3.0 t \hat{ i }+2.0 t^2 \hat{ j }+5.0 \hat{ k }\right) \\
& =3.0 \hat{ i }+4.0 t \hat{ j } \\
a (t) & =\frac{ d v }{ d t}=+4.0 \hat{ j } \\
a & =4.0 m s ^{-2} \text { along } y \text { - direction }
\end{aligned}
$
At $t=1.0 s , \quad v =3.0 \hat{ i }+4.0 \hat{ j }$
It's magnitude is $v=\sqrt{3^2+4^2}=5.0 m s ^{-1}$ and direction is
$\theta=\tan ^{-1}\left(\frac{v_y}{v_x}\right)=\tan ^{-1}\left(\frac{4}{3}\right) \cong 53^{\circ}$ with $x$-axis.

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Question 52 Marks

Rain is falling vertically with a speed of $35 m s ^{-1}$. Winds starts blowing after sometime with a speed of $12 m s ^{-1}$ in east to west direction. In which direction should a boy waiting at a bus stop hold his umbrella?

Answer

The velocity of the rain and the wind are represented by the vectors $v _{ r }$ and $v _{ w }$ in Fig. 3.7 and are in the direction specified by the problem. Using the rule of vector addition, we see that the resultant of $v _{ r }$ and $v _{ w }$ is $R$ as shown in the figure. The magnitude of $R$ is
$
R=\sqrt{v_r^2+v_w^2}=\sqrt{35^2+12^2} m s ^{-1}=37 m s ^{-1}
$
The direction $\theta$ that $R$ makes with the vertical is given by
$
\tan \theta=\frac{v_w}{v_r}=\frac{12}{35}=0.343
$
Or, $\quad \theta=\tan ^{-1}(0.343)=19^{\circ}$
Therefore, the boy should hold his umbrella in the vertical plane at an angle of about $19^{\circ}$ with the vertical towards the east.

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Question 62 Marks

If $\vec{\text{A}}=(-2\hat{\text{i}}+3\hat{\text{j}}-4\hat{\text{k}})$ and $\vec{\text{B}}=(3\hat{\text{i}}-4\hat{\text{j}}+5\hat{\text{k}})$ find $\vec{\text{A}}\times\vec{\text{B}}$ and $\vec{\text{A}}.\vec{\text{B}}$

Answer

$\vec{\text{A}}\times\vec{\text{B}}=\begin{vmatrix}\hat{\text{i}} & \hat{\text{j}}&\hat{\text{k}} \\-2 & 3&-4\\3&-4&5 \end{vmatrix}$

$=\hat{\text{i}}\begin{vmatrix}3&-4\\-4&5\end{vmatrix}-\hat{\text{j}}\begin{vmatrix}-2&-4\\3&5\end{vmatrix}+\hat{\text{k}}\begin{vmatrix}-2&3\\3&-4\end{vmatrix}$

$=\hat{\text{i}}(15-16)-\hat{\text{j}}(-10+12)+\hat{\text{k}}(8-9)$

$=-\hat{\text{i}}-2\hat{\text{j}}-\hat{\text{k}}$

$\vec{\text{A}}.\vec{\text{B}}=(-2\hat{\text{i}}+3\hat{\text{j}}-4\hat{\text{k}}).(3\hat{\text{i}}-4\hat{\text{j}}+5\hat{\text{k}})$

$=-6-12-20=-38$

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Question 72 Marks

At what point of projectile path the speed is minimum?
At which point, the speed is maximum?

Answer

At the highest point.
At the projection point.

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Question 82 Marks

Prove the following statement, “For Elevation which exceed or fall short of 45° by equal amount, the range is equal.”

Answer

At 45°, the projectile has maximum range.
At $(45^\circ-\theta)$ the range is,
$\text{R}_1=\frac{\text{u}^2\sin[2(45^\circ-\theta)]}{\text{g}}$
$=\frac{\text{u}^2\sin(90^\circ-2\theta)}{\text{g}}=\frac{\text{u}^2\cos2\theta}{\text{g}}$
At $(45^\circ+\theta)$ the range is,
$\text{R}_2=\frac{\text{u}^2\sin[2(45^\circ+\theta)]}{\text{g}}$
$=\frac{\text{u}^2\sin(90^\circ+2\theta)}{\text{g}}=\frac{\text{u}^2\cos2\theta}{\text{g}}$
$\text{R}_1=\text{R}_2$

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Question 92 Marks

Give the geometrical meaning of (i) scalar product (ii) cross product of two vectors.

