4.12 Exercise Problems

Type: Short Questions 

S1: Is a vector necessarily changed if it is rotated through an angle?

Answer: Yes. A vector is defined by its magnitude and direction, so a vector can be changed by changing its magnitude and direction. If we rotate it through an angle, its direction changes and we can say that the vector has changed.

S2: Is it possible to add two vectors of unequal magnitudes and get zero? Is it possible to add three, vectors of equal magnitudes and get zero?

Answer: No, it is not possible to obtain zero by adding two vectors of unequal magnitudes.
Example: Let us add two vectors \(\vec{A}\) and \(\vec{B}\) of unequal magnitudes acting in opposite directions. The resultant vector is given by
\(
R=\sqrt{A^2+B^2+2 A B \cos \theta}
\)
If two vectors are exactly opposite to each other, then
\(
\begin{aligned}
& \theta=180^{\circ}, \cos 180^{\circ}=-1 \\
& R=\sqrt{A^2+B^2-2 A B} \\
& \Rightarrow R=\sqrt{(A-B)^2} \\
& \Rightarrow R=(A-B) \text { or }(B-A)
\end{aligned}
\)
From the above equation, we can say that the resultant vector is zero \((R=0)\) when the magnitudes of the vectors \(\vec{A}\) and \(\vec{B}\) are equal \((A=B)\) and both are acting in the opposite directions.
Yes, it is possible to add three vectors of equal magnitudes and get zero. \(\vec{A}, \vec{B} \text { and } \vec{C} \text {, given these three vectors make an angle of } 120^{\circ} \text { with each other. }\)

\(
\begin{aligned}
& A_x=A \\
& A_y=0 \\
& B_x=-B \cos 60^{\circ} \\
& B_y=B \sin 60^{\circ} \\
& C_x=-C \cos 60^{\circ} \\
& C_y=-C \sin 60^{\circ} \\
& \text { Here, } \mathrm{A}=\mathrm{B}=\mathrm{C}
\end{aligned}
\)
Here, \(A=B=C\)
So, along the \(x\) – axis, we have:
\(
\begin{aligned}
& A-\left(2 A \cos 60^{\circ}\right)=0, \text { as } \cos 60^{\circ}=\frac{1}{2} \\
& \Rightarrow B \sin 60^{\circ}-C \sin 60^{\circ}=0
\end{aligned}
\)
Hence, proved.

S3:  Does the phrase “direction of zero vector” have physical significance? Discuss in terms of velocity, force, etc.

Answer: A zero vector has physical significance in physics, as the operations on the zero vector gives us a vector.
For any vector \(\vec{A}\), assume that
\(
\vec{A}+\overrightarrow{0}=\vec{A}
\)
\(
\vec{A}-\overrightarrow{0}=\vec{A}
\)
\(
\vec{A} \times \overrightarrow{0}=\overrightarrow{0}
\)
Again, for any real number \(\lambda\) we have:
\(
\lambda \overrightarrow{0}=\overrightarrow{0}
\)
The significance of a zero vector can be better understood through the following examples:
The displacement vector of a stationary body for a time interval is a zero vector.
Similarly, the velocity vector of the stationary body is a zero vector.
When a ball, thrown upward from the ground, falls to the ground, the displacement vector is a zero vector, which defines the displacement of the ball.

S4: Can you add three unit vectors to get a unit vector? Does your answer change if two unit vectors are along the coordinate axes?

Answer: Yes, we can add three unit vectors to get a unit vector.
No, the answer does not change if two unit vectors are along the coordinate axes. Assume three unit vectors \(\hat{i},-\hat{i}\) and \(\hat{j}\) along the positive \(x\)-axis, negative \(x\)-axis and positive \(y\)-axis, respectively. Consider the figure given below:

The magnitudes of the three unit vectors \((\hat{i},-\hat{i}\) and \(\hat{j})\) are the same, but their directions are different. So, the resultant of \(\hat{i}\) and \(-\hat{i}\) is a zero vector.
Now, \(\hat{j}+\overrightarrow{0}=\hat{j}\) (Using the property of zero vector)
\(\therefore\) The resultant of three unit vectors \((\hat{i},-\hat{i}\) and \(\hat{j})\) is a unit vector \((\hat{j})\).

S5: Can we have physical quantities having magnitude and direction which are not vectors?

Answer: Yes, there are physical quantities like electric current and pressure which have magnitudes and directions, but are not considered as vectors because they do not follow vector laws of addition.

S6: Which of the following two statements is more appropriate?
(a) Two forces are added using the triangle rule because force is a vector quantity.
(b) Force is a vector quantity because two forces are added using the triangle rule.

Answer: Two forces are added using the triangle rule because force is a vector quantity. This statement is more appropriate, because we know that force is a vector quantity and only vectors are added using the triangle rule.

S7: Can you add two vectors representing physical quantities having different dimensions? Can you multiply two vectors representing physical quantities having different dimensions?

Answer: No, we cannot add two vectors representing physical quantities of different dimensions. However, we can multiply two vectors representing physical quantities with different dimensions.
Example: Torque,
\(
\vec{\tau}=\vec{r} \times \vec{F}
\)

S8: Can a vector have zero components along a line and still have nonzero magnitude?

Answer: Yes, a vector can have zero components along a line and still have a nonzero magnitude.
Example: Consider a two-dimensional vector \(2 \hat{i}+0 \hat{j}\). This vector has zero components along a line lying along the \(Y\)-axis and a nonzero component along the \(\mathrm{X}\)-axis. The magnitude of the vector is also nonzero.
Now, magnitude of \(2 \hat{i}+0 \hat{j}=\sqrt{2^2+0^2}=2\)

S9: Let \(\varepsilon_1\) and \(\varepsilon_2\) be the angles made by \(\vec{A}\) and \(-\vec{A}\) with the positive \(X\)-axis. Show that \(\tan \varepsilon_1=\tan \varepsilon_2\). Thus, giving \(\tan \varepsilon\) does not uniquely determine the direction of \(\vec{A}\).

Answer: The direction of \(-\vec{A}\) is opposite to \(\vec{A}\).So, if vector \(\vec{A}\) and \(-\vec{A}\) make the angles \(\varepsilon_1\) and \(\varepsilon_2\) with the X-axis, respectively, then \(\varepsilon_1\) is equal to \(\varepsilon_2\) as shown in the figure:

Here, \(\tan \varepsilon_1=\tan \varepsilon_2\)
Because these are alternate angles.
Thus, giving \(\tan \varepsilon\) does not uniquely determine the direction of \(-\vec{A}\).

S10: Is the vector sum of the unit vectors \(\vec{i}\) and \(\vec{j}\) a unit vector? If no, can you multiply this sum by a scalar number to get a unit vector?

Answer: No, the vector sum of the unit vectors \(\vec{i}\) and \(\vec{i}\) is not a unit vector, because the magnitude of the resultant of \(\vec{i}\) and \(\vec{j}\) is not one.
The magnitude of the resultant vector is given by
\(
R=\sqrt{1^2+1^2+\cos 90^{\circ}}=\sqrt{2}
\)
Yes, we can multiply this resultant vector by a scalar number \(\frac{1}{\sqrt{2}}\) to get a unit vector.

S11: Let \(\vec{A}=3 \vec{i}+4 \vec{j}\). Write four-vector \(\vec{B}\) such that \(\vec{A} \neq \vec{B}\) but \(A=B\)

Answer: A vector \(\vec{B}\) such that \(\vec{A} \neq \vec{B}\), but \(A=B\) are as follows:
(i) \(\vec{B}=3 \vec{i}-4 \vec{j}\)
(ii) \(\vec{B}=3 \vec{j}+4 \vec{k}\)
(iii) \(\vec{B}=3 \vec{k}+4 \vec{i}\)
(i v) \(\vec{B}=3 \vec{j}-4 \vec{k}\)

S12: Can you have \(\vec{A} \times \vec{B}=\vec{A} \cdot \vec{B}\) with \(A \neq 0\) and \(B \neq 0\) ? What if one of the two vectors is zero?

Answer: No, we cannot have \(\vec{A} \times \vec{B}=\vec{A} \cdot \vec{B}\) with \(\mathrm{A} \neq 0\) and \(\mathrm{B} \neq 0\). This is because the left-hand side of the given equation gives a vector quantity, while the right-hand side gives a scalar quantity. However, if one of the two vectors is zero, then both sides will be equal to zero and the relation will be valid.

S13: \(\text { If } \vec{A} \times \vec{B}=0 \text {, can you say that (a) } \vec{A}=\vec{B} \text {, (b) } \vec{A} \neq \vec{B} \text { ? }\)

Answer: If \(\vec{A} \times \vec{B}=0\), then both the vectors are either parallel or antiparallel, i.e., the angle between the vectors is either \(0^{\circ}\) or \(180^{\circ}\). \(\vec{A} \vec{B} \sin \theta \hat{n}=0 \ldots \ldots\left(\because \sin 0^{\circ}=\sin 180^{\circ}=0\right)\)
Both the conditions can be satisfied:
(a) \(\vec{A}=\vec{B}\), i.e., the two vectors are equal in magnitude and parallel to each other
(b) \(\vec{A} \neq \vec{B}\), i.e., the two vectors are unequal in magnitude and parallel or anti-parallel to each other.

S14: Let \(\vec{A}=5 \vec{i}-4 \vec{j}\) and \(\vec{B}=-7 \cdot 5 \vec{i}+6 \vec{j}\). Do we have \(\vec{B}=k \vec{A}\) ? Can we say \(\frac{\vec{B}}{\vec{A}}=k\) ?

Answer: If \(\vec{A}=5 \vec{i}-4 \vec{j}\) and \(\vec{B}=-7 \cdot 5 \vec{i}+6 \vec{j}\), then we have \(\vec{B}=k \vec{A}\) by putting the value of scalar \(\mathrm{k}\) as -1.5 .
However, we cannot say that \(\frac{\vec{B}}{\vec{A}}=k\), because a vector cannot be divided by other vectors, as vector division is not possible.

S15: Refer to figure (2-E1). Find (a) the magnitude, (b) \(x\) and \(y\) components and (c) the angle with the \(X\)-axis of the resultant of \(\overrightarrow{O A}, \overrightarrow{B C}\) and \(\overrightarrow{D E}\).

Answer: 

\(
\begin{aligned}
& \text { x component of } \overrightarrow{\mathrm{OA}}=2 \cos 30^{\circ}=\sqrt{3} \\
& \text { x component of } \overrightarrow{\mathrm{BC}}=1.5 \cos 120^{\circ}=-0.75 \\
& \text { x component of } \overrightarrow{\mathrm{DE}}=1 \cos 270^{\circ}=0 \\
& \text { y component of } \overrightarrow{\mathrm{OA}}=2 \sin 30^{\circ}=1 \\
& \text { y component of } \overrightarrow{\mathrm{BC}}=1.5 \sin 120^{\circ}=1.3 \\
& \text { y component of } \overrightarrow{\mathrm{DE}}=1 \sin 270^{\circ}=-1
\end{aligned}
\)
\(
\begin{aligned}
& \mathrm{R}_{\mathrm{x}}=\mathrm{x} \text { component of resultant }=\sqrt{3}-0.75+0=0.98 \mathrm{~m} \\
& \mathrm{R}_{\mathrm{y}}=\text { resultant } \mathrm{y} \text { component }=1+1.3-1=1.3 \mathrm{~m} \\
& \text { So, } \mathrm{R}=\text { Resultant }=1.6 \mathrm{~m}
\end{aligned}
\)
If it makes and angle \(\alpha\) with positive \(x\)-axis
\(
\begin{aligned}
& \text { Tan } \alpha=\frac{\text { y component }}{x \text { component }}=1.32 \\
& \Rightarrow \alpha=\tan ^{-1} 1.32
\end{aligned}
\)

S16: Two vectors have magnitudes 3 unit and 4 unit respectively. What should be the angle between them if the magnitude of the resultant is (a) 1 unit, (b) 5 unit and (c) 7 unit.

Answer:
\(
|\vec{a}|=3 m|~ \vec{b}|=4
\)
a) If \(R=1\) unit \(\Rightarrow \sqrt{3^2+4^2+2 \cdot 3 \cdot 4 \cdot \cos \theta}=1\)
\(
\theta=180^{\circ}
\)
b)
\(
\begin{aligned}
& \sqrt{3^2+4^2+2 \cdot 3 \cdot 4 \cdot \cos \theta}=5 \\
& \theta=90^{\circ}
\end{aligned}
\)
c)
\(
\begin{aligned}
& \sqrt{3^2+4^2+2 \cdot 3 \cdot 4 \cdot \cos \theta}=7 \\
& \theta=0^{\circ}
\end{aligned}
\)
The angle between them is \(0^{\circ}\).

S17: A carrom board ( \(4 \mathrm{ft} \times 4 \mathrm{ft}\) square) has the queen at the centre. The queen, hit by the striker moves to the front edge, rebounds and goes in the hole behind the striking line. Find the magnitude of displacement of the queen (a) from the centre to the front edge, (b) from the front edge to the hole and (c) from the centre to the hole.

Answer:

\(
\begin{aligned}
& \text { In } \triangle A B C, \tan \theta=\frac{x}{2} \\
& \text { and in } \triangle D C E, \tan \theta=\frac{2-x}{4} \\
& \tan \theta=\frac{x}{2}=\frac{(2-x)}{4} \\
& \Rightarrow 2(2-x)=4 x \\
& \Rightarrow 4-2 x=4 x \\
& \Rightarrow 6 x=4 \\
& \Rightarrow x=\frac{2}{3} \mathrm{ft} \\
& \text { (a) } \ln \triangle A B C, A C=\sqrt{A B^2+B C^2} \\
& =\sqrt{\left(\frac{2^2}{3}\right)}+2^2 \\
& =\sqrt{\frac{4}{9}+4}=\sqrt{\frac{40}{9}} \\
& =\frac{2}{3} \sqrt{10} \mathrm{ft}
\end{aligned}
\)
(b) In \(\triangle C D E, D E=2-\frac{2}{3}=\frac{6-2}{3}=\frac{4}{3} \mathrm{ft}\) since \(C D=4 \mathrm{ft}\), therefore \(C E\)
\(
\begin{aligned}
& =\sqrt{C D^2+D E^2} \\
& =\sqrt{4^2+\left(\frac{4^2}{3}\right)}=\frac{4}{3} \sqrt{10} \mathrm{ft}
\end{aligned}
\)
(c) In \(\triangle A G E, A E=\sqrt{A G^2+G E^2}\)
\(
\begin{aligned}
& =\sqrt{2^2+2^2} \\
& =\sqrt{8}=2 \sqrt{2} \mathrm{ft}
\end{aligned}
\)

S18: A mosquito net over a \(7 \mathrm{ft} \times 4 \mathrm{ft}\) bed is \(3 \mathrm{ft}\) high. The net has a hole at one corner of the bed through which a mosquito enters the net. It flies and sits at the diagonally opposite upper corner of the net. (a) Find the magnitude of the displacement of the mosquito. (b) Taking the hole as the origin, the length of the bed as the \(X\)-axis, its width as the \(Y\)-axis, and vertically up as the \(Z\)-axis, write the components of the displacement vector.

Answer: 

Displacement vector
\(
\vec{r}=7 \hat{i}+4 \hat{i}+3 \hat{k}
\)
(a) Magnitude of displacement
\(
\begin{aligned}
& =\sqrt{7^2+4^2+3^2} \\
& =\sqrt{74} \mathrm{ft}
\end{aligned}
\)
(b) Components of the displacement vector are \(7 \mathrm{ft}, 4 \mathrm{ft}\) and \(3 \mathrm{ft}\).

S19: Two vectors have magnitudes \(2 \mathrm{~m}\) and \(3 \mathrm{~m}\). The angle between them is \(60^{\circ}\). Find (a) the scalar product of the two vectors, (b) the magnitude of their vector product. 