Answer

Geometrically,

$\vec{\text{A}}.\vec{\text{B}}$ refers to the product of magnitude of $\vec{\text{A}}$ and projection of $\vec{\text{B}}$ on $\vec{\text{A}}.$
$\vec{\text{A}}\times\vec{\text{B}}$ refers to the area of the parallelogram formed by the vectors.

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Question 102 Marks

The dot product of two vectors vanishes when vectors are orthogonal and has maximum value when vectors are parallel to each other. Explain.

Answer

We know that $\text{A}.\text{B}=\text{AB}\cos\theta,$ when vectors are or thogonal, $\theta=90^\circ.$
So, $\text{A}.\text{B}=\text{AB}\cos90^\circ=0,$ when vectors are parallel, then, $\theta=0^\circ.$
So, $\text{A}.\text{B}=\text{AB}\cos0^\circ=\text{AB}$ (maximum)

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Question 112 Marks

Prove the following statement "For Elevations which exceed or fall short of 45° by equal amounts, the ranges are equal”.

Answer

Case I: When angle of projection
$\theta=45^\circ+\alpha$
Range, $\text{R}_1=\frac{\text{u}^2\sin2(45^\circ+\alpha)}{\text{g}}$
$=\frac{\text{u}^2\cos2\alpha}{\text{g}}$
Case II: When angle of projection
$\theta=45^\circ-\alpha$
Range, $\text{R}_2=\frac{\text{u}^2\sin2(45^\circ-\alpha)}{\text{g}}$
$=\frac{\text{u}^2}{\text{g}}\sin(90^\circ-2\alpha)$
$=\frac{\text{u}^2\cos2\alpha}{\text{g}}=\text{R}_1$

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Question 122 Marks

Prove that the maximum horizontal range is four times the maximum height attained by a projectile which is fired along the required oblique direction.

Answer

The required angle of projection for max. horizontal range is 45°.
$\text{R}_\text{max}=\frac{\text{u}^2}{\text{g}}$
Max. height, $\text{H}=\frac{\text{u}^2\sin^2\theta}{2\text{g}}$
$=\frac{\text{u}^2\sin^245^\circ}{2\text{g}}=\frac{\text{u}^2}{4\text{g}}$
$\frac{\text{R}_\text{max}}{\text{H}}=\frac{\frac{\text{u}^2}{\text{g}}}{\frac{\text{u}^2}{4\text{g}}}=4$
or $\text{R}_\text{max}=4\text{H}$

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Question 132 Marks

Prove that the horizontal range is same when angle of projection is:

Greater than 45° by certain value.
Less than 45° by the same value.

Answer

The horizontal range of the projectile is given by

$\text{R}=\frac{\text{u}^2\sin2\theta}{\text{g}}$

If angle of projection $\theta=45^\circ+\alpha$ and R = R₁

then, $\text{R}_1=\frac{\text{u}^2\sin2(45^\circ+\alpha)}{\text{g}}=\frac{\text{u}^2}{\text{g}}\cos2\alpha\dots(\text{i})$

If angle of projection $\theta=45^\circ-\alpha$ and R = R₂

$\therefore\ \text{R}_2=\frac{\text{u}^2\sin2(45^\circ-\alpha)}{\text{g}}=\frac{\text{u}^2}{\text{g}}\cos2\alpha\dots(\text{ii})$

Comparing eq. (i) and (ii), we have

R₁ = R₂

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Question 142 Marks

Prove that the vectors $(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}})$ and $(2\hat{\text{i}}-\hat{\text{j}})$ are perpendicular to each other.