Answer: Let the two vectors be \(|\vec{a}|=2 m\) and \(|\vec{b}|=3 m\).
Angle between the vectors, \(\theta=60^{\circ}\)
(a) The scalar product of two vectors is given by \(\vec{a} \cdot \vec{b}=|\vec{a}| \cdot|\vec{b}| \cos \theta\)
\(
\begin{aligned}
& \therefore \vec{a} \cdot \vec{b}=|\vec{a}| \cdot|\vec{b}| \cos 60^{\circ} \\
& =2 \times 3 \times \frac{1}{2}=3 m^2
\end{aligned}
\)
(b) The vector product of two vectors is given by \(|\vec{a} \times \vec{b}|=|\vec{a}||\vec{a}| \sin \theta\).
\(
\begin{aligned}
& \therefore|\vec{a} \times \vec{b}|=|\vec{a}||\vec{a}| \sin 60^{\circ} \\
& =2 \times 3 \times \frac{\sqrt{3}}{2} \\
& =3 \sqrt{3} m^2
\end{aligned}
\)

S20: Let \(A_1 A_2 A_3 A_4 A_5 A_6 A_1\) be a regular hexagon. Write the \(x\)-components of the vectors represented by the six sides taken in order. Use the fact that the resultant of these six vectors is zero, to prove that \(\cos 0+\cos \pi / 3+\cos 2 \pi / 3+\cos 3 \pi / 3+\cos 4 \pi / 3+\cos 5 \pi / 3=0\)

Answer: According to the polygon law of vector addition, the resultant of these six vectors is zero.
Here, \(a=b=c=d=e=f\) (magnitudes), as it is a regular hexagon. A regular polygon has all sides equal to each other.
So, \(R_x=A \cos 0+A \cos \frac{\pi}{3}+A \cos \frac{2 \pi}{3}+A \cos \frac{3 \pi}{3}+A \cos \frac{4 \pi}{3}+A \cos \frac{5 \pi}{3}=0\)
[As the resultant is zero, the \(x\)-component of resultant \(R_X\) is zero]
\(
\Rightarrow \cos 0+\cos \frac{\pi}{3}+\cos \frac{2 \pi}{3}+\cos \frac{3 \pi}{3}+\cos \frac{4 \pi}{3}+\cos \frac{5 \pi}{5}=0
\)
Note: Similarly, it can be proven that
\(
\sin 0+\sin \frac{\pi}{3}+\sin \frac{2 \pi}{3}+\sin \frac{3 \pi}{3}+\sin \frac{4 \pi}{3}+\sin \frac{5 \pi}{3}=0
\)

S21: Prove that } \(\vec{A} \cdot(\vec{A} \times \vec{B})=0\)

Answer:

\(
\vec{A} \cdot(\vec{A} \times \vec{B})=0 \text { (claim) }
\)
As, \(\vec{A} \times \vec{B}=A B \sin \theta \hat{n}\)
\(A B \sin \theta\) n̂ is a vector which is perpendicular to the plane containing \(\vec{A}\) and \(\vec{B}\), this implies that it is also perpendicular to \(\vec{A}\). As the dot product of two perpendicular vector is zero.
Thus \(\vec{A} \cdot(\vec{A} \times \vec{B})=0\).

S22: If \(\vec{A}, \vec{B}, \vec{C}\) are mutually perpendicular, show that \(\vec{C} \times(\vec{A} \times \vec{B})=0\). Is the converse true?

Answer:

Given that \(\vec{A}, \vec{B}\) and \(\vec{C}\) are mutually perpendicular \(\vec{A} \times \vec{B}\) is a vector which direction is perpendicular to the plane containing \(\vec{A}\) and \(\vec{B}\).
Also \(\vec{C}\) is perpendicular to \(\vec{A}\) and \(\vec{B}\)
\(\therefore\) Angle between \(\overrightarrow{\mathrm{C}}\) and \(\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}}\) is \(0^{\circ}\) or \(180^{\circ}\) (fig. 1 )
So, \(\vec{C} \times(\vec{A} \times \vec{B})=0\)
The converse is not true.
For example, if two of the vector are parallel, (fig.2), then also
\(
\overrightarrow{\mathrm{C}} \times(\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}})=0
\)
So, they need not be mutually perpendicular.

S23: A particle moves on a given straight line with a constant speed \(v\). At a certain time it is at a point \(P\) on its straight line path. \(O\) is a fixed point. Show that \(\overrightarrow{O P} \times \vec{v}\) is independent of the position \(P\).

Answer:

The particle moves on the straight line PP’ at speed \(v\). From the figure,
\(
\overrightarrow{O P} \times v=(O P) v \sin \theta \hat{n}=v(O P) \sin \theta \hat{n}=v(O Q) \hat{n}
\)
It can be seen from the figure, \(O Q=O P \sin \theta=O P^{\prime} \sin \theta\)
So, whatever may be the position of the particle, the magnitude and direction of \(\overrightarrow{O P} \times \vec{V}\) remain constant.
\(\therefore \overrightarrow{\mathrm{OP}} \times \overrightarrow{\mathrm{V}}\) is independent of the position \(\mathrm{P}\).

S24: The force on a charged particle due to electric and magnetic fields is given by \(\vec{F}=q \vec{E}+q \vec{v} \times \vec{B}\). Suppose \(\vec{E}\) is along the \(X\)-axis and \(\vec{B}\) along the \(Y\)-axis. In what direction and with what minimum speed \(v\) should a positively charged particle be sent so that the net force on it is zero?

Answer: 

According to the problem, the net electric and magnetic forces on the particle should be zero.
\(
\begin{aligned}
& \vec{F}=q \vec{E}+q(\vec{V} \times \vec{B})=0 \\
& \Rightarrow E=-(\vec{V} \times \vec{B})
\end{aligned}
\)
So, the direction of \(\vec{V} \times \vec{B}\) should be opposite to the direction of \(\vec{E}\) Hence, \(\vec{V}\) should be along the positive \(z\)-direction.
Again, \(E=V B \sin \theta\)
\(
\Rightarrow V=\frac{E}{B} \sin \theta
\)
For \(V\) to be minimum, \(\theta=90\) and , thus, \(V_{\min }=\frac{E}{B}\)
So, the particle must be projected at a minimum speed of \(\frac{E}{B}\) along the \(z\)-axis.
So, the particle must be projected at a minimum speed of \(E / B\) along \(+v e z\)-axis \(\left(\theta=90^{\circ}\right)\) as shown in the figure, so that the force is zero.

S25: Give an example for which \(\vec{A} \cdot \vec{B}=\vec{C} \cdot \vec{B}\) but \(\vec{A} \neq \vec{C}\).

Answer:

For example, as shown in the figure,
\(\overrightarrow{\mathrm{A}} \perp \overrightarrow{\mathrm{B}} \quad \overrightarrow{\mathrm{B}}\) along west
\(\overrightarrow{\mathrm{B}} \perp \overrightarrow{\mathrm{C}} \quad \overrightarrow{\mathrm{A}}\) along south
\(\vec{C}\) along north
\(\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}=0 \quad \therefore \overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}=\overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{C}}\)
\(\overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{C}}=0 \quad\) But \(\overrightarrow{\mathrm{B}} \neq \overrightarrow{\mathrm{C}}\)

S26: Draw a graph from the following data. Draw tangents at \(x=2,4,6\) and 8. Find the slopes of these tangents. Verify that the curve drawn is \(y=2 x^2\) and the slope of the tangent is \(\tan \theta=\frac{d y}{d x}=4 x\).
\(
\begin{array}{ccccccccccc}
x & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\
y & 2 & 8 & 18 & 32 & 50 & 72 & 98 & 128 & 162 & 200
\end{array}
\)

Answer: The graph \(y=2 x^2\) should be drawn by the student on a graph paper for exact results.

To find the slope at any point, draw a tangent at the point and extend the line to meet \(x\)-axis. Then find \(\tan \theta\) as shown in the figure.
It can be checked that,
Slope \(=\tan \theta=\frac{d y}{d x}=\frac{d}{d x}\left(2 x^2\right)=4 x\)
Where \(x=\) the \(x\)-coordinate of the point where the slope is to be measured.

S27: A curve is represented by \(y=\sin x\). If \(x\) is changed from \(\frac{\pi}{3}\) to \(\frac{\pi}{3}+\frac{\pi}{100}\), find approximately the change in \(y\).

Answer: 

\(
\begin{aligned}
& y=\sin x \\
& \text { So, } y+\Delta y=\sin (x+\Delta x) \\
& \Delta y=\sin (x+\Delta x)-\sin x \\
& =\left(\frac{\pi}{3}+\frac{\pi}{100}\right)-\sin \frac{\pi}{3}=0.0157
\end{aligned}
\)

S28: The electric current in a charging \(R-C\) circuit is given by \(i=i_0 e^{-t / R C}\) where \(i_0, R\) and \(C\) are constant parameters of the circuit and \(t\) is time. Find the rate of change of current at (a) \(t=0\), (b) \(t=R C\), (c) \(t=10 R C\).

Answer: Given: \(i=i_0 e^{-t / R C}\)
\(\therefore\) Rate of change of current
\(
\begin{aligned}
& =\frac{d i}{d t} \\
& =i_0\left(\frac{-1}{R C}\right) e^{-t / R C} \\
& =\frac{-i_0}{R C} \times e^{-t / R C}
\end{aligned}
\)
On applying the conditions given in the questions, we get:
(a) At \(t=0, \frac{d i}{d t}=\frac{-i_0}{R C} \times e^0=\frac{-i_0}{R C}\)
(b) At \(t=R C, \frac{d i}{d t}=\frac{-i_0}{R C} \times e^{-1}=\frac{-i_0}{R C e}\)
(c) At \(t=10 R C, \frac{d t}{d i}=\frac{-i_0}{R C} \times e^{10}=\frac{-i_0}{R C e^{10}}\)

S29: The electric current in a discharging \(\mathrm{R}-\mathrm{C}\) circuit is given by \(i=i_0 e^{-t / R C}\) where \(i_0, R\) and \(C\) are constant parameters and \(t\) is time. Let \(i_0=2.00 \mathrm{~A}, \quad R=6.00 \times 10^5 \Omega\) and \(C=0.500 \mu \mathrm{F}\). (a) Find the current at \(t=0.3 \mathrm{~s}\).
(b) Find the rate of change of current at \(t=0.3 \mathrm{~s}\).
(c) Find approximately the current at \(t=0.31 \mathrm{~s}\).

Answer: Electric current in a discharging R-C circuit is given by the below equation:
\(
\mathrm{i}=\mathrm{i}_0 \cdot \mathrm{e}^{-\mathrm{t} / \mathrm{RC}} \dots(i)
\)
Here, \(\mathrm{i}_0=2.00 \mathrm{~A}\)
\(
\begin{aligned}
\mathrm{R} & =6 \times 10^5 \Omega \\
C & =0.0500 \times 10^{-6} \mathrm{~F} \\
& =5 \times 10^{-7} \mathrm{~F}
\end{aligned}
\)
On substituting the values of \(R, C\) and \(i_0\) in equation (i), we get:
\(
\mathrm{i}=2.0 \mathrm{e}^{-\mathrm{t} / 0.3} \quad \ldots \text { (ii) }
\)
According to the question, we have:
(a) current at \(\mathrm{t}=0.3 \mathrm{~s}\)
\(
i=2 \times e^{-1}=\frac{2}{e} A
\)
(b) rate of change of current at \(t=0.3 \mathrm{~s}\)
\(
\frac{d i}{d t}=\frac{-i_0}{R C} \cdot e^{-t / R C}
\)
When \(\mathrm{t}=0.3 \mathrm{~s}\), we have:
\(
\frac{d i}{d t}=\frac{2}{0.30} \cdot e^{\left(\frac{-0.3}{0.3}\right)}=\frac{-20}{3 e} A / s
\)
(c) approximate current at \(\mathrm{t}=0.31 \mathrm{~s}\)
\(
\begin{aligned}
& i=2 e^{\left(\frac{-0.3}{0.3}\right)} \\
& =\frac{5.8}{3 e} A(\text { approx } .)
\end{aligned}
\)

S30: Round the following numbers to 2 significant digits.
(a) 3472,
(b) \(84 \cdot 16\),
(c) 2.55 and
(d) \(28 \cdot 5\).

Answer: (a) In the value 3472, after the digit 4,7 is present. Its value is greater than 5 . So, the next two digits are neglected and the value of 4 is increased by 1.
\(\therefore\) value becomes 3500
(b) value 84
(c) value is 2.6
(d) value is 28 

S31: State, for each of the following physical quantities, if it is a scalar or a vector: volume, mass, speed, acceleration, density, number of moles, velocity, angular frequency, displacement, angular velocity.

Answer:

Scalar: Volume, mass, speed, density, number of moles, angular frequency
Vector: Acceleration, velocity, displacement, angular velocity
A scalar quantity is specified by its magnitude only. It does not have any direction associated with it. Volume, mass, speed, density, number of moles, and angular frequency are some of the scalar physical quantities.
A vector quantity is specified by its magnitude as well as the direction associated with it. Acceleration, velocity, displacement, and angular velocity belong to this category.

S32: Pick out the two scalar quantities in the following list:
force, angular momentum, work, current, linear momentum, electric field, average velocity, magnetic moment, and relative velocity.

Answer: Work and current are scalar quantities.
Work done is given by the dot product of force and displacement. Since the dot product of two quantities is always a scalar, work is a scalar physical quantity.
Current is described only by its magnitude. Its direction is not taken into account. Hence, it is a scalar quantity.

S33: Pick out the only vector quantity in the following list:
Temperature, pressure, impulse, time, power, total path length, energy, gravitational potential, coefficient of friction, and charge.

Answer: Impulse
Impulse is given by the product of force and time. Since force is a vector quantity, its product with time (a scalar quantity) gives a vector quantity.

S34: State with reasons, whether the following algebraic operations with scalar and vector physical quantities are meaningful:
(a) adding any two scalars, (b) adding a scalar to a vector of the same dimensions, (c) multiplying any vector by any scalar, (d) multiplying any two scalars, (e) adding any two vectors, (f) adding a component of a vector to the same vector.

Answer: a. Not meaningful. Two scalars of different dimensions/units cannot be added.
b. Not meaningful. A scalar and a vector cannot be added.
c. Meaningful. A scalar and vector can be multiplied irrespective of their dimensions/units.
d. Meaningful. Two scalars can be multiplied irrespective of their dimensions/units.
e. Not meaningful. Two vectors of different dimensions/units cannot be added.
f. Not meaningful. A component of a vector cannot be added to it.

Note: Only (c) and (d) are permissible

S35: Read each statement below carefully and state with reasons, if it is true or false :
(a) The magnitude of a vector is always a scalar, (b) each component of a vector is always a scalar, (c) the total path length is always equal to the magnitude of the displacement vector of a particle. (d) the average speed of a particle (defined as total path length divided by the time taken to cover the path) is either greater or equal to the magnitude of the average velocity of the particle over the same interval of time, (e) Three vectors not lying in a plane can never add up to give a null vector.

Answer: \(\text { (a) } \mathrm{T} \text {, (b) } \mathrm{F} \text {, (c) } \mathrm{F} \text {, (d) } \mathrm{T} \text {, (e) } \mathrm{T}\)

(a) True, the magnitude of the velocity of a body moving in a straight line may be equal to the speed of the body.
(b) False, each component of a vector is always a vector, not scalar.
(c) False, Total path length is a scalar quantity, whereas displacement is a vector quantity. Hence, the total path length is always greater than the magnitude of displacement. It becomes equal to the magnitude of displacement only when a particle is moving in a straight line.
(d) True, because the total path length is either greater than or equal to the magnitude of the displacement vector.
(e) True, this is because the resultant of two vectors will not lie in the plane of third vector and hence cannot cancel its effect to give null vector.

S36: Establish the following vector inequalities geometrically or otherwise :
(a) \(|\mathbf{a}+\mathbf{b}| \leq|\mathbf{a}|+|\mathbf{b}|\)
(b) \(|\mathbf{a}+\mathbf{b}| \geq|| \mathbf{a}|-| \mathbf{b}||\)
(c) \(|\mathbf{a}-\mathbf{b}| \leq|\mathbf{a}|+|\mathbf{b}|\)
(d) \(|\mathbf{a}-\mathbf{b}| \geq|| \mathbf{a}|-| \mathbf{b}||\)
When does the equality sign above apply?

Answer:

(a) The figure below shows the parallelogram, whose adjacent sides are two vectors, say \(a\) and \(b\), the diagonal of the parallelogram represents the vector \(a+b\).

From parallelogram OMNP,
\(
O M=P N=|a|, O P=M N=|b|, O N=|a+b|
\)
From the properties of the triangle, the length of each side of the triangle is less than the sum of the other two sides of the triangle.
In \(\triangle \mathrm{OMN}\)
\(
O N<(O M+O N)
\)
Substitute the values in the above expression.
\(
|a+b|<|a|+|b|(1)
\)
If \(a\) and \(b\) acts along the straight line in same direction, then
\(
|a+b|=|a|+|b|(I I)
\)
From equation (I) and equation (II),
\(
|a+b| \leq|a|+|b|
\)
Thus, it is proved that \(|a+b| \leq|a|+|b|\).

(b) The figure below shows the parallelogram, whose adjacent sides are two vectors, say \(a\) and \(b\), the diagonal of the parallelogram represents the vector \(a+b\).

From parallelogram OMNP,
\(
O M=P N=|a|, O P=M N=|b|, O N=|a+b|
\)
From the properties of the triangle, the length of each side of the triangle is less than the sum of the other two sides of the triangle.
In \(\triangle O M N\),
\(
O N+O M>M N|O N|>|M N-O M|
\)
Substitute the values in the above expression.
\(
|| a+b||>|| a|-| b|||a+b|>|| a|-| b|| \text { (III) }
\)
If \(a\) and \(b\) acts along a straight line in the same direction, then \(|a+b|=|| a|-| b||(I V)\)
From equation (III) and equation (IV),
\(|a+b| \geq|| a|-| b||\)
Thus, it is proved that \(|a+b| \geq|| a|-| b||\).

(c) The figure below shows the parallelogram, whose adjacent sides are two vectors, say \(a\) and \(-b\), the diagonal of the parallelogram represents the vector \(a-b\).

From parallelogram OPSR,
\(
O R=P S=|b|, O P=|a|, O S=|a-b|
\)
From the properties of the triangle, the length of each side of the triangle is less than the sum of the other two sides of the triangle.
In \(\triangle \mathrm{OPS}\)
\(O S<O P+P S\)
Substitute the values in the above expression.
\(
|a-b|<|a|+|b|(V)
\)
If \(a\) and \(b\) acts along a straight line in the same direction, then \(|a+b|=|| a|-| b||(V I)\)
From equation ( \(\mathrm{V}\) ) and equation (VI),
\(
|a-b| \leq|a|+|b|
\)
Thus, it is proved that \(|a-b| \leq|a|+|b|\).

(d) The figure below shows the parallelogram, whose adjacent sides are two vectors, say \(a\) and \(-b\), the diagonal of the parallelogram represents the vector \(a-b\).

From parallelogram OPSR,
\(
O R=P S=|b|, O P=|a|, O S=|a-b|
\)
From the properties of the triangle, the length of each side of the triangle is less than the sum of the other two sides of the triangle.
In \(\triangle \mathrm{OPS}\)
\(
O S+P S>O P O S>O P-P S|O S|>|O P-P S|
\)
Substitute the values in the above expression.
\(
|| a-b||>|| a|-| b|||a-b|>|| a|-| b||(V I I)
\)
If \(a\) and \(b\) acts along the straight line in same direction, then
\(
|a-b|=|| a|-| b||(V I I I)
\)
From equation (VII) and equation (VIII),
\(|a-b| \geq|| a|-| b||\)
Thus, it is proved that \(|a-b| \geq|| a|-| b||\).

S37: Three girls skating on a circular ice ground of radius \(200 \mathrm{~m}\) start from a point \(P\) on the edge of the ground and reach a point Q diametrically opposite to P following different paths as shown in Fig. 4.20. What is the magnitude of the displacement vector for each? For which girl is this equal to the actual length of the path skate?

Answer: Displacement is given by the minimum distance between the initial and final positions of a particle. In the given case, all the girls start from point \(P\) and reach point \(Q\). The magnitudes of their displacements will be equal to the diameter of the ground.
The radius of the ground \(=200 \mathrm{~m}\)
Diameter of the ground \(=2 \times 200=400 \mathrm{~m}\)
Hence, the magnitude of the displacement for each girl is \(400 \mathrm{~m}\). This is equal to the actual length of the path skated by a girl \(\mathrm{B}\).

S38: A passenger arriving in a new town wishes to go from the station to a hotel located \(10 \mathrm{~km}\) away on a straight road from the station. A dishonest cabman takes him along a circuitous path \(23 \mathrm{~km}\) long and reaches the hotel in \(28 \mathrm{~min}\). What is (a) the average speed of the taxi, (b) the magnitude of average velocity? Are the two equal?

Answer: Here, actual path length traveled, \(\mathrm{s}=23 \mathrm{~km}\); Displacement \(=10 \mathrm{~km}\);
Time taken, \(\mathrm{t}=28 \mathrm{~min}=\frac{28}{60} h\)
(a) Average speed of taxi \(=\frac{\text { actual path length }}{\text { time taken }}=\frac{23}{\frac{28}{60}} k \frac{m}{h}=49.3 \mathrm{~km} / \mathrm{h}\)
(b)Magnitude of average velocity \(=\frac{\text { displacement }}{\text { time taken }}=\frac{10}{\frac{28}{60}} \mathrm{~km} / \mathrm{h}=21.4 \mathrm{~km} / \mathrm{h}\)
No, the average speed equals average velocity magnitude only for a straight path.

S39: Read each statement below carefully and state, with reasons, if it is true or false :
(a) The net acceleration of a particle in circular motion is always along the radius of the circle toward the centre.
(b) The velocity vector of a particle at a point is always along the tangent to the path of the particle at that point.
(c) The acceleration vector of a particle in uniform circular motion averaged over one cycle is a null vector.

Answer: (a) False, the acceleration is towards the centre only in a uniform circular motion. In non Uniform motion, acceleration has a tangential component as well.
(b) True, the velocity is always tangential, if the motion is uniform or not.
(c) True, When a particle returns to its original position after completing one cycle, its displacement is zero. Hence, the average velocity and acceleration over one complete cycle is zero, which is a null vector.

S40: \(\hat{\mathbf{i}}\) and \(\hat{\mathbf{j}}\) are unit vectors along \(x\) – and \(y\)-axis respectively. What is the magnitude and direction of the vectors \(\hat{\mathbf{i}}+\hat{\mathbf{j}}\), and \(\hat{\mathbf{i}}-\hat{\mathbf{j}}\)? What are the components of a vector \(\mathbf{A}=2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}\) along the directions of \(\hat{\mathbf{i}}+\hat{\mathbf{j}}\) and \(\hat{\mathbf{i}}-\hat{\mathbf{j}}\)? [You may use the graphical method]

Answer:

1) \(\hat{i}+\hat{j}=\sqrt{(1)^2+(1)^2+2 \times 1 \times 1 \times \cos 90^{\circ}}=\sqrt{2}=1.414\) unit \(\tan \theta=\frac{1}{1}=1 \therefore \theta=45^{\circ}\)
So the vector \(\hat{i}+\hat{j}\) makes an angle \(45^{\circ}\) with \(\mathrm{x}\)-axis
2) \(|\hat{i}-\hat{j}|=\sqrt{(1)^2+(2)^2-2 \times 1 \times 1 \times \cos 90^{\circ}}\) \(=\sqrt{2}=1.414\) units

The vector \(\hat{i}-\hat{j}\) makess an angle \(-45^{\circ}\) with x-axis
3)Let us now determine the component of \(\vec{A}=2 \hat{i}+3 \hat{j}\) in the direction of \(\hat{i}+\hat{j}\)
Let \(\vec{B}=\hat{i}+\hat{j}\)
\(\vec{A} \cdot \vec{B}=A B \cos \theta=(A \cos \theta) B\)
So the component of \(\vec{A}\) in the direction of \(\vec{B}=\frac{\vec{A} \cdot \vec{B}}{B}\)
\(
=\frac{(2 \hat{i}+3 \hat{j}) \cdot(\hat{i}+\hat{j})}{\sqrt{(1)^2+(1)^2}}=\frac{2 \hat{i} \cdot \hat{i}+2 \hat{i} \cdot \hat{j}+3 \hat{j} \cdot \hat{i}+3 \hat{j} \cdot \hat{j}}{\sqrt{2}}=\frac{5}{\sqrt{2}} \text { units }
\)
4) Component of \(\vec{A}\) in the direction of \(\hat{i}-\hat{j}=\frac{(2 \hat{i}+3 \hat{j}) \cdot(\hat{i}-\hat{j})}{\sqrt{2}}=-\frac{1}{\sqrt{2}}\) units.

S41: For any arbitrary motion in space, which of the following relations are true :
(a) \(\mathbf{v}_{\text {average }}=(1 / 2)\left(\mathbf{v}\left(t_1\right)+\mathbf{v}\left(t_2\right)\right)\)
(b) \(\mathbf{v}_{\text {average }}=\left[\mathbf{r}\left(t_2\right)-\mathbf{r}\left(t_1\right)\right] /\left(t_2-t_1\right)\)
(c) \(\mathbf{v}(t)=\mathbf{v}(0)+\mathbf{a} t\)
(d) \(\mathbf{r}(t)=\mathbf{r}(0)+\mathbf{v}(0) t+(1 / 2) \mathbf{a} t^2\)
(e) \(\mathbf{a}_{\text {average }}\) \(=\left[\mathbf{v}\left(t_2\right)-\mathbf{v}\left(t_1\right)\right] /\left(t_2-t_1\right)\)
(The ‘average’ stands for the average of the quantity over the time interval \(t_1\) to \(t_2\) )

Answer: (a) This statement is False. Explanation: It is given that the motion of the particle is arbitrary. Therefore, the average velocity of the particle cannot be given by this equation.
(b) True, The arbitrary motion of the particle can be represented by this equation.
(c) This statement is false. Explanation: The motion of the particle is arbitrary. The acceleration of the particle may also be non-uniform. Hence, this equation cannot represent the motion of the particle in space.
(d) This statement is false. Explanation: The motion of the particle is arbitrary; the acceleration of the particle may also be non-uniform. Hence, this equation cannot represent the motion of a particle in space.
(e)This statement is true. Explanation: The arbitrary motion of the particle can be represented by this equation.

S42: Read each statement below carefully and state, with reasons and examples, if it is true or false :
A scalar quantity is one that
(a) is conserved in a process
(b) can never take negative values
(c) must be dimensionless
(d) does not vary from one point to another in space
(e) has the same value for observers with different orientations of axes.

Answer: (a) This statement is false. Explanation: Despite being a scalar quantity, energy is not conserved in inelastic collisions.
(b) This statement is false. Explanation: Despite being a scalar quantity, the temperature can take negative values.
(c) This statement is false. Explanation: Total path length is a scalar quantity. Yet it has the dimension of length.
(d) This statement is false. Explanation: A scalar quantity such as gravitational potential can vary from one point to another in space.
(e) This statement is true. Explanation: The value of a scalar does not vary for observers with different orientations of axes.

S43: A vector has magnitude and direction. Does it have a location in space? Can it vary with time? Will two equal vectors \(\mathbf{a}\) and \(\mathbf{b}\) at different locations in space necessarily have identical physical effects? Give examples in support of your answer.

Answer: (i) Besides having magnitude and direction, each vector has also a location in space.
(ii) A vector can vary with time. As an example, velocity and acceleration vectors may vary with time.
(iii) Two equal vectors a and b having different locations may not have the same physical effect. As an example, two balls thrown with the same force, one from the earth and the other from the moon will attain different ‘maximum heights’.

S44: A vector has both magnitude and direction. Does it mean that anything that has magnitude and direction is necessarily a vector? The rotation of a body can be specified by the direction of the axis of rotation, and the angle of rotation about the axis. Does that make any rotation a vector?

Answer: No; A physical quantity having both magnitude and direction need not be considered a vector. For example, despite having magnitude and direction, the current is a scalar quantity. The essential requirement for a physical quantity to be considered a vector is that it should follow the law of vector addition.

Generally speaking, the rotation of a body about an axis is not a vector quantity as it does not follow the law of vector addition. However, a rotation by a certain small angle follows the law of vector addition and is therefore considered a vector.

S45: Can you associate vectors with (a) the length of a wire bent into a loop, (b) a plane area, and (c) a sphere? Explain.

Answer: (a) We cannot associate a vector with the length of a wire bent into a loop. This is because the length of the loop does not have a definite direction.
(b) We can associate a vector with a plane area. Such a vector is called area vector and its direction is represented by a normal drawn outward to the area.
(c) The area of a sphere does not point in any definite direction. However, we can associate a null vector with the area of the sphere. We cannot associate a vector with the volume of a sphere.

S46: A cyclist starts from centre \(O\) of a circular park of radius \(1 \mathrm{~km}\) and moves along the path OPRQO as shown Fig. 4.3. If he maintains constant speed of \(10 \mathrm{~ms}^{-1}\), what is his acceleration at point \(R\) in magnitude and direction?

Answer: According to the problem the path of the cyclist is O-P-R-Q-O.
The cyclist is in a uniform circular motion and it is given that linear velocity \(=10 \mathrm{~m} / \mathrm{s}, R=1 \mathrm{~km}=1000 \mathrm{~m}\). As we know whenever an object is performing a circular motion, acceleration is called centripetal acceleration and is always directed toward the centre. So cyclist experiences a centripetal force (acceleration) at point \(R\) towards centre.
The centripetal acceleration at \(R\) is given by the relation, \(a_c=\frac{v^2}{r}\) \(\Rightarrow \quad a_c=\frac{(10)^2}{1000}=\frac{100}{10^3}=0.1 \mathrm{~m} / \mathrm{s}^2\) along \(R O\).

S47: A particle is projected in the air at some angle to the horizontal, and moves along the parabola as shown in Fig. 4.4, where \(x\) and \(y\) indicate horizontal and vertical directions, respectively. Show in the diagram, the direction of velocity and acceleration at points \(\mathrm{A}, \mathrm{B}\), and \(\mathrm{C}\).

Answer: In projectile motion horizontal component of velocity will always be constant and acceleration is always vertically downward and is equal to \(\mathrm{g}\). The direction of velocity will always be tangential to the curve in the direction of motion.
As shown in the diagram in which a particle is projected at an angle \(\theta\).

\(
\begin{aligned}
v_x & =\text { Horizontal component of velocity } \\
& =v \cos \theta=\text { constant } \\
v_y & =\text { Vertical component of velocity } \\
& =v \sin \theta
\end{aligned}
\)

S48: A ball is thrown from a rooftop at an angle of \(45^{\circ}\) above the horizontal. It hits the ground a few seconds later. At what point during its motion, does the ball have
(a) greatest speed?
(b) smallest speed?
(c) greatest acceleration?
Explain

Answer:

In this problem, the total mechanical energy of the ball is conserved. As the ball is projected from point \(\mathrm{O}\), and covering the path \(\mathrm{OABC}\).
At point A it has both kinetic and potential energy.
But at point \(\mathrm{C}\) it has only kinetic energy, (keeping the ground as a reference where \(\mathrm{PE}\) is zero.)
(a) At the point \(B\), it will gain the same speed \(u\) and after that speed increases and will be maximum just before reaching \(C\).
(b) During the upward journey from OtoA speed decreases and the smallest speed attained by it is at the highest point, i.e., at point A.
(c) Acceleration is always constant throughout the journey and is vertically downward equal to \(\mathrm{g}\).

S49: A football is kicked into the air vertically upwards. What is its
(a) acceleration, and (b) velocity at the highest point?

Answer:

(a) The situation is shown in the diagram below in which a football is kicked into the air vertically upwards. Acceleration of the football will always be vertical downward and is called acceleration due to gravity \((\mathrm{g})\).
(b) When the football reaches the highest point it is momentarily at rest and at that moment its velocity will be zero as it is continuously retarded by acceleration due to gravity \((\mathrm{g})\).

S50: \(\mathbf{A}\), \(\mathbf{B}\) and \(\mathbf{C}\) are three non-collinear, non co-planar vectors. What can you say about direction of \(\mathbf{A} \times(\mathbf{B} \times \mathbf{C})\)?

Answer: Key concept: Collinear vectors: When the vectors under consideration can share the same support or have a common support then the considered vectors are collinear.