Answer

$\vec{\text{A}}=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}$
$\vec{\text{B}}=2\hat{\text{i}}-\hat{\text{j}}$
$\vec{\text{A}}$ is perpendicular to $\vec{\text{B}},$
if $\vec{\text{A}}.\vec{\text{B}}=0$
$\therefore\ (\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}).(2\hat{\text{i}}-\hat{\text{j}})=2-2=0$
Thus, $\vec{\text{A}}$ is perpendicular to $\vec{\text{B}}.$

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Question 152 Marks

Find the angle of projection at which horizontal range and maximum height are equal.

Answer

Horizontal range = Maximum height (given)

$\therefore\ \frac{\text{u}^2}{\text{g}}\sin2\theta=\frac{\text{u}^2\sin^2\theta}{2\text{g}}$

$\Rightarrow\ 2\sin\theta\cos\theta=\frac{\sin^2\theta}{2}$ $[\therefore\ \sin2\theta=2\cos\theta\sin\theta]$

$\tan\theta=4$

$\Rightarrow\ \theta=75^\circ58'$

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Question 162 Marks

Determine that vector which when added to the resultant of $\text{A}=3\hat{\text{i}}-5\hat{\text{j}}+7\hat{\text{k}}$ and $\text{B}=2\hat{\text{i}}+4\hat{\text{j}}-3\hat{\text{k}}$ gives unit vector along y-direction.

Answer

We are given, $\text{A}=3\hat{\text{i}}-5\hat{\text{j}}+7\hat{\text{k}}$ and $\text{B}=2\hat{\text{i}}+4\hat{\text{j}}-3\hat{\text{k}}$
Thus, the resultant vector is given by,
$\text{R}=\text{A}+\text{B}=(3\hat{\text{i}}-5\hat{\text{j}}+7\hat{\text{k}})+(2\hat{\text{i}}+4\hat{\text{j}}-3\hat{\text{k}})$
$=5\hat{\text{i}}-\hat{\text{j}}+4\hat{\text{k}}$
But the unit vector along y-direction $=\hat{\text{j}}$
$\therefore$ Required vector $=\hat{\text{j}}-(5\hat{\text{i}}-\hat{\text{j}}+4\hat{\text{k}})$
$=-5\hat{\text{i}}+2\hat{\text{j}}-4\hat{\text{k}}$

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Question 172 Marks

Water from a sprinkler comes out with a constant velocity u in all the directions. What is the maximum area of the grass-land that can be watered at any time?

Answer

Water can go to a maximum distance equal to maximum range (i.e.), $\frac{\text{u}^2}{\text{g}}.$
$\therefore$ Area that can be watered $=\pi\text{r}^2$
$=\pi\Big(\frac{\text{u}^2}{\text{g}}\Big)^2=\frac{\pi\text{u}^4}{\text{g}^2}$

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Question 182 Marks

Two forces 5kg-wt. and 10kg-wt. are acting with an inclination of 120° between them. Find the angle when the resultant makes with 10kg-wt.

Answer

Given, A = 5kg-wt, B = 10kg-wt, $\theta=120^\circ$ then $\beta=?$
$\tan\beta=\frac{\text{B}\sin\theta}{\text{A}+\text{B}\cos\theta}=\frac{10\sin120^\circ}{5+10\cos120^\circ}$
$=\frac{5\sin60^\circ}{10-5\cos60^\circ}=\frac{5\times\frac{\sqrt{3}}{2}}{10-\frac{5}{2}}=\frac{1}{\sqrt{3}}=\tan30^\circ$
$\therefore\ \beta=30^\circ$

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Question 192 Marks

What is the speed of an aircraft if the pilot remains in contact with the seat, even while looping in vertical plane?

Answer

For the pilot to stay at rest, centripetal force should be provided by his weight, i.e., $\frac{\text{mv}^2}{\text{r}}=\text{mg}$ $\therefore\ \text{v}=\sqrt{\text{rg}}$

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Question 202 Marks

What is the distance travelled by a point during the time, if it moves in x - y plane, according to the relation $\text{x}=\text{a}\sin\omega\text{t}$ and $\text{y}=\text{a}(1-\cos\omega\text{t})?$

Answer

The motion will be Simple Harmonic with amplitude of $\sqrt{\text{a}^2+\text{a}^2}=\text{a}\sqrt{2}$ if in the same direction.