Coplanar vectors: Three (or more) vectors are called coplanar vector if they lie in the same plane. Two (free) vectors are always coplanar.
\(
\vec{B} \times \vec{C}=B C \sin \theta \hat{n}
\)
where \(\hat{n}=\) unit vector perpendicular to the plane containing \(\vec{B}\) and \(\vec{C}\)
\(
\begin{aligned}
\vec{A} \times(\vec{B} \times \vec{C}) & =\vec{A} \times \hat{n}(B C \sin \theta) \\
& =(B C \sin \theta) A \sin \alpha \hat{p}
\end{aligned}
\)
where \(\hat{p}=\) unit vector perpendicular to both \(\hat{n}\) and \(\vec{A}\).
Here perpendicular to \(\hat{n}\) and \(\vec{A}\) means in the plane of \(\vec{B}\) and \(\vec{C}\).
Hence \(\vec{A} \times(\vec{B} \times \vec{C})\) will lie in the plane of \(\vec{B}\) and \(\vec{C}\), and is perpendicular to \(\vec{A}\).

S51: A boy travelling in an open car moving on a levelled road with constant speed tosses a ball vertically up in the air and catches it back. Sketch the motion of the ball as observed by a boy standing on the footpath. Give an explanation to support your diagram.

Answer: With respect to the observer standing on the footpath ball is thrown with velocity \(u\) at an angle \(\theta\) with the horizontal, hence it seems as a projectile. So the path of the ball will be parabolic. The horizontal speed of the ball is the same as that of the car, therefore, the ball as well car travels an equal horizontal ult distance. Due to its vertical speed, the ball follows a parabolic path.

Important point: We must be very clear that we are working with respect to the ground. When we observe with respect to the car, the motion will be along vertical direction only.

S52: A boy throws a ball in the air at \(60^{\circ}\) to the horizontal along a road with a speed of \(10 \mathrm{~m} / \mathrm{s}(36 \mathrm{~km} / \mathrm{h})\). Another boy sitting in a passing by car observes the ball. Sketch the motion of the ball as observed by the boy in the car, if the car has a speed of \((18 \mathrm{~km} / \mathrm{h})\). Give an explanation to support your diagram.

Answer: The situation is shown in the below diagram.

According to the problem, the boy standing on the ground throws the ball at an angle of \(60^{\circ}\) with horizontal at a speed of \(10 \mathrm{~m} / \mathrm{s}\).
\(\therefore\) Horizontal component of velocity, \(u_x=10 \cos \theta\)
\(
u_x=(10 \mathrm{~m} / \mathrm{s}) \cos 60^{\circ}=10 \times \frac{1}{2}=5 \mathrm{~m} / \mathrm{s}
\)
Vertical component of velocity, \(u_y=10 \sin \theta\)
\(
u_y=(10 \mathrm{~m} / \mathrm{s}) \sin 60^{\circ}=10 \times \frac{\sqrt{3}}{2}=5 \sqrt{3} \mathrm{~m} / \mathrm{s}
\)
Speed of the car \(=18 \mathrm{~km} / \mathrm{h}=5 \mathrm{~m} / \mathrm{s}\)
As the horizontal speed of the ball and the car is the same, hence the relative velocity of the ball w.r.t car in the horizontal direction will be zero. Only the vertical motion of the ball will be observed by the boy in the car, as shown in the above diagram.

S53: In dealing with the motion of the projectile in air, we ignore the effect of air resistance on motion. This gives trajectory as a parabola as you have studied. What would the trajectory look like if air resistance is included? Sketch such a trajectory and explain why you have drawn it that way.

Answer: When we are dealing with projectile motion generally we neglect the air resistance. But if air resistance is included the horizontal component of velocity will not be constant and obviously trajectory will change. Due to air resistance, particle energy as well as the horizontal component of velocity keep on decreasing making the fall steeper than the rise as shown in the figure. When we are neglecting the air resistance path was symmetric parabola (OAC). When air resistance is considered path is an asymmetric parabola (OAB).

S54: A fighter plane is flying horizontally at an altitude of \(1.5 \mathrm{~km}\) with speed \(720 \mathrm{~km} / \mathrm{h}\). At what angle of sight (w.r.t. horizontal) when the target is seen, should the pilot drop the bomb in order to attack the target?

Answer: Key concept:
1. Displacement of Projectile \((\vec{r})\) : After time \(t\), horizontal displacement \(x=u t\) and vertical displacement \(y=\frac{1}{2} g t^2\).
So, the position vector \(\vec{r}=u t \hat{i}+\frac{1}{2} g t^2 \hat{j}\)
Therefore \(r=u t \sqrt{1+\left(\frac{g t}{2 u}\right)^2}\) and \(\alpha=\tan ^{-1}\left(\frac{g t}{2 u}\right)\)
\(
\alpha=\tan ^{-1}\left(\sqrt{\frac{g y}{2}} / u\right)\left(\text { as } t=\sqrt{\frac{2 y}{g}}\right)
\)

2. When an object is dropped/released by any moving vehicle. The initial velocity of the object is same as the moving vehicle.

When the bomb is dropped from a Plane the plane which is moving horizontally. So, the bomb will have the same initial velocity as that of the plane along the horizontal direction.
The situation is shown in the diagram below. Let a fighter plane, when it be plane at position P, drops a bomb to hit a target \(T\).
Let the target is seen at an angle \(\theta\) with horizontal.

And the speed of the plane \(=720 \mathrm{~km} / \mathrm{h}=720 \times \frac{5}{18} \mathrm{~m} / \mathrm{s}=200 \mathrm{~m} / \mathrm{s}\)
Altitude of the plane \(y=1.5 \mathrm{~km}=1500 \mathrm{~m}\)
If a bomb hits the target after time \(t\), then the horizontal distance travelled by the bomb is
\(
x=u \times t=200 t \dots(i)
\)
Vertical distance travelled by the bomb
\(
\begin{aligned}
y & =\frac{1}{2} g t^2 \Rightarrow 1500=\frac{1}{2} \times 9.8 t^2 \\
\Rightarrow \quad t^2 & =\frac{1500}{4.9} \Rightarrow t=\sqrt{\frac{1500}{49}}=17.49 \mathrm{~s}
\end{aligned}
\)
Using value of \(t\) in Eq. (i),
\(
x=200 \times 17.49 \mathrm{~m}
\)
Now, \(\tan \theta=\frac{y}{x}=\frac{1500}{200 \times 17.49}=49287=\tan 23^{\circ} 12^{\prime}\)
\(
\Rightarrow \theta=23^{\circ} 12^{\prime}
\)
Important point: Angle is with respect to the target. As seen by the observer in the plane, the motion of the bomb will be vertically downward below the plane.

S55:(a) Earth can be thought of as a sphere of radius \(6400 \mathrm{~km}\). Any object (or a person) is performing circular motion around the axis of the earth due to the earth’s rotation (period 1 day). What is the acceleration of an object on the surface of the earth (at the equator) towards its centre? what is it at latitude \(\theta\)? How does these accelerations compare with \(g=9.8 \mathrm{~m} / \mathrm{s}^2\)?

(b) Earth also moves in a circular orbit around the sun once every year with on orbital radius of \(1.5 \times 10^{11} \mathrm{~m}\). What is the acceleration of Earth (or any object on the surface of the Earth) towards the centre of the sun? How does this acceleration compare with \(g=9.8 \mathrm{~m} / \mathrm{s}^2?\)
\(\left(\right.\) Hünt : acceleration \(\left.\frac{V^2}{R}=\frac{4 \pi^2 R}{T^2}\right)\)

Answer: (a) According to the problem,
Radius of the earth \((R)=6400 \mathrm{~km}=6.4 \times 10^6 \mathrm{~m}\).
Time period \((T)=1\) day \(=24 \times 60 \times 60 \mathrm{~s}=86400 \mathrm{~s}\)
Centripetal acceleration,
\(
\begin{aligned}
\left(a_c\right)=\omega^2 R=\frac{4 \pi^2 R}{T} & =\frac{4 \times(22 / 7)^2 \times 6.4 \times 10^6}{(24 \times 60 \times 60)^2} \\
& =\frac{4 \times 484 \times 64 \times 10^6}{49 \times(24 \times 3600)^2}=0.034 \mathrm{~m} / \mathrm{s}^2
\end{aligned}
\)
At equator, latitude \(\theta=0^{\circ}\)
\(
\therefore \quad \frac{a_c}{g}=\frac{0.034}{9.8}=\frac{1}{288}
\)
(b) Orbital radius of the earth around the sun \((R)=1.5 \times 10^{11} \mathrm{~m}\) Time period \(=1 \mathrm{yr}=365\) days \(=365 \times 24 \times 60 \times 60 \mathrm{~s}=3.15 \times 10^7 \mathrm{~s}\) Centripetal acceleration
\(
\begin{aligned}
& \left(a_c\right)=R \omega^2=\frac{4 \pi^2 R}{T^2}=\frac{4 \times(22 / 7)^2 \times 1.5 \times 10^{11}}{\left(3.15 \times 10^7\right)^2}=5.97 \times 10^{-3} \mathrm{~m} / \mathrm{s}^2 \\
& \frac{a_c}{g}=\frac{5.97 \times 10^{-3}}{9.8}=\frac{1}{1642}
\end{aligned}
\)

S56: Given below in column I are the relations between vectors \(\mathbf{a}, \mathbf{b}\) and \(\mathbf{c}\) and in column II are the orientations of \(\mathbf{a}, \mathbf{b}\) and \(\mathbf{c}\) in the XY plane. Match the relation in column I to the correct orientations in column II.

Answer: We apply triangular law of addition
Triangular law of vector addition: Two vectors are considered as two sides of a triangle taken in the same order. The third side or completing side of the triangle is the resultant taken in the opposite order. or
We can say that vectors are arranged head to tail, this graphical method is called the head-to-tail method. The two vectors and their resultant form three sides of a triangle, so this method is also known as the triangle method of vector addition.
As shown in the diagram below in which vectors \(\mathrm{A}\) and \(\mathrm{B}\) are corrected by head and tail. Resultant vector \(\mathrm{C}\) \(=A+B\)
(a) from (iv), it is clear that \(c=a+b\)
(b) from (iii), \(c+b=a=>a-c=b\)
(c) from (i), b \(=a+c=>b-a=c\)
(d) from (ii), \(-c=a+b \Rightarrow a+b+c=0\)

S57: If \(|\mathbf{A}|=2\) and \(|\mathbf{B}|=4\), then match the relations in column I with the angle \(\theta_\theta\) between \(\mathbf{A}\) and \(\mathbf{B}\) in column II.

Answer: a) Matches with (ii)
\(
\begin{aligned}
& |\vec{A}||\vec{B}| \cos \theta=0 \\
& \cos \theta=0 \\
& \theta=90
\end{aligned}
\)
b) Matches with (i)
\(
\begin{aligned}
& |\vec{A}||\vec{B}| \cos \theta=8 \\
& 2 \times 4 \cos \theta=8 \\
& \cos \theta=1 \\
& \theta=0
\end{aligned}
\)
c) Matches with (iv)
\(
\begin{aligned}
& |\vec{A}||\vec{B}| \cos \theta=4 \\
& 2 \times 4 \cos \theta=84 \\
& \cos \theta=\frac{1}{2} \\
& \theta=60
\end{aligned}
\)
d) Matches with (iii)
\(
\begin{aligned}
& |\vec{A}||\vec{B}| \cos \theta=-8 \\
& 2 \times 4 \cos \theta=-8 \\
& \cos \theta=-1 \\
& \theta=180
\end{aligned}
\)

S58: If \(|\mathbf{A}|=2\) and \(|\mathbf{B}|=4\), then match the relations in column I with the angle \(\theta\) between \(A\) and \(B\) in column II.

Answer: \(
\text { Given : }|\vec{a}|=2 \text { and }|\vec{B}|=4
\)
a) Matches with (iv)
\(
\begin{aligned}
& |\vec{A} \times \vec{B}|=0 \\
& |\vec{A}||\vec{B}| \sin \theta=0 \\
& 2 \times 4 \sin \theta=0 \\
& \sin \theta=\sin 0 \\
& \theta=0
\end{aligned}
\)
b) matches with (iii)
\(
\begin{aligned}
& |\vec{A} \times \vec{B}|=8 \\
& |\vec{A}||\vec{B}| \sin \theta=8 \\
& 2 \times 4 \sin \theta=8 \\
& \sin \theta=1 \\
& \theta=90
\end{aligned}
\)
c) matches with (i)
\(
\begin{aligned}
& |\vec{A} \times \vec{B}|=4 \\
& |\vec{A}||\vec{B}| \sin \theta=4 \\
& 2 \times 4 \sin \theta=4 \\
& \sin \theta=\frac{1}{2} \\
& \Theta=30
\end{aligned}
\)
d) matches with (ii)
\(
\begin{aligned}
& |\vec{A} \times \vec{B}|=4 \sqrt{2} \\
& |\vec{A}||\vec{B}| \sin \theta=4 \sqrt{2} \\
& 2 \times 4 \sin \theta=4 \sqrt{2} \\
& \sin \theta=\frac{1}{\sqrt{2}} \\
& \theta=45
\end{aligned}
\)

Type: Long Questions 

Q1. A vector has a component along the \(X\)-axis equal to 25 units and along the \(Y\)-axis equal to 60 units. Find the magnitude and direction of the vector.

Answer: The given vector is the resultant of two perpendicular vectors, one along the \(X\)-axis of magnitude 25 units and the other along the \(Y\)-axis of magnitude 60 units. The resultant has a magnitude \(A\) given by
\(
\begin{aligned}
A & =\sqrt{(25)^2+(60)^2+2 \times 25 \times 60 \cos 90^{\circ}} \\
& =\sqrt{(25)^2+(60)^2}=65 .
\end{aligned}
\)
The angle \(\alpha\) between this vector and the \(X\)-axis is given by
\(
\tan \alpha=\frac{60}{25}
\)

Q2: A bomb is dropped from a plane flying horizontally with uniform speed. Show that the bomb will explode vertically below the plane. Is the statement true if the plane flies with uniform speed but not horizontally?

Answer: The plane is flying horizontally with a uniform speed. Therefore, the bomb also has the same speed.
Let the speed of the plane be represented by \(u\).
Now, let \(\mathrm{t}\) be the time taken by the bomb to reach the ground.
Distance travelled by the bomb in horizontal direction \(=\mathrm{ut}\)
Both the plane and bomb are travelling in the same direction.
Distance travelled by the plane in the same time = ut
Hence, the bomb will explode vertically below the plane.

When the plane is flying with a uniform speed but not horizontally:
Let us consider it will make an angle of projection \(\theta\) along the horizontal direction.
So, both the plane and the bomb will be flying with the same angle of projection.
Therefore, both will have the same horizontal speed \(u \cos \theta\), where \(u\) is the initial speed of the plane and the bomb.
When the bomb is released, the time taken by the bomb to reach the ground is \(\mathrm{t}\).
The distance travelled by the bomb and the plane will be \(\mathrm{u} \cos \theta \mathrm{t}\).
Hence, again the bomb will explode vertically below the plane.
(i) During the motion of the bomb, its horizontal velocity u remains constant and is the same as that of the plane at every point of its path.
Let the bomb reach the ground in time t.
Distance travelled in the horizontal direction by the bomb \(=\) ut
Distance travelled in the horizontal direction by the bomb is the same as that travelled by the plane.
So, the bomb will explode vertically below the plane.
(ii) Let the plane move making an angle \(\alpha\) with the horizontal.
Horizontal distance for both the bomb and the plane \(=u \cos \alpha \mathrm{t}^{\prime}\), \(\mathrm{t}^{\prime}\) = Time taken by the bomb to reach the ground
So, in this case, also, the bomb will explode vertically below the plane.

Q3: Consider the situation of the previous problem. The man has to reach the other shore at the point directly opposite to his starting point. If he reaches the other shore somewhere else, he has to walk down to this point. Find the minimum distance that he has to walk.

Answer: Given Width of the river \(=500 \mathrm{~m}\)
Rate of flow of the river \(=5 \mathrm{~km} / \mathrm{h}\)
Swimmer’s speed with respect to water \(=3 \mathrm{~km} / \mathrm{h}\)
As per the question, the man has to reach the other shore at the point directly opposite to his starting point.

Horizontal distance is BD for the resultant velocity \(v_r\).
\(x\)-component of the resultant velocity, \(R=5-3 \cos \theta\)
Vertical component of velocity \(=3 \sin \theta \mathrm{km} / \mathrm{h}\)
\(
\text { Time }=\frac{\text { Distance }}{\text { Velocity }}=\frac{0.5}{3 \sin \theta} h
\)
This is the same as the horizontal component of velocity.
Thus, we have:
\(
\begin{aligned}
& B D=(5-3 \cos \theta)\left(\frac{0.5}{3 \sin \theta}\right) \\
& =\frac{5-3 \cos \theta}{6 \sin \theta}
\end{aligned}
\)
For \(\mathrm{H}\) (horizontal distance) to be minimum,
\(
\begin{aligned}
& \left(\frac{d H}{d \theta}\right)=0 \\
& \Rightarrow \frac{d}{d \theta}\left(\frac{5+3 \cos \theta}{6 \sin \theta}\right)=0 \\
& \Rightarrow 18\left(\sin ^2 \theta+\cos ^2 \theta\right)+30 \cos \theta=0 \\
& \Rightarrow-30 \cos \theta=18 \\
& \Rightarrow \cos \theta=-\frac{18}{30}=-\frac{3}{5}
\end{aligned}
\)
The negative sign shows that \(\theta\) lies in the 2nd Quadrant.
And,
\(
\begin{aligned}
& \sin \theta=\sqrt{1-\cos ^2 \theta}=\frac{4}{5} \\
& \therefore H=\frac{5-3 \cos \theta}{6 \sin \theta} \\
& =\frac{5-3\left(\frac{3}{5}\right)}{6 \times \frac{4}{5}}=\frac{25-9}{24} \\
& =\frac{16}{24}=\frac{2}{3} \mathrm{~km}
\end{aligned}
\)

\(
\text { The minimum distance that the man has to walk is }=2 / 3 \mathrm{~km}
\)

Q4: Suppose \(A\) and \(B\) in the previous problem change their positions in such a way that the line joining them becomes perpendicular to the direction of wind while maintaining the separation \(x\). What will be the time lag \(B\) finds between seeing and hearing the drum beating by \(A\)?

Answer: Given Distance between \(A\) and \(B=x\)

Let \(v\) be the velocity of sound in the direction along line \(A C\). Let \(u\) be the velocity of air in the direction along line \(A B\).
Angle between \(v\) and \(u=\boldsymbol{\theta}>\frac{\pi}{2}\)
\(
\text { Resultant velocity of sound and air that will reach } \mathrm{B}=\overrightarrow{A D}=\sqrt{\left(v^2-u^2\right)}
\)
Here, the time taken by light to reach B is neglected.
\(\therefore\) Time lag between seeing and hearing \(=\) Time taken to hear the sound of the drum
\(
t=\frac{\text { Displacement }}{\text { Velocity }}=\frac{x}{\sqrt{\left(v^2-u^2\right)}}
\)

Q5: Find the resultant of the three vectors shown in Figure (2-W 1).

Answer: Take the axes as shown in the figure.
The \(x\)-component of the \(5.0 \mathrm{~m}\) vector \(=5.0 \mathrm{~m} \cos 37^{\circ}=4.0 m\)
the \(x\)-component of the \(3.0 \mathrm{~m}\) vector \(=3.0 \mathrm{~m}\)
and the \(x\)-component of the \(2.0 \mathrm{~m}\) vector \(=2.0 \mathrm{~m} \cos 90^{\circ}=0\)
Hence, the \(x\)-component of the resultant
\(
=4.0 \mathrm{~m}+3.0 \mathrm{~m}+0=7.0 \mathrm{~m} \text {. }
\)
The \(y\)-component of the \(5.0 \mathrm{~m}\) vector \(=5^{\circ} 0 \mathrm{~m} \sin 37^{\circ}=3.0 m.\)
the \(y\)-component of the \(3.0 \mathrm{~m}\) vector \(=0\) and the \(y\)-component of the \(2.0 \mathrm{~m}\) vector \(=2.0 \mathrm{~m}\). Hence, the \(y\)-component of the resultant \(=3.0 \mathrm{~m}+0+2.0 \mathrm{~m}=5.0 \mathrm{~m}\)
The magnitude of the resultant vector
\(
\begin{aligned}
& =\sqrt{(7 \cdot 0 \mathrm{~m})^2+(5 \cdot 0 \mathrm{~m})^2} \\
& =8.6 \mathrm{~m} .
\end{aligned}
\)
If the angle made by the resultant with the \(X\)-axis is \(\theta\), then
\(
\tan \theta=\frac{y \text {-component }}{x \text {-component }}=\frac{5 \cdot 0}{7 \cdot 0} \text { or, } \theta=355^{\circ} \text {. }
\)

Q6: The sum of the three vectors shown in figure \((2-W 2)\) is zero. Find the magnitudes of the vectors \(\overrightarrow{O B}\) and \(\overrightarrow{O C}\).

Answer: The \(x\)-component of \(\overrightarrow{O A}=(O A) \cos 90^{\circ}=0\)
The \(x\)-component of \(\overrightarrow{O B}=(O B) \cos 0^{\circ}=O B\).
The \(x\)-component of \(\overrightarrow{O C}=(O C) \cos 135^{\circ}=-\frac{1}{\sqrt{2}} O C\).
Hence, the \(x\)-component of the resultant
\(
=O B-\frac{1}{\sqrt{2}} O C \text {. } \dots(i)
\)
It is given that the resultant is zero and hence its \(x\)-component is also zero. From (i),
\(
O B=\frac{1}{\sqrt{2}} O C \dots(ii)
\)
The \(y\)-component of \(\overrightarrow{O A}=O A \cos 180^{\circ}=-O A\).
The \(y\)-component of \(\overrightarrow{O B}=O B \cos 90^{\circ}=0\).
The \(y\)-component of \(\overrightarrow{O C}=O C \cos 45^{\circ}=\frac{1}{\sqrt{2}} O C\).
Hence, the \(y\)-component of the resultant
\(
=\frac{1}{\sqrt{2}} O C-O A \dots(iii)
\)
As the resultant is zero, so is its \(y\)-component. From (iii),
\(
\frac{1}{\sqrt{2}} O C=O A, \text { or, } O C=\sqrt{2} O A=5 \sqrt{2} \mathrm{~m} .
\)
From (ii), \(O B=\frac{1}{\sqrt{2}} O C=5 \mathrm{~m}\).

Q7: The magnitudes of vectors \(\overrightarrow{O A}, \overrightarrow{O B}\) and \(\overrightarrow{O C}\) in figure (2-W3) are equal. Find the direction of \(\overrightarrow{O A}+\overrightarrow{O B}-\overrightarrow{O C}\).

Answer: Let \(O A=O B=O C=F\).
\(x\)-component of \(\overrightarrow{O A}=F \cos 30^{\circ}=F \frac{\sqrt{3}}{2}\).
\(x\)-component of \(\overrightarrow{O B}=F \cos 60^{\circ}=\frac{F}{2}\).
\(x\)-component of \(\overrightarrow{O C}=F \cos 135^{\circ}=-\frac{F}{\sqrt{ } 2}\).
\(x\)-component of \(\overrightarrow{O A}+\overrightarrow{O B}-\overrightarrow{O C}\)
\(
\begin{aligned}
& =\left(\frac{F \sqrt{ } 3}{2}\right)+\left(\frac{F}{2}\right)-\left(-\frac{F}{\sqrt{ } 2}\right) \\
& =\frac{F}{2}(\sqrt{ } 3+1+\sqrt{ } 2) .
\end{aligned}
\)
\(y\)-component of \(\overrightarrow{O A}=F \cos 60^{\circ}=\frac{F}{2}\).
\(y\)-component of \(\overrightarrow{O B}=F \cos 150^{\circ}=-\frac{F \sqrt{ } 3}{2}\).
\(y\)-component of \(\overrightarrow{O C}=F \cos 45^{\circ}=\frac{F}{\sqrt{2}}\).
\(
y \text {-component of } \overrightarrow{O A}+\overrightarrow{O B}-\overrightarrow{O C}
\)
\(
\begin{aligned}
& =\left(\frac{F}{2}\right)+\left(-\frac{F \sqrt{ } 3}{2}\right)-\left(\frac{F}{\sqrt{ } 2}\right) \\
& =\frac{F}{2}(1-\sqrt{ } 3-\sqrt{ } 2) .
\end{aligned}
\)
\(
\text { Angle of } \overrightarrow{O A}+\overrightarrow{O B}-\overrightarrow{O C} \text { with the } X \text {-axis }
\)
\(
=\tan ^{-1} \frac{\frac{F}{2}(1-\sqrt{3}-\sqrt{2})}{\frac{F}{2}(1+\sqrt{3}+\sqrt{2})}=\tan ^{-1} \frac{(1-\sqrt{ } 3-\sqrt{2})}{(1+\sqrt{ } 3+\sqrt{2})}
\)

Q8: Find the resultant of the three vectors \(\overrightarrow{O A}, \overrightarrow{O B}\) and \(\overrightarrow{O C}\) shown in figure (2-W4). Radius of the circle is \(R\).

Answer: \(O A=O C\).
\(\overrightarrow{O A}+\overrightarrow{O C}\) is along \(\overrightarrow{O B}\) (bisector) and its magnitude is \(2 R \cos 45^{\circ}=R \sqrt{ } 2\).
\((\overrightarrow{O A}+\overrightarrow{O C})+\overrightarrow{O B}\) is along \(\overrightarrow{O B}\) and its magnitude is \(R \sqrt{ } 2+R=R(1+\sqrt{2})\).

Q9: The resultant of vectors \(\overrightarrow{O A}\) and \(\overrightarrow{O B}\) is perpendicular to \(\overrightarrow{O A}\) (figure 2-W5). Find the angle \(A O B\).

Answer: Take the dotted lines as \(X, Y\) axes. \(x\)-component of \(\overrightarrow{O A}=4 \mathrm{~m}, x\)-component of \(\overrightarrow{O B}=6 \mathrm{~m} \cos \theta\)
\(x\)-component of the resultant \(=(4+6 \cos \theta) \mathrm{m}\).
But it is given that the resultant is along \(Y\)-axis. Thus, the \(x\)-component of the resultant \(=0\)
\(
4+6 \cos \theta=0 \text { or, } \cos \theta=-2 / 3 .
\)

Q10: Write the unit vector in the direction of \(\vec{A}=5 \vec{i}+\vec{j}-2 \vec{k}\).

Answer: \(\quad|\vec{A}|=\sqrt{5^2+1^2+(-2)^2}=\sqrt{30}\).
The required unit vector is \(\frac{\vec{A}}{|\vec{A}|}\)
\(
=\frac{5}{\sqrt{30}} \vec{i}+\frac{1}{\sqrt{30}} \vec{j}-\frac{2}{\sqrt{30}} \vec{k} .
\)

Q11: \(\text { If }|\vec{a}+\vec{b}|=|\vec{a}-\vec{b}| \text { show that } \vec{a} \perp \vec{b} \text {. }\)

Answer: We have
\(
\begin{aligned}
|\vec{a}+\vec{b}|^2 & =(\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b}) \\
& =\vec{a} \cdot \vec{a}+\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{a}+\vec{b} \cdot \vec{b} \\
& =a^2+b^2+2 \vec{a} \cdot \vec{b} .
\end{aligned}
\)
Similarly,
\(
\begin{aligned}
|\vec{a}-\vec{b}|^2 & =(\vec{a}-\vec{b}) \cdot(\vec{a}-\vec{b}) \\
& =a^2+b^2-2 \vec{a} \cdot \vec{b} .
\end{aligned}
\)
If
\(
|\vec{a}+\vec{b}|=|\vec{a}-\vec{b}|,
\)
\(
a^2+b^2+2 \vec{a} \cdot \vec{b}=a^2+b^2-2 \vec{a} \cdot \vec{b}
\)
or,
\(
\vec{a} \cdot \vec{b}=0
\)
or, \(\quad \vec{a} \perp \vec{b}\).

Q12: If \(\vec{a}=2 \vec{i}+3 \vec{j}+4 \vec{k}\) and \(\vec{b}=4 \vec{i}+3 \vec{j}+2 \vec{k}\), find the angle between \(\vec{a}\) and \(\vec{b}\)

Answer: We have \(\vec{a} \cdot \vec{b}=a b \cos \theta\)
or,
\(
\cos \theta=\frac{\vec{a} \cdot \vec{b}}{a b}
\)
where \(\theta\) is the angle between \(\vec{a}\) and \(\vec{b}\).
Now
\(
\begin{aligned}
\vec{a} \cdot \vec{b} & =a_x b_x+a_y b_y+a_z b_z \\
& =2 \times 4+3 \times 3+4 \times 2=25 .
\end{aligned}
\)
Also
\(
\begin{aligned}
a & =\sqrt{a_x^2+a_y^2+a_z^2} \\
& =\sqrt{4+9+16}=\sqrt{29}
\end{aligned}
\)
and
\(
b=\sqrt{b_x^2+b_y^2+b_z^2}=\sqrt{16+9+4}=\sqrt{29} .
\)
Thus, \(\quad \cos \theta=\frac{25}{29}\)
or,
\(\theta=\cos ^{-1}\left(\frac{25}{29}\right)\)

Q13: If \(\vec{A}=2 \vec{i}-3 \vec{j}+7 \vec{k}, \quad \vec{B}=\vec{i}+2 \vec{k} \quad\) and \(\quad \vec{C}=\vec{j}-\vec{k} \quad\) find \(\vec{A} \cdot(\vec{B} \times \vec{C})\)

Answer:

\(
\begin{aligned}
\vec{B} \times \vec{C} & =(\vec{i}+2 \vec{k}) \times(\vec{j}-\vec{k}) \\
& =\vec{i} \times(\vec{j}-\vec{k})+2 \vec{k} \times(\vec{j}-\vec{k}) \\
& =\vec{i} \times \vec{j}-\vec{i} \times \vec{k}+2 \vec{k} \times \vec{j}-2 \vec{k} \times \vec{k} \\
& =\vec{k}+\vec{j}-2 \vec{i}-0=-2 \vec{i}+\vec{j}+\vec{k} . \\
\vec{A} \cdot(\vec{B} \times \vec{C}) & =(2 \vec{i}-3 \vec{j}+7 \vec{k}) \cdot(-2 \vec{i}+\vec{j}+\vec{k}) \\
& =(2)(-2)+(-3)(1)+(7)(1) \\
& =0 .
\end{aligned}
\)

Q14: The volume of a sphere is given by
\(
V=\frac{4}{3} \pi R^3
\)
where \(R\) is the radius of the sphere. (a) Find the rate of change of volume with respect to \(R\). (b) Find the change in volume of the sphere as the radius is increased from \(20.0 \mathrm{~cm}\) to \(20.1 \mathrm{~cm}\). Assume that the rate does not appreciably change between \(R=20.0 \mathrm{~cm}\) to \(R=20.1 \mathrm{~cm}\).

Answer: (a) \(\quad V=\frac{4}{3} \pi R^3\)
or, \(\quad \frac{d V}{d R}=\frac{4}{3} \pi \frac{d}{d R}(R)^3=\frac{4}{3} \pi \cdot 3 R^2=4 \pi R^2\).
(b) At \(R=20 \mathrm{~cm}\), the rate of change of volume with the radius is
\(
\begin{aligned}
\frac{d V}{d R} & =4 \pi R^2=4 \pi\left(400 \mathrm{~cm}^2\right) \\
& =1600 \pi \mathrm{cm}^2 .
\end{aligned}
\)
The change in volume as the radius changes from \(20^{\circ} 0 \mathrm{~cm}\) to \(20^{\circ} 1 \mathrm{~cm}\) is
\(
\begin{aligned}
\Delta V & =\frac{d V}{d R} \Delta R \\
& =\left(1600 \pi \mathrm{cm}^2\right)(0.1 \mathrm{~cm}) \\
& =160 \pi \mathrm{cm}^3 .
\end{aligned}
\)

Q15: Find the derivative of the following functions with respect to \(x\). (a) \(y=x^2 \sin x\),
(b) \(y=\frac{\sin x}{x}\) and
(c) \(y=\sin \left(x^2\right)\).