Because directions are perpendicular to X and Y, it may be interpreted for a circular motion with radius a.

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Question 212 Marks

Two equal forces have their resultant equal to either. What is the inclination between them?

Answer

Here, A = F; B = F, R = F

$\text{R}=\sqrt{\text{A}^2+\text{B}^2+2\text{AB}\cos\theta}$

$\text{F}=\sqrt{\text{F}^2+\text{F}^2+2\text{F}.\text{F}\cos\theta}$

$\text{F}=\text{F}\sqrt{2(1+\cos\theta)}$

$1=2(1+\cos\theta)$

$\cos\theta=-\frac{1}{2}=\cos120^\circ$

$\theta=120^\circ$

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Question 222 Marks

The angle between vector $\vec{\text{A}}$ and $\vec{\text{B}}$ is 60°. What is the ratio of $\vec{\text{A}}.\vec{\text{B}}$ and $|\vec{\text{A}}\times\vec{\text{B}}|.$

Answer

$\vec{\text{A}}.\vec{\text{B}}=\text{AB}\cos\theta$
$|\vec{\text{A}}\times\vec{\text{B}}|=\text{AB}\sin\theta$
$\therefore\ \frac{\vec{\text{A}}.\vec{\text{B}}}{|\vec{\text{A}}\times\vec{\text{B}}|}=\cot\theta$
$=\cot60^\circ=\frac{1}{\sqrt{3}}$

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Question 232 Marks

A body is projected with a speed v at an angle $\theta$ with horizontal to have maximum range. What is the velocity at the highest point?

Answer

For maximum range $\theta=45^\circ,$ velocity at the highest point $=\text{v}\cos45^\circ=\frac{\text{v}}{\sqrt{2}}.$

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Question 242 Marks

Can there be a two dimensional motion even though acceleration is present in only one direction?

Answer

Yes, the motion of projectile is an example for the same.

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Question 252 Marks

Instantaneous velocity and tangential acceleration of two particles moving in circular paths of radius r are shown in figure I and II. Which of the particles is speeding up and which one is slowing down?

Answer

Particle moving in a circular path I is speeding up because the direction of both $\vec{\text{v}}$ and $\vec{\text{a}}_\text{t}$ are same. Particle moving in a circtrlar path II is slowing down because the directions of both $\vec{\text{v}}$ and $\vec{\text{a}}_\text{t}$ are opposite to each other.

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Question 262 Marks

Two cars are going in two concentric circular orbits of radius r₁ and r₂ with angular velocities $\omega_1$ and $\omega_2.$ What is the ratio of their linear velocities?

Answer

We know, $\text{v}=\text{r}\omega,\ \therefore\ \frac{\text{v}_1}{\text{v}_2}=\frac{\text{r}_1\omega_1}{\text{r}_2\omega_2}.$

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Question 272 Marks

Can there be two vectors where the resultant is equal to either of them?

Answer

Yes, when two vectors of same magnitude inclined at an angle of 120°, then resultant is equal to either of them.
Let x be the magnitude of each of two vectors which make an angle of 120°, then resultant is equal to either of them.
Now, $\text{R}=\sqrt{\text{x}^2+\text{x}^2+2\text{x.x}\cos120^\circ}$
$\Rightarrow\ \text{R}=\sqrt{2\text{x}^2-\text{x}^2}=\sqrt{\text{x}^2}=\text{x}$ $\Big[\because\ \cos120^\circ=-\frac{1}{2}\Big]$

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Question 282 Marks

The angle between vectors A and B is 60°. What is the ratio of A.B and |A × B|?

Answer

The dot product, $\text{A}.\text{B}=\text{AB}\cos\theta$ cross product $|\text{A}\times\text{B}|=\text{AB}\sin\theta.$

$\therefore$ Ratio is $\frac{\text{A.B}}{|\text{A}\times\text{B}|}=\frac{\text{AB}\cos\theta}{\text{AB}\sin\theta}=\cot\theta$

$=\cot60^\circ=\frac{1}{\sqrt{3}}$

As, $\theta=60^\circ,\ \cot60^\circ=\frac{1}{\sqrt{3}}$

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Question 292 Marks

A cyclist has to bend a little inwards from his vertical position while turning. Why?