Answer: (a)
\(
\begin{aligned}
y & =x^2 \sin x \\
\frac{d y}{d x} & =x^2 \frac{d}{d x}(\sin x)+(\sin x) \frac{d}{d x}\left(x^2\right) \\
& =x^2 \cos x+(\sin x)(2 x) \\
& =x(2 \sin x+x \cos x) .
\end{aligned}
\)
(b)
\(
\begin{aligned}
y & =\frac{\sin x}{x} \\
\frac{d y}{d x} & =\frac{x \frac{d}{d x}(\sin x)-\sin x\left(\frac{d x}{d x}\right)}{x^2} \\
& =\frac{x \cos x-\sin x}{x^2} .
\end{aligned}
\)
(c)
\(
\begin{aligned}
\frac{d y}{d x} & =\frac{d}{d x^2}\left(\sin x^2\right) \cdot \frac{d\left(x^2\right)}{d x} \\
& =\cos x^2(2 x) \\
& =2 x \cos x^2 .
\end{aligned}
\)

Q16: Find the maximum or minimum values of the function \(y=x+\frac{1}{x}\) for \(x>0\)

Answer:

\(
\begin{aligned}
y & =x+\frac{1}{x} \\
\frac{d y}{d x} & =\frac{d}{d x}(x)+\frac{d}{d x}\left(x^{-1}\right) \\
& =1+\left(-x^{-2}\right) \\
& =1-\frac{1}{x^2}
\end{aligned}
\)
For \(y\) to be maximum or minimum,
\(
\begin{aligned}
& \frac{d y}{d x}=0 \\
& -\frac{1}{x^2}=0
\end{aligned}
\)
or,
\(
1-\frac{1}{x^2}=0
\)
Thus,
\(
x=1 \text { or }-1
\)
For \(x>0\) the only possible maximum or minimum is at \(x=1\). At \(x=1, y=x+\frac{1}{x}=2\).

Near \(x=0, y=x+\frac{1}{x}\) is very large because of the term \(\frac{1}{x}\). For very large \(x\), again \(y\) is very large because of the \(\operatorname{term} x\). Thus \(x=1\) must correspond to a minimum. Thus, \(y\) has only a minimum for \(x>0\). This minimum occurs at \(x=1\) and the minimum value of \(y\) is \(y=2\).

Q17: Figure (2-W6) shows the curve \(y=x^2\). Find the area of the shaded part between \(x=0\) and \(x=6\).

Answer: The area can be divided into strips by drawing ordinates between \(x=0\) and \(x=6\) at a regular interval of \(d x\). Consider the strip between the ordinates at \(x\) and \(x+d x\). The height of this strip is \(y=x^2\). The area of this strip is \(d A=y d x=x^2 d x\).
The total area of the shaded part is obtained by summing up these strip areas with \(x\) varying from 0 to 6. Thus
\(
\begin{aligned}
A & =\int_0^6 x^2 d x \\
& =\left[\frac{x^3}{3}\right]_0^6=\frac{216-0}{3}=72 .
\end{aligned}
\)

Q18: \(\text { Evaluate } \int_0^t A \sin \omega t d t \text { where } A \text { and } \omega \text { are constants. }\)

Answer:

\(
\begin{aligned}
& \int_0^t A \sin \omega t d t \\
& =A\left[\frac{-\cos \omega t}{\omega}\right]_0^t=\frac{A}{\omega}(1-\cos \omega t) .
\end{aligned}
\)

Q19: The velocity \(v\) and displacement \(x\) of a particle executing simple harmonic motion are related as
\(
v \frac{d v}{d x}=-\omega^2 x .
\)
At \(x=0, v=v_0\). Find the velocity \(v\) when the displacement becomes \(x\).

Answer: We have
\(
\begin{aligned}
& v \frac{d v}{d x}=-\omega^2 x \\
& v d v=-\omega^2 x d x
\end{aligned}
\)
or,
\(
v d v=-\omega^2 x d x
\)
or,
\(
\int_{v_0}^v v d v=\int_0^x-\omega^2 x d x \dots(i)
\)
When summation is made on \(-\omega^2 x d x\) the quantity to be varied is \(x\). When summation is made on \(v d v\) the quantity to be varied is \(v\). As \(x\) varies from 0 to \(x\) the velocity varies from \(v_0\) to \(v\). Therefore, on the left, the limits of integration are from \(v_0\) to \(v\) and on the right they are from 0 to \(x\). Simplifying (i),
\(
\left[\frac{1}{2} v^2\right]_{v_0}^v=-\omega^2\left[\frac{x^2}{2}\right]_0^x
\)
\(
\begin{aligned}
& \text { or, } \quad \frac{1}{2}\left(v^2-v_0^2\right)=-\omega^2 \frac{x^2}{2} \\
& \text { or, } \quad v^2=v_0^2-\omega^2 x^2 \\
& v=\sqrt{v_0^2-\omega^2 x^2} . \\
&
\end{aligned}
\)

Q20: The charge flown through a circuit in the time interval between \(t\) and \(t+d t\) is given by \(d q=e^{-t / \tau} d t\), where \(\tau\) is a constant. Find the total charge flown through the circuit between \(t=0\) to \(t=\tau\).

Answer: The total charge flown is the sum of all the \(d q\) ‘s for \(t\) varying from \(t=0\) to \(t=\tau\). Thus, the total charge flown is
\(
\begin{aligned}
Q & =\int_0^\tau e^{-t / \tau} d t \\
& =\left[\frac{e^{-t / \tau}}{-1 / \tau}\right]_0^\tau=\tau\left(1-\frac{1}{e}\right) .
\end{aligned}
\)

Q21: \(\text { Evaluate }(21 \cdot 6002+234+2732 \cdot 10) \times 13 \text {. }\)

Answer:

The three numbers are arranged with their decimal points aligned (shown on the left part above). The column just left to the decimals has 4 as the doubtful digit. Thus, all the numbers are rounded to this column. The rounded numbers are shown on the right part above. The required expression is \(2988 \times 13=38844\). As 13 has only two significant digits the product should be rounded off after two significant digits. Thus the result is 39000.

Q22: A helicopter on a flood relief mission, flying horizontally with a speed \(u\) at an altitude \(H\), has to drop a food packet for a victim standing on the ground. At what distance from the victim should the packet be dropped? The victim stands in the vertical plane of the helicopter’s motion.

Answer: The velocity of the food packet at the time of release is \(u\) and is horizontal. The vertical velocity at the time of release is zero.

Vertical motion: If \(t\) be the time taken by the packet to reach the victim, we have for vertical motion,
\(
H=\frac{1}{2} g t^2 \quad \text { or, } \quad t=\sqrt{\frac{2 H}{g}} \text {. } \dots(i)
\)
Horizontal motion: If \(D\) be the horizontal distance travelled by the packet, we have \(D=u t\). Putting \(t\) from (i),
\(
D=u \sqrt{\frac{2 H}{g}}
\)
The distance between the victim and the packet at the time of release is
\(
\sqrt{D^2+H^2}=\sqrt{\frac{2 u^2 H}{g}+H^2}
\)

Q23: A particle is projected horizontally with a speed \(u\) from the top of a plane inclined at an angle \(\theta\) with the horizontal. How far from the point of projection will the particle strike the plane?

Answer: Take \(X, Y\)-axes as shown in figure (3-W8). Suppose that the particle strikes the plane at a point \(P\) with coordinates \((x, y)\). Consider the motion between \(A\) and \(P\).

Motion in \(x\)-direction:
Initial velocity \(=u\)
Acceleration \(=0\)
\(
x=u t .\dots(i)
\)
Motion in \(y\)-direction:
Initial velocity \(=0\)
Acceleration \(=g\)
\(
y=\frac{1}{2} g t^2 .\dots(ii)
\)
Eliminating \(t\) from (i) and (ii)
\(
y=\frac{1}{2} g \frac{x^2}{u^2}
\)
Also
\(
y=x \tan \theta .
\)
Thus, \(\frac{g x^2}{2 u^2}=x \tan \theta\) giving \(x=0\), or, \(\frac{2 u^2 \tan \theta}{g}\).
Clearly the point \(P\) corresponds to \(x=\frac{2 u^2 \tan \theta}{g}\),
then \(y=x \tan \theta=\frac{2 u^2 \tan ^2 \theta}{g}\).
The distance \(A P=l=\sqrt{x^2+y^2}\)
\(
\begin{aligned}
& =\frac{2 u^2}{g} \tan \theta \sqrt{1+\tan ^2 \theta} \\
& =\frac{2 u^2}{g} \tan \theta \sec \theta .
\end{aligned}
\)

Q24: A projectile is fired with a speed \(u\) at an angle \(\theta\) with the horizontal. Find its speed when its direction of motion makes an angle \(\alpha\) with the horizontal.

Answer: Let the speed be \(v\) when it makes an angle \(\alpha\) with the horizontal. As the horizontal component of velocity remains constant,
\(
\begin{aligned}
v \cos \alpha & =u \cos \theta \\
v & =u \cos \theta \sec \alpha .
\end{aligned}
\)

Q25: A bullet is fired horizontally aiming at an object which starts falling at the instant the bullet is fired. Show that the bullet will hit the object.

Answer: The situation is shown in Figure (3-W10). The object starts falling from the point \(B\). Draw a vertical line \(B C\) through \(B\). Suppose the bullet reaches the line \(B C\) at a point \(D\) and it takes a time \(t\) in doing so.

Consider the vertical motion of the bullet. The initial vertical velocity \(=0\). The distance travelled vertically \(=B D=\frac{1}{2} g t^2\). In time \(t\) the object also travels a distance \(\frac{1}{2} g t^2=B D\). Hence at time \(t\), the object will also be at the same point \(D\). Thus, the bullet hits the object at point \(D\).

Q26: A man can swim in still water at a speed of \(3 \mathrm{~km} / \mathrm{h}\). He wants to cross a river that flows at \(2 \mathrm{~km} / \mathrm{h}\) and reach the point directly opposite to his starting point. (a) In which direction should he try to swim (that is, find the angle his body makes with the river flow)? (b) How much time will he take to cross the river if the river is \(500 \mathrm{~m}\) wide?

Answer: (a) The situation is shown in figure (3-W11). The \(X\)-axis is chosen along the river flow and the origin at the starting position of the man. The direction of the velocity of man with respect to ground is along the \(Y\)-axis (perpendicular to the river). We have to find the direction of velocity of the man with respect to water.
Let \(\vec{v}_{r, g}=\) velocity of the river with respect to the ground \(=2 \mathrm{~km} / \mathrm{h}\) along the \(X\)-axis

\(\vec{v}_{m, r}=\) velocity of the man with respect to the river \(=3 \mathrm{~km} / \mathrm{h}\) making an angle \(\theta\) with the \(Y\)-axis and \(\vec{v}_{m, g}=\) velocity of the man with respect to the ground along the \(Y\)-axis.
We have
\(
\vec{v}_{m, g}=\vec{v}_{m, r}+\vec{v}_{r, g} \dots(i)
\)
Takingcomponentsalongthe \(X\)-axis
\(
0=-(3 \mathrm{~km} / \mathrm{h}) \sin \theta+2 \mathrm{~km} / \mathrm{h}
\)
or,
\(
\sin \theta=\frac{2}{3} \text {. }
\)
(b) Taking components in equation (i) along the \(Y\)-axis,
\(
\begin{aligned}
& v_{m, g}=(3 \mathrm{~km} / \mathrm{h}) \cos \theta+0 \\
& v_{m, g}=\sqrt{ } 5 \mathrm{~km} / \mathrm{h} .
\end{aligned}
\)
or,
\(
v_{m, g}=\sqrt{5} \mathrm{~km} / \mathrm{h} \text {. }
\)
\(
\begin{aligned}
\text { Time } & =\frac{\text { Displacement in } y \text { direction }}{\text { Velocity in } y \text { direction }} \\
& =\frac{0.5 \mathrm{~km}}{\sqrt{5} \mathrm{~km} / \mathrm{h}}=\frac{\sqrt{ } 5}{10} \mathrm{~h} .
\end{aligned}
\)

Q27: A man can swim at a speed of \(3 \mathrm{~km} / \mathrm{h}\) in still water. He wants to cross a \(500 \mathrm{~m}\) wide river flowing at \(2 \mathrm{~km} / \mathrm{h}\). He keeps himself always at an angle of \(120^{\circ}\) with the river flow while swimming.
(a) Find the time he takes to cross the river. (b) At what point on the opposite bank will he arrive?

Answer: The situation is shown in figure (3-W12).

Here \(\vec{v}_{r, g}=\) velocity of the river with respect to the ground
\(\vec{v}_{m, r}=\) velocity of the man with respect to the river
\(\vec{v}_{m, g}=\) velocity of the man with respect to the ground.
(a) We have,
\(
\vec{v}_{m, g}=\vec{v}_{m, r}+\vec{v}_{r, g} \dots(i)
\)
Hence, the velocity with respect to the ground is along \(A C\). Taking \(y\)-components in equation (i), \(\vec{v}_{m, g} \sin \theta=3 \mathrm{~km} / \mathrm{h} \cos 30^{\circ}+2 \mathrm{~km} / \mathrm{h} \cos 90^{\circ}=\frac{3 \sqrt{ } 3}{2} \mathrm{~km} / \mathrm{h}\).
Time taken to cross the river
\(
\begin{aligned}
& =\frac{\text { displacement along the } Y_{\text {-axis }}}{\text { velocity along the } Y_{\text {-axis }}} \\
& =\frac{1 / 2 \mathrm{~km}}{3 \sqrt{ } 3 / 2 \mathrm{~km} / \mathrm{h}}=\frac{1}{3 / 3} \mathrm{~h} .
\end{aligned}
\)
(b) Taking \(x\)-components in equation (i),
\(
\begin{aligned}
\vec{v}_{m, g} \cos \theta & =-3 \mathrm{~km} / \mathrm{h} \sin 30^{\circ}+2 \mathrm{~km} / \mathrm{h} \\
& =\frac{1}{2} \mathrm{~km} / \mathrm{h} .
\end{aligned}
\)
Displacement along the \(X\)-axis as the man crosses the river
\(
\begin{aligned}
& =(\text { velocity along the } X \text {-axis }) \cdot(\text { time) } \\
& =\left(\frac{1 \mathrm{~km}}{2 \mathrm{~h}}\right) \times\left(\frac{1}{3 \sqrt{3}} \mathrm{~h}\right)=\frac{1}{6 \sqrt{3}} \mathrm{~km} .
\end{aligned}
\)

Q28: A man standing on a road has to hold his umbrella at \(30^{\circ}\) with the vertical to keep the rain away. He throws the umbrella and starts running at \(10 \mathrm{~km} / \mathrm{h}\). He finds that raindrops are hitting his head vertically. Find the speed of raindrops with respect to (a) the road, (b) the moving man.

Answer: When the man is at rest with respect to the ground, the rain comes to him at an angle \(30^{\circ}\) with the vertical. This is the direction of the velocity of raindrops with respect to the ground. The situation when the man runs is shown in the figure (3-W13b).

Here \(\vec{v}_{r, g}=\) velocity of the rain with respect to the ground \(\vec{v}_{m, g}=\) velocity of the man with respect to the ground and \(\vec{v}_{r, m}=\) velocity of the rain with respect to the man. We have, \(\quad \vec{v}_{r, g}=\vec{v}_{r, m}+\vec{v}_{m, g} \dots(i)\).
Taking horizontal components, equation (i) gives
\(
\begin{aligned}
& v_{r, g} \sin 30^{\circ}=v_{m, g}=10 \mathrm{~km} / \mathrm{h} \\
& \text { or, } v_{r, g}=\frac{10 \mathrm{~km} / \mathrm{h}}{\sin 30^{\circ}}=20 \mathrm{~km} / \mathrm{h},
\end{aligned}
\)
Taking vertical components, equation (i) gives
\(
v_{r, g} \cos 30^{\circ}=v_{r, m}
\)
or,
\(
\begin{aligned}
v_{r, m} & =(20 \mathrm{~km} / \mathrm{h}) \frac{\sqrt{3}}{2} \\
& =10 \sqrt{ } 3 \mathrm{~km} / \mathrm{h} .
\end{aligned}
\)

Q29: A man running on a horizontal road at \(8 \mathrm{~km} / \mathrm{h}\) finds the rain falling vertically. He increases his speed to \(12 \mathrm{~km} / \mathrm{h}\) and finds that the drops make angle \(30^{\circ}\) with the vertical. Find the speed and direction of the rain with respect to the road.