Answer

By bending, a component of normal reaction of the ground is spared to provide him the necessary centripetal force for turning.

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Question 302 Marks

Two forces whose magnitudes are in the ratio 3 : 5 give a resultant of 28N. If the angle of their inclination is 60^o. Find the magnitude of each force.

Answer

Let A and B be the two forces.
Then, A = 3x, B = 5x, R = 28N and $\theta=60^\circ$
Thus, $\frac{\text{A}}{\text{B}}=\frac{3}{5}$
Now, $\text{R}=\sqrt{\text{A}^2+\text{B}^2+2\text{AB}\cos\theta}$
$\Rightarrow\ 28=\sqrt{9\text{x}^2+25\text{x}^2+30\text{x}^2\cos60^\circ}=7\text{x}$
$\Rightarrow\ \text{x}=4$
$\therefore$ Forces are A = 12N and B = 20N.

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Question 312 Marks

What is the angular acceleration of a particle moving in a circle of radius 'r' with a angular speed $'\omega'?$

Answer

Since $\omega$ is constant, angular acceleration will be zero.

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Question 322 Marks

Calculate the area of a parallelogram whose adjacent sides are given by the vectors.
$\vec{\text{A}}=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}};\ \vec{\text{B}}=2\hat{\text{i}}-3\hat{\text{j}}+\hat{\text{k}}.$

Answer

Area of the parallelogram $=|\vec{\text{A}}\times\vec{\text{B}}|$
$\vec{\text{A}}\times\vec{\text{B}}=\begin{vmatrix} \hat{\text{i}}& \hat{\text{j}}&\hat{\text{k}}\\1 & 2&3\\2&-3&1 \end{vmatrix}$
$=11\hat{\text{i}}+5\hat{\text{j}}-7\hat{\text{k}}$
$\therefore\ |\vec{\text{A}}\times\vec{\text{B}}|=\sqrt{121+25+49}$
$=\sqrt{195}\text{ sq. units}$

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Question 332 Marks

Explain the property of two vectors A and B if |A + B| = |A - B|.

Answer

As we know that, $|\text{A}+\text{B}|=\sqrt{\text{A}^2+\text{B}^2+2\text{AB}\cos\theta}$
and $|\text{A}-\text{B}|=\sqrt{\text{A}^2+\text{B}^2-2\text{AB}\cos\theta}$
But as per question, we have
$\sqrt{\text{A}^2+\text{B}^2+2\text{AB}\cos\theta}=\sqrt{\text{A}^2+\text{B}^2-2\text{AB}\cos\theta}$
Squaring both sides, we have $(4\text{ AB}\cos\theta)=0$
$\Rightarrow\ \cos\theta=0\ \text{or }\theta=90^\circ$
Hence, the two vectors A and B are perpendicular to each other.

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Question 342 Marks

Find the angle of projection for a projectile motion whose range R is n times the maximum height H.

Answer

Given R = nH
$\Rightarrow\ \frac{\text{u}^2\sin2\theta}{\text{g}}=\text{n}\times\frac{\text{u}^2\sin^2\theta}{2\text{g}}$
$\Rightarrow\ \theta=\tan^{-1}\Big(\frac{4}{\text{n}}\Big)$

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Question 352 Marks

Find the scalar product of two vectors $\vec{\text{a}}=(3\hat{\text{i}}-4\hat{\text{j}}+5\hat{\text{k}})$ and $\vec{\text{b}}=(-2\hat{\text{i}}+\hat{\text{j}}-3\hat{\text{k}}).$

Answer

$\vec{\text{a}}=(3\hat{\text{i}}-4\hat{\text{j}}+5\hat{\text{k}})$
$\vec{\text{b}}=(-2\hat{\text{i}}+\hat{\text{j}}-3\hat{\text{k}})$
$\vec{\text{a}}.\vec{\text{b}}=3(-2)-4(1)+5(-3)$
$=-6-4-15=-25$

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Question 362 Marks

The sum and difference of two vectors are perpendicular to each other. Prove that the ectors are equal in magnitude.