Answer: We have \(\quad \vec{v}_{\text {rain, road }}=\vec{v}_{\text {rain, man }}+\vec{v}_{\text {man, road }} \quad \ldots\) (i)
The two situations given in the problem may be represented by the following figure.

\(v_{\text {rain, road }}\) is the same in magnitude and direction in both the figures.
Taking horizontal components in equation (i) for the figure (3-W14a),
\(
v_{\text {rain, road }} \sin \alpha=8 \mathrm{~km} / \mathrm{h} \text {. }\dots(ii)
\)
Now consider Figure (3-W14b). Draw a line \(O A \perp v_{\text {rain, man }}\) as shown.
Taking components in equation (i) along the line \(O A\). \(v_{\text {rain, } \text { road }} \sin \left(30^{\circ}+\alpha\right)=12 \mathrm{~km} / \mathrm{h} \cos 30^{\circ} \dots(iii)\).
From (ii) and (iii),
\(
\begin{aligned}
\frac{\sin \left(30^{\circ}+\alpha\right)}{\sin \alpha} & =\frac{12 \times \sqrt{3}}{8 \times 2} \\
\text { or, } \quad \frac{\sin 30^{\circ} \cos \alpha+\cos 30^{\circ} \sin \alpha}{\sin \alpha} & =\frac{3 \sqrt{ } 3}{4}
\end{aligned}
\)
\(
\text { or, } \quad \frac{1}{2} \cot \alpha+\frac{\sqrt{ } 3}{2}=\frac{3 \sqrt{ } 3}{4}
\)
or, \(
\cot \alpha=\frac{\sqrt{3}}{2}
\)
or, \(
\alpha=\cot ^{-1} \frac{\sqrt{3}}{2} .
\)
From (ii),
\(
v_{\text {rain, road }}=\frac{8 \mathrm{~km} / \mathrm{h}}{\sin \alpha}=4 \sqrt{ } 7 \mathrm{~km} / \mathrm{h} \text {. }
\)

Q30: Three particles \(A, B\) and \(C\) are situated at the vertices of an equilateral triangle \(A B C\) of side \(d\) at \(t=0\). Each of the particles moves with constant speed \(v\). A always has its velocity along \(A B, B\) along \(B C\) and \(C\) along \(C A\). At what time will the particles meet each other?

Answer: The motion of the particles is roughly sketched in Figure (3-W15). By symmetry, they will meet at the centroid \(O\) of the triangle. At any instant, the particles will form an equilateral triangle \(A B C\) with the same centroid \(O\). Concentrate on the motion of any one particle, say A. At any instant, its velocity makes angle \(30^{\circ}\) with \(A O\).

The component of this velocity along \(A O\) is \(v \cos 30^{\circ}\). This component is the rate of decrease of the distance \(A O\). Initially,
\(
A O=\frac{2}{3} \sqrt{d^2-\left(\frac{d}{2}\right)^2}=\frac{d}{\sqrt{3}} .
\)
Therefore, the time taken for \(A O\) to become zero
\(
=\frac{d / \sqrt{3}}{v \cos 30^{\circ}}=\frac{2 d}{\sqrt{3} v \times \sqrt{3}}=\frac{2 d}{3 v} \text {. }
\)

Alternative: Velocity of \(A\) is \(v\) along \(A B\). The velocity of \(B\) is along \(B C\). Its component along \(B A\) is \(v \cos 60^{\circ}\) \(=v / 2\). Thus, the separation \(A B\) decreases at the rate
\(
v+\frac{v}{2}=\frac{3 v}{2}
\)
Since this rate is constant, the time taken in reducing the separation \(A B\) from \(d\) to zero is
\(
t=\frac{d}{\frac{3 v}{2}}=\frac{2 d}{3 v}
\)

Q31: A hill is \(500 \mathrm{~m}\) high. Supplies are to be sent across the hill using a canon that can hurl packets at a speed of \(125 \mathrm{~m} / \mathrm{s}\) over the hill. The canon is located at a distance of \(800 \mathrm{~m}\) from the foot of the hill and can be moved on the ground at a speed of \(2 \mathrm{~m} / \mathrm{s}\); so that its distance from the hill can be adjusted. What is the shortest time in which a packet can reach on the ground across the hill? Take \(g=10 \mathrm{~m} / \mathrm{s}^2\).

Answer: According to the problem, the speed of packets \(=125 \mathrm{~m} / \mathrm{s}\), the height of the hill \(=500 \mathrm{~m}\), the distance between the cannon and the foot of the hill, \(d=800 \mathrm{~m}\)
To cross the hill in the shortest time, then the vertical component of the velocity should be minimum so that it just crosses the height of the hill.

Distance through which canon has to be moved \(=800-750=50 \mathrm{~m}\) Speed with which canon can move \(=2\)
\(\mathrm{m} / \mathrm{s}\)
\(
u_y=\sqrt{2 g h} \geq \sqrt{2 \times 10 \times 500} \geq 100 \mathrm{~m} / \mathrm{s}
\)
But \(u^2=u_x^2+u_y^2\)
\(\therefore\) Horizontal component of initial velocity,
\(
u_x=\sqrt{u^2-u_y^2}=\sqrt{(125)^2-(100)^2}=75 \mathrm{~m} / \mathrm{s}
\)
Time taken by the packet to reach the top of the hill,
\(
t=\sqrt{\frac{2 h}{g}}=\sqrt{\frac{2 \times 500}{10}}=10 \mathrm{~s}
\)
Time taken to reach the ground from the top of the hill \(t^{\prime}=t=10 \mathrm{~s}\). Horizontal distance travelled in \(10 \mathrm{~s}\),
\(
x=u_x \times t=75 \times 10=750 \mathrm{~m}
\)
so the distance between Canon and hill is \(750 \mathrm{~m}\)
Distance for which the canon needs to move \(=800-750=50 \mathrm{~m}\)
Time taken for the canon to move \(50 \mathrm{~m}=\frac{50}{2}=25 \mathrm{sec}\)
so the total time taken by the packet \(=25+10+10=45\) seconds

Q32: A gun can fire shells with maximum speed \(v_o\) and the maximum horizontal range that can be achieved is \(R=\frac{v_o{ }^2}{g}\).
If a target farther away by distance \(\Delta x\) (beyond R) has to be hit with the same gun (Fig 4.5), show that it could be achieved by raising the gun to a height at least \(h=\Delta x\left[1+\frac{\Delta x}{R}\right]\)

Answer: The solution to this problem can be given in two ways:
i) the target is present at the horizontal distance of \((R+\Delta x)\) and is \(h\) meters below the projection point. \(Y=-h\)
ii) the motion of projectile starting from point \(\mathrm{P}\) till reaching point \(\mathrm{T}\). vertical height covered is -h and horizontal range is \(\delta x\)
\(
R=\frac{v^2}{g}
\)
max range of a projectile is given by here, \(\theta=45\)
we assume that the gun is raised to a height \(\mathrm{h}\) to hit the target \(\mathrm{T}\) total range \(=(R+\Delta x)\)
the horizontal component of velocity is, \(v \cos \theta\)
horizontal velocity at \(P=V_x=-V \cos \theta\)
vertical velocity at \(P=V_y=V \sin \theta\)
now, \(h=u t+\frac{1}{2} a t^2\)
So, \(h=-V \sin \theta t+\frac{1}{2} g t^2 \quad————(1)\)
\(V \cos \theta . t=(R+\Delta x)\)
\(t=\frac{(R+\Delta x)}{V \cos \theta}\)
Substituting the value of \(\mathrm{t}\) in (1) we get,
\(
\begin{aligned}
& h=-V \sin \theta\left(\frac{(R+\Delta x)}{V \cos \theta}\right)+\frac{1}{2} g\left(\frac{(R+\Delta x)}{V \cos \theta}\right)^2 \\
& h=-(R+\Delta x)+\frac{1}{R}\left(R^2+\Delta x^2+2 R \Delta x\right)=-R-\Delta x+R+\frac{\Delta x^2}{R}+2 \Delta x \\
& h=\Delta x\left[1+\frac{\Delta x}{R}\right]
\end{aligned}
\)
hence proved.

Q33: A particle is projected in the air at an angle \(\beta\) to a surface which itself is inclined at an angle \(\alpha\) to the horizontal (Fig below).
(a) Find an expression of range on the plane surface (the distance on the plane from the point of projection at which the particle will hit the surface).
(b) Time of flight.
(c) \(\beta\) at which range will be maximum.

Answer: (a) expression of range on the plane surface
Now, \(a_y=-g \cos \alpha\) and \(a_x=g \sin \alpha\), \(y=0\) at \(\mathrm{O}\) and \(\mathrm{P}\),
So
\(u_y=v \sin \beta\) where \(\mathrm{t}=\mathrm{T}\)
We calculate the Time of Flight Part (b) before Part (a)
(b) the motion of a particle along \(Y\) axis
\(
\begin{aligned}
& s=u t+\frac{1}{2} g t^2 \\
& s=0, u=v \sin \beta, g=a=-g \cos \alpha, t=T \\
& 0=v \sin \beta \cdot T+\frac{1}{2}(-g \cos \alpha) T^2 \\
& T\left[v \sin \beta-T \cdot \frac{g}{2} \cos \alpha\right]=0 \\
& \frac{T g}{2} \cdot \cos \alpha=v \sin \beta \\
& \text { So, } T=2 v \sin \beta / g \cdot \cos \alpha
\end{aligned}
\)
(a) Now we continue the part a
\(
\begin{aligned}
& T=2 v \sin \beta / g \cdot \cos \alpha \\
& s=u t+\frac{1}{2} g t^2 \\
& L=v \cos \beta(T)+\frac{1}{2}(-g \sin \alpha) T^2 \\
& L=2 v^2 \sin \beta[\cos \beta \cos \alpha-\sin \beta \sin \alpha] / g \cos ^2 \alpha \\
& L=2 v^2 \sin \beta \times \cos [\alpha+\beta] / g \cos ^2 \alpha
\end{aligned}
\)
(c) on the axis \(\mathrm{X}, \mathrm{L}\) will be maximum when \(\sin \beta \cos [\alpha+\beta]\) will be maximum
\(
\begin{aligned}
& \text { let } z=\sin \beta \cos [\alpha+\beta] \\
& =\sin \beta[\cos \beta \cos \alpha-\sin \beta \sin \alpha] \\
& =\frac{1}{2}[\cos \alpha 2 \sin \beta+\sin \alpha \cos 2 \beta-\sin \alpha] \\
& =\frac{1}{2}[\sin (2 \beta+\alpha)-\sin \alpha]
\end{aligned}
\)
In order to make Z maximum, we put \([\sin (2 \beta+\alpha)-\sin \alpha]=1\)
Opening the brackets, we get \((2 \beta+\alpha)=90\)
\(
2 \beta=90-\alpha
\)
\(
\beta=\frac{\pi}{4}-\frac{\alpha}{2}
\)

Q34: A particle falling vertically from a height hits a plane surface inclined to horizontal at an angle \(\theta[latex] with speed [latex]v_o\) and rebounds elastically (Fig 4.7). Find the distance along the plane where if will hit second time.

Answer: \(\text { Considering } x \text { and } y \text {-axis as shown in the diagram. }\)

For the motion of the projectile from \(O\) to \(A\)
\(
\begin{aligned}
& y=0, u_y=v_0 \cos \theta \\
& a_y=-g \cos \theta, t=T
\end{aligned}
\)
Applying the equation of kinematics,
\(
\begin{aligned}
& \mathrm{y}=u_y t+\frac{1}{2} a_y t^2 \\
& \Rightarrow 0=v_0 \cos \theta T+\frac{1}{2}(-g \cos \theta) T^2 \\
& \Rightarrow T\left[v_0 \cos \theta-\frac{g \cos \theta T}{2}\right]=0 \\
& \mathrm{~T}=\frac{2 v_0 \cos \theta}{g \cos \theta}
\end{aligned}
\)
As \(T=0\), corresponds to point \(O\)
Hence, \(\mathrm{T}=\frac{2 v_0}{g}\)
Now considering motion along \(O X\).
\(
\begin{aligned}
& x=L \\
& u_x=v_0 \sin \theta \\
& a_x=g \sin \theta \\
& t=T=\frac{2 v_0}{g}
\end{aligned}
\)
Applying the equation of kinematics,
\(
x=u_x t+\frac{1}{2} a_x t^2
\)
\(
\begin{aligned}
& \Rightarrow L=v_0 \sin \theta t+\frac{1}{2} g \sin \theta t^2 \\
& =\left(v_0 \sin \theta\right)(T)+\frac{1}{2} g \sin \theta T^2 \\
& =\left(v_0 \sin \theta\right)\left(\frac{2 v_0}{g}\right)+\frac{1}{2} g \sin \theta \times\left(\frac{2 v_0}{g}\right)^2 \\
& =\frac{2 v_0^2}{g} \sin \theta+\frac{1}{2} g \sin \theta \times \frac{4 v_0^2}{g^2} \\
& =\frac{2 v_0^2}{g}[\sin \theta+\sin \theta] \\
& \Rightarrow L=\frac{4 v_0^2}{g} \sin \theta
\end{aligned}
\)

Q35: A girl riding a bicycle with a speed of \(5 \mathrm{~m} / \mathrm{s}\) towards the north direction, observes rain falling vertically down. If she increases her speed to \(10 \mathrm{~m} / \mathrm{s}\), rain appears to meet her at \(45^{\circ}\) to the vertical. What is the speed of the rain? In what direction does rainfall as observed by a ground-based observer?

Answer: \(V_{\text {rg}}\) is the velocity of rain that appears to the girl.
We must draw all vectors in the reference frame of the ground-based observer.
Assume north to be \(\hat{i}\) direction and vertically downward to be \((-\hat{j})\).
Let the rain velocity
\(
v_r=a \hat{i}+b \hat{j}
\)
Case I:

According to the problem, the velocity of girl \(=v_g=(5 \mathrm{~m} / \mathrm{s}) \hat{i}\)
Let \(v_{r g}=\) velocity of rain w.r.t girl
\(
=v_r-v_g=(a \hat{i}+b \hat{j})-5 \hat{i}=(a-5) \hat{i}+b \hat{j}
\)
According to the question, rain appears to fall vertically downward.
Hence, \(\quad a-5=0 \Rightarrow a=5\)

Case II:

Now velocity of the girl after increasing her speed,
\(
v_g=(10 \mathrm{~m} / \mathrm{s}) \hat{i}
\)
\(
\therefore \quad \begin{aligned}
v_{r g} & =v_r-v_g \\
& =(a \hat{i}+b \hat{j})-10 \hat{i}=(a-10) \hat{i}+b \hat{j}
\end{aligned}
\)
According to question rain appears to fall at \(45^{\circ}\) to the vertical, hence \(\tan 45^{\circ}=\frac{b}{a-10}=1\)
\(
\Rightarrow \quad b=a-10=5-10=-5
\)
Hence, velocity of rain \(=a \hat{i}+b \hat{j}\)
\(
\Rightarrow \quad v_r=5 \hat{i}-5 \hat{j}
\)
Speed of rain
\(
=\left|v_r\right|=\sqrt{(5)^2+(-5)^2}=\sqrt{50}=5 \sqrt{2} \mathrm{~m} / \mathrm{s}
\)

Q36: A river is flowing due east with a speed \(3 \mathrm{~m} / \mathrm{s}\). A swimmer can swim in still water at a speed of \(4 \mathrm{~m} / \mathrm{s}\) (Fig. 4.8).
(a) If the swimmer starts swimming due north, what will be his resultant velocity (magnitude and direction)?
(b) If he wants to start from point \(\mathrm{A}\) on the south bank and reach opposite point \(B\) on the north bank,
(a) which direction should he swim?
(b) what will be his resultant speed?
(c) From two different cases as mentioned in (a) and (b) above, in which case will he reach the opposite bank in a shorter time?

Answer: Key concept: Analysis of Relative Velocity in Different Situations:
The motion of a boat (or a swimmer) in running water, say, a river.

(a) If the swimmer starts swimming due north, what will be his resultant velocity
\(V_s=4 \mathrm{~m} / \mathrm{s}\) due north
\(V_r=4 \mathrm{~m} / \mathrm{s}\) due east
Now since both directions are perpendicular,
\(
\left|V_r\right|^2=4^2+3^2=5 \mathrm{~m} / \mathrm{s}
\)
\(\tan \theta=\frac{V_r}{V_s}=0.75=36^{\circ} 54^{\prime}\) in the North direction.

(b) The swimmer wants to start from point \(A\) on the south bank and reaches the opposite point B on the north bank.
The swimmer makes an angle \(\theta\) with the north.

From the figure we have the relation,
\(
V^2=v_s^2-v_r^2=16-9=7
\)
Hence \({ }^v=\sqrt{7} \mathrm{~m} / \mathrm{s}\)
Now we calculate the value of \(\theta\) through the below formula,
\(
\tan \theta=\frac{v_r}{v}=\frac{3 \sqrt{7}}{7}=1.13
\)
So, \(\theta=48^{\circ} 29^{\prime} 30^{\prime \prime}\) in th edirection from North to West
c) we need to find from the above two scenarios that for the swimmer to reach the opposite bank in the shorter time
we know that the velocity component perpendicular to the river is \(4 \mathrm{~m} / \mathrm{s}\)
let us assume the width of the river to be ‘ \(w\) ‘
Time taken – North
\(
\frac{w}{4}=t 1
\)
time taken in part b) when \(v=\sqrt{7} \mathrm{~m} / \mathrm{s}\)
\(
\frac{w}{\sqrt{7}}=t 2
\)
taking ratio,
\(
\begin{aligned}
& \frac{t 1}{t 2}=\frac{\left(\frac{w}{4}\right)}{\left(\frac{w}{\sqrt{7}}\right)} \\
& 4 t 1=\sqrt{7} t 2
\end{aligned}
\)
Now as, \(4>\sqrt{7}\)
\(
t 1<t 2
\)
So, the swimmer will take a shorter time in case (a)

Q37: A cricket fielder can throw the cricket ball with a speed \(v_0\). If he throws the ball while running with speed \(u\) at an angle \(\theta\) to the horizontal, find
(a) the effective angle to the horizontal at which the ball is projected in air as seen by a spectator.
(b) what will be time of flight?
(c) what is the distance (horizontal range) from the point of projection at which the ball will land?
(d) find \(\theta\) at which he should throw the ball that would maximise the horizontal range as found in (iii).
(e) how does \(\theta\) for maximum range change if \(u>v_0, u=v_0, u<v_0\) ?
(f) how does \(\theta\) in (v) compare with that for \(u=0\) (i.e. \(45^{\circ}\) )?

Answer: The observer on the ground (spectator) observes that the \(\mathrm{x}\)-component of the ball is more because of the speed of the fielder. As shown in the adjacent diagram,

So, initial velocity in \(x\)-direction
So, initial velocity in \(x\)-direction,
\(
u_x=u+v_0 \cos \theta
\)
Initial velocity in \(y\)-direction,
\(
u_y=v_0 \sin \theta
\)
where the angle of projection is \(\theta\).
(a) Now, the angle of projection with horizontal seen by the spectator will be
\(
\begin{array}{r}
\tan \theta=\frac{u_y}{u_x}=\frac{u_0 \sin \theta}{u+u_0 \cos \theta} \\
\Rightarrow \quad \theta=\tan ^{-1}\left(\frac{v_0 \sin \theta}{u+v_0 \cos \theta}\right)
\end{array}
\)

(b) As net displacement along \(y\)-axis is zero over time period \(T\) (time of flight).
\(
y=0, u_y=v_0 \sin \theta, a_y=-g, t=T
\)
We know that \(y=u_y t+\frac{1}{2} a_y t^2\)
\(
\begin{aligned}
& \Rightarrow \quad 0=v_0 \sin \theta T+\frac{1}{2}(-g) T^2 \\
& \Rightarrow \quad T\left[v_0 \sin \theta-\frac{g}{2} T\right]=0 \Rightarrow T=0, \frac{2 v_0 \sin \theta}{g}
\end{aligned}
\)
\(T=0\), corresponds to point \(O\).
Hence, \(T=\frac{2 u_0 \sin \theta}{g}\)

(c) Horizontal range,
\(
\begin{aligned}
R=\left(u+v_0 \cos \theta\right) T & =\left(u+v_0 \cos \theta\right) \frac{2 v_0 \sin \theta}{g} \\
& =\frac{v_0}{g}\left[2 u \sin \theta+v_0 \sin 2 \theta\right]
\end{aligned}
\)
(d) For the horizontal range to be maximum, \(\frac{d R}{d \theta}=0\)
\(
\begin{aligned}
& \Rightarrow \quad \frac{v_0}{g}\left[2 u \cos \theta+v_0 \cos 2 \theta \times 2\right]=0 \\
& \Rightarrow \quad 2 u \cos \theta+2 v_0\left[2 \cos ^2 \theta-1\right]=0 \\
& \Rightarrow \quad 4 v_0 \cos ^2 \theta+2 u \cos \theta-2 v_0=0 \\
& \Rightarrow \quad 2 v_0 \cos ^2 \theta+u \cos \theta-v_0=0 \\
& \Rightarrow \quad \cos \theta=\frac{-u \pm \sqrt{u^2+8 v_0^2}}{4 v_0} \\
& \Rightarrow \quad \theta_{\max }=\cos ^{-1}\left[\frac{-u \pm \sqrt{u^2+8 v_0^2}}{4 v_0}\right]=\cos ^{-1}\left[\frac{-u+\sqrt{u^2+8 v_0^2}}{4 v_0}\right]
\end{aligned}
\)
(e) If \(u=v_0\),
\(
\begin{aligned}
& \cos \theta=\frac{-v_0 \pm \sqrt{v_0^2+8 v_0^2}}{4 v_0}=\frac{-1+3}{4}=\frac{1}{2} \\
& \Rightarrow \theta=60^{\circ} \\
&
\end{aligned}
\)
-If \(u \ll<v_0\), then \(8 v_0^2+u^2 \approx 8 v_0^2\)
\(
\theta_{\max }=\cos ^{-1}\left[\frac{-u \pm 2 \sqrt{2} v_0}{4 v_0}\right]=\cos ^{-1}\left[\frac{1}{\sqrt{2}}-\frac{u}{4 v_0}\right]
\)
If \(u<v_0\) then \(\theta_{\max } \cos ^{-1}\left(\frac{1}{\sqrt{2}}\right)=\frac{\pi}{4}\)
If \(u>u_0\) and \(u \gg v_0\)
\(
\begin{aligned}
\theta_{\max } & =\cos ^{-1}\left[\frac{-u \pm u}{4 v_0}\right]=0 \\
\Rightarrow \quad & \theta_{\max }=\frac{\pi}{2}
\end{aligned}
\)
(f) when \(u=0\)
\(
\begin{aligned}
& \cos \theta=\frac{-u \pm \sqrt{u^2+8 v_0^2}}{4 v_0} \\
& \cos \theta=\frac{2 \sqrt{2} v}{4 v}=\frac{1}{\sqrt{2}}
\end{aligned}
\)
\(
\begin{aligned}
& \cos \theta=45 \\
& \cos \theta=\frac{\pi}{4}
\end{aligned}
\)

Q38: Motion in two dimensions. in a plane can be studied by expressing position. velocity and acceleration as vectors in Cartesian co-ordinates \(\mathbf{A}=A_x \hat{\mathbf{i}}+A_y \hat{\mathbf{j}}\) where \(\hat{\mathbf{i}}\) and \(\hat{\mathbf{j}}\) are unit vector along \(x\) and \(y\) directions. respectively and \(\mathrm{A}_{\mathrm{x}}\) and \(\mathrm{A}_{\mathrm{y}}\) are corresponding components of \(\mathbf{A}\) (Fig. 4.9). Motion can also be studied by expressing vectors in circular polar co-ordinates as \(\mathbf{A}=A_r \hat{\mathbf{r}}+A_8 \hat{\boldsymbol{\theta}}\) where \(\hat{\mathbf{r}}=\frac{\mathbf{r}}{r}=\cos \theta \hat{\mathbf{i}}+\sin \theta \hat{\mathbf{j}}\) and \(\hat{\theta}=-\sin \theta \hat{\mathbf{i}}+\cos \theta \hat{\mathbf{j}}\) are unit vectors along the direction in which ‘ \(r\) ‘ and ‘ \(\theta\) ‘ are increasing.
(a) Express \(\hat{\mathbf{i}}\) and \(\hat{\mathbf{j}}\) in terms of \(\hat{\mathbf{r}}\) and \(\hat{\theta}\).
(b) Show that both \(\hat{\mathbf{r}}\) and \(\hat{\theta}\) are unit vectors and are perpendicular to each other.
(c) Show that \(\frac{d}{d t}(\hat{\mathbf{r}})=\omega \hat{\theta}\) where
\(
\omega=\frac{d \theta}{d t} \text { and } \frac{d}{d t}(\hat{\theta})=-\omega \hat{\mathbf{r}}
\)
(d) For a particle moving along a spiral given by \(\mathbf{r}=a \theta \hat{\mathbf{r}}\), where \(a=1\) (unit), find dimensions of ‘ \(a\) ‘.
(e) Find velocity and acceleration in polar vector representation for particle moving along the spiral described in (d) above.

Answer: (a) According to the problem, the unit vector
\(
\begin{aligned}
& \hat{r}=\cos \theta \hat{i}+\sin \theta \hat{j} \dots(i)\\
& \hat{\theta}=-\sin \theta \hat{i}+\cos \theta \hat{j} \dots(ii)
\end{aligned}
\)
Multiplying Eq. (i) by \(\sin \theta\) and Eq. (ii) with \(\cos \theta\) and adding
\(
\begin{aligned}
\Rightarrow \hat{r} \sin \theta+\hat{\theta} \cos \theta & =\sin \theta \cdot \cos \theta \hat{i}+\sin ^2 \theta \hat{j}+\cos ^2 \theta \hat{j}-\sin \theta \cdot \cos \theta \hat{i} \\
& =\hat{j}\left(\cos ^2 \theta+\sin ^2 \theta\right)=\hat{j} \\
\hat{r} \sin \theta+\hat{\theta} \cos \theta & =\hat{j} \text { CBSELabs.com }
\end{aligned}
\)
By Eq. (i) \(x \cos \theta\) – Eq.(ii) \(x \sin \theta\)
\(
n(\hat{r} \cos \theta-\hat{\theta} \sin \theta)=\hat{i}
\)
(b)
\(
\begin{aligned}
\hat{r} \cdot \hat{\theta} & =(\cos \theta \hat{i}+\sin \theta \hat{j}) \cdot(-\sin \theta \hat{i}+\cos \theta \hat{j}) \\
& =-\cos \theta \cdot \sin \theta+\sin \theta \cdot \cos \theta=0 \\
\Rightarrow \quad \theta & =90^{\circ}, \text { angle between } \hat{r} \text { and } \hat{\theta} .
\end{aligned}
\)
(c) Given,
\(
\begin{aligned}
\hat{r} & =\cos \theta \hat{i}+\sin \theta \hat{j} \\
\frac{d \hat{r}}{d t} & =\frac{d}{d t}(\cos \theta \hat{i}+\sin \theta \hat{j}) \\
& =-\sin \theta \cdot \frac{d \theta}{d t} \hat{i}+\cos \theta \cdot \frac{d \theta}{d t} \hat{j} \\
& =\omega[-\sin \theta \hat{i}+\cos \theta \hat{j}]\left[\because \theta=\frac{d \theta}{d t}\right]
\end{aligned}
\)
(d)
\(
\begin{aligned}
& \text { Since, } \omega=\frac{d \theta}{d t} \\
& \frac{d \hat{r}}{d t}=\omega \hat{\theta} \\
& \hat{r}=|\hat{a}||\hat{\theta}| \hat{r}
\end{aligned}
\)
Now looking at the dimensions of the quantities on the LHS and the RHS,
\(
[a]=\frac{[\hat{r}]}{[\hat{\theta}][\hat{r}]} 
\)
\(
\begin{aligned}
& =\frac{\left[M^0 L^1 T^0\right]}{\left[M^0 L^0 T^0\right]\left[M^0 L^0 T^0\right]} \\
& =\left[M^0 L^1 T^0\right]
\end{aligned}
\)
\(
\begin{aligned}
& \text { (e) } a=1 \\
& \hat{r}=\hat{\theta}[\cos \theta \hat{i}+\sin \theta \hat{J}] \\
& V=\frac{d \hat{r}}{d t}=\frac{d \theta}{d t} \hat{r}+\theta(-\sin \theta \hat{i}+\cos \theta \hat{J}) \frac{d \theta}{d t} \\
& V=\omega \hat{r}+\theta \cdot \hat{\theta} \omega \\
& \vec{a}=\frac{d v}{d t} \\
& =\frac{d(\omega \hat{r}+\theta \cdot \hat{\theta} \omega)}{d t}
\end{aligned}
\)
\(
\begin{aligned}
& =d^2 \frac{\theta}{d t^2} \hat{r}+d \frac{\theta}{d t} d \hat{r} \frac{\hat{r}}{d t}+d^2 \frac{\theta}{d t^2}(\theta . \hat{\theta})+d \frac{\theta}{d t}(\theta . \hat{\theta}) \\
& =d^2 \frac{\theta}{d t^2} \hat{r}+\omega\left(\frac{-\sin \theta \hat{i} d}{d t}+\frac{\cos \theta \hat{J} d \theta}{d t}\right)+\frac{d^2 \theta}{d t^2}(\theta . \hat{\theta})+\frac{\omega d \theta}{d t}(\theta . \hat{\theta}) \\
& =\frac{d^2 \theta}{d t^2 \hat{r}}+\omega^2 \hat{\theta}+\frac{d^2 \theta}{d t^2}(\theta . \hat{\theta})+\omega^2 \hat{\theta}+\omega^2(-\hat{r}) \\
& \vec{a}=\left(\frac{d^2 \theta}{d t^2-\omega^2}\right) \hat{r}+\left(\frac{2 \omega^2+d^2 \theta}{d t^2 \theta}\right) \hat{\theta}
\end{aligned}
\)

Q39: A man wants to reach from A to the opposite corner of the square C (Fig. 4.10). The sides of the square are \(100 \mathrm{~m}\). A central square of \(50 \mathrm{~m} 50 \mathrm{~m}\) is filled with sand. Outside this square, he can walk at a speed \(1 \mathrm{~m} / \mathrm{s}\). In the central square, he can walk only at a speed of \(v \mathrm{~m} / \mathrm{s}(v<1)\). What is the smallest value of \(v\) for which he can reach faster via a straight path through the sand than any path in the square outside the sand?

Answer: 

Consider the straight line path APQC through the sand. Time taken to go from A to C via this path
\(
\begin{aligned}
& T_{\text {sand }}=\frac{\mathrm{AP}+\mathrm{QC}}{1}+\frac{\mathrm{PQ}}{v} \\
& =\frac{25 \sqrt{2}+25 \sqrt{2}}{1}+\frac{50 \sqrt{2}}{v} \\
& =50 \sqrt{2}\left[\frac{1}{v}+1\right]
\end{aligned}
\)
The shortest path outside the sand will be ARC.
Time taken to go from A to C via this path
\(
\begin{aligned}
&= T_{\text {outside }}=\frac{\mathrm{AR}+\mathrm{RC}}{1} \mathrm{~s} \\
&= 2 \sqrt{75^2+25^2} \mathrm{~s} \\
&= 2 \times 25 \sqrt{10} \mathrm{~s} \\
& \text { For } T_{\text {sand }}<\mathrm{T}_{\text {outside }}, 50 \sqrt{2}\left[\frac{1}{v}+1\right]<2 \times 25 \sqrt{10} \\
& \Rightarrow \frac{1}{v}+1<\sqrt{5} \\
& \Rightarrow \frac{1}{v}<\sqrt{5}-1 \text { or } v>\frac{1}{\sqrt{5}-1} \approx 0.81 \mathrm{~m} / \mathrm{s} .
\end{aligned}
\)