Answer

As the vectors A + B and A - B are perpendicular to each other, therefore

$(\text{A}+\text{B}).(\text{A}-\text{B})=0$

$\text{A}.\text{A}-\text{A}.\text{B}+\text{B}.\text{A}-\text{B}.\text{B}=0$

$\text{A}-\text{B}=0\ \ [\because\ \text{A}.\text{B}=\text{B}.\text{A}]$

$\Rightarrow\ \text{A}=\text{B}$

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Question 372 Marks

What is meant by resolving a force into rectangular components? Resolve a force of 10N into two components, if it acts at an angle 30° with the horizontal.

Answer

Let F be the force at 30° to horizontal.
The horizontal component is,
$\text{F}\cos30^\circ=\frac{\sqrt{3}\text{F}}{2}$
The vertical component is, $\text{F}\sin30^\circ=\frac{\text{F}}{2}$

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Question 382 Marks

If two vectors of equal magnitude add to either of them by magnitude, what is the angle between them?

Answer

$\sqrt{\text{p}^2+\text{p}^2+2\text{p}^2\cos\theta}=\text{p}$
Squaring, $2\text{p}^2(1+\cos\theta)=\text{p}^2$
$1+\cos\theta=\frac{1}{2}\cos\theta=\frac{-1}{2},$
$\theta=120^\circ$

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Question 392 Marks

Determine $\lambda$ such that:
$\vec{\text{A}}=2\hat{\text{i}}+\lambda\hat{\text{j}}+\hat{\text{k}};\ \vec{\text{B}}=4\hat{\text{i}}-2\hat{\text{j}}-2\hat{\text{k}}$ are perpendicular to each other.

Answer

$\vec{\text{A}}=2\hat{\text{i}}+\lambda\hat{\text{j}}+\hat{\text{k}}$
$\vec{\text{B}}=4\hat{\text{i}}-2\hat{\text{j}}-2\hat{\text{k}}$
$\vec{\text{A}}.\vec{\text{B}}=(2\hat{\text{i}}+\lambda\hat{\text{j}}+\hat{\text{k}}).(4\hat{\text{i}}-2\hat{\text{j}}-2\hat{\text{k}})$
$=8-2\lambda-2$
$0=6-2\lambda$ or $\lambda=3$

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Question 402 Marks

What is the projection of $\hat{\text{i}}+\hat{\text{j}}$ on $(\hat{\text{i}}-\hat{\text{j}})?$

Answer

Let $\vec{\text{A}}=\hat{\text{i}}+\hat{\text{j}}$ and $\vec{\text{B}}=\hat{\text{i}}-\hat{\text{j}}$
Projection of
$\vec{\text{A}}$ on $\vec{\text{B}}=|\vec{\text{A}}||\vec{\text{B}}|\cos\theta$
$\cos\theta=\frac{\vec{\text{A}}.\vec{\text{B}}}{|\vec{\text{A}}||\vec{\text{B}}|}=0$ $(\theta=90^\circ)$
$\theta=90^\circ,$
$\therefore$ Projection = 0.

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Question 412 Marks

Can a flight of a bird, an example of composition of vectors. Why?

Answer

Yes, the flight of a bird is an example of composition of vectors. As the bird flies, it strikes the air with its wings W, W along WO. According to Newton's third law of motion, air strikes the wings in opposite directions with the same force in reaction. The reactions are OA and OB. From law of parallelogram vectors, OC is the resultant of OA and OB. This resultant upwards force OC is responsible for the flight of the bird.

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Question 422 Marks

A string can withstand a tension of 25N. What is the greatest speed at which a body of mass 1kg can be whirled in a horizontal circle using a 1m length of the string?

Answer

As the body is whirled in a horizontal circle, the force of gravity, acting vertically downwards, has no effect on its motion.
If v is the greatest speed with which the body can be whirled, the centripetal forces in the string must balance the maximum tension that the string can withstand.
$\therefore\ \text{Tension, T}=\frac{\text{mv}^2}{\text{r}}$ or $25=\frac{1\times\text{v}^2}{1}$
or, $\text{v}=5\text{ ms}^{-1}$

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Question 432 Marks

Derive $\text{x}_2=\text{x}_1+\text{v}_1(\text{t}_2-\text{t}_1)+\frac{1}{2}\text{a}(\text{t}_2-\text{t}_1)^2.$

Answer

Consider a body starting from x₁ and reaching x₂ at time t₂. Starting with a velocity v₁ at time t₁ with a uniform acceleration ‘a'. The motion can be described by the v-t graph as shown.

Area below v-t graph is PMWYNP

$=\text{WM}\times\text{WY}\times\frac{1}{2}\text{WY}\times\text{NP}$

$=\text{v}_1(\text{t}_2-\text{t}_1)+\frac{1}{2}(\text{t}_2-\text{t}_1)\times(\text{v}_2-\text{v}_1)$

Since area below v-t graph is displacement and v₂ - v₁ = a(t₂ - t₁), we get

$\text{x}_2-\text{x}_1=\text{v}_1(\text{t}_2-\text{t}_1)+\frac{1}{2}(\text{t}_2-\text{t}_1)\times\text{a}(\text{t}_2-\text{t}_1)$

$\therefore\ \text{x}_2=\text{x}_1+\text{v}_1(\text{t}_2-\text{t}_1)+\frac{1}{2}\text{a}(\text{t}_2-\text{t}_1)^2$

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Question 442 Marks

What are co-initial and collinear vectors?

Answer

Two vectors having the same initial point are called co-initial vectors. Two vectors which either act along the same line or along parallel lines are called collinear vectors.

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Question 452 Marks

A skilled gun man always keeps his gun slightly tilted above the line of sight while shooting. Why?

Answer

When a bullet is fired from a gun with its barrel directed towards the target, it starts falling downwards on account of acceleration due to gravity.

Due to which the bullet hits below the target. Just to avoid it, the barrel of the gun is lined up little above the target, so that the bullet after travelling in parabolic path hits the distant target.

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Question 462 Marks

A football is kicked 20m/ s at a projection angle of 45°. A receiver on the goal line 25m away in the direction of the kick runs the same instant to meet the ball. What must be his speed, if he has to catch the ball before it hits the ground?

Answer

Given, u = 20m/ s, $\theta=45^\circ,$ d = 25m

Horizontal range is given by

$\text{R}=\frac{\text{u}^2}{\text{g}}\sin2\theta=\frac{(20)^2}{9.8}\sin2(45^\circ)$

$=\frac{400}{9.8}\times1=40.82\text{m}$

Time of flight, $\text{T}=\frac{2\text{u}\sin\theta}{\text{g}}=\frac{2\times20}{9.8}\sin45^\circ$

$=2.886\text{s}$

The goal man is 25m away in the direction of the ball, so to catch the ball, he is to cover a distance

= 40.82 - 25 = 15.82m in time 2.886s.

$\therefore$ Velocity of the goal man to catch the ball

$\text{v}=\frac{15.82}{2.886}=5.48\text{m/ s}$

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Question 472 Marks

We can order events in time and there is no sense of time, distinguishing past, present and future. Is time a vector?

Answer

We know that time always flows on and on i.e. from past to present and then to future.
Therefore, a direction can be assigned to time. Since, the direction of time is unique and it is unspecified or unstated. That is why, time cannot be a vector though it has a direction.

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Question 482 Marks

Pick out the only vector quantity in the following list:
Temperature, pressure, impulse, time, power, total path length, energy, gravitational potential, coefficient of friction, charge.

Answer

Since, Impulse = change in momentum = force × time. As momentum and force are vector quantities, hence impulse is a vector quantity.

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Question 492 Marks

Keeping the angle of projection same, what is the effect on horizontal range of a particle when its velocity is doubled?

Answer

$\text{R}=\frac{\text{v}_0^2\sin2\theta}{\text{g}},\text{R }\alpha\text{ v}_0^2$
On doubling v₀, R becomes 4 times.

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Question 502 Marks

Two bombs of 20kg and 30kg are thrown from a cannon with the same velocity in the same direction. Which bomb will reach the ground first?

Answer

Time of flight $\text{T}=\frac{2\text{u}\sin\theta}{\text{g}}.$ Since time of flight does not depend upon mass, both will reach simultaneously.

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