Text Book Back Questions and Answers - Chapter 2 Kinematics 11th Science Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Kinematics
Text Book Questions Solved
Multiple Choice Questions

Question  1 .

Answer:

Question 2.
Identify the unit vector in the following:
(b) $\frac{\hat{i}}{\sqrt{2}}$
(c) $\hat{k}-\frac{\hat{j}}{\sqrt{2}}$
(d) $\frac{\hat{i}+\hat{j}}{\sqrt{2}}$
Answer:
(d) $\frac{\hat{i}+\hat{j}}{\sqrt{2}}$
Question 3.
Which one of the following physical quantities cannot be represented by a scalar?
(a) Mass
(b) length
(c) momentum
(d) magnitude of acceleration
Answer:
(c) momentum
Question 4.
Two objects of masses $m_1$ and $m_2$, fall from the heights $h_1$ and $h_2$ respectively. The ratio of the magnitude of their momenta when they hit the ground is
(a) $\sqrt{\frac{h_1}{h_2}}$
(b) $\sqrt{\frac{m_1 h_1}{m_2 h_2}}$
(c) $\frac{m_1}{m_2} \sqrt{\frac{h_1}{h_2}}$
(d) $\frac{m_1}{m_2}$
Answer:
(c) $\frac{m_1}{m_2} \sqrt{\frac{h_1}{h_2}}$

Question 5.
If a particle has negative velocity and negative acceleration, its speed
(a) increases
(b) decreases
(c) remains same
(d) zero
Answer:
(a) increases
Question 6.
If the velocity is $\overrightarrow{\mathrm{v}} 2 \hat{\mathrm{t} t} \quad 2 \hat{j}-9 \overrightarrow{\mathrm{k}}$, then the magnitude of acceleration at $\mathrm{t}=0.5 \mathrm{~s}$ is
(a) $1 \mathrm{~m} \mathrm{~s}^{-2}$
(b) $1 \mathrm{~m}$
(c) zero
(d) $-1 \mathrm{~m} \mathrm{~s} \mathrm{~s}^{-2}$
Answer:
(a) $1 \mathrm{~m} \mathrm{~s}^{-2}$
Question 7.
If an object is dropped from the top of a building and it reaches the ground at $t=4 \mathrm{~s}$, then the height of the building is (ignoring air resistance) $\left(\mathrm{g}=9.8 \mathrm{~m} \mathrm{~s}^{-2}\right)$.
(a) $77.3 \mathrm{~m}$
(b) $78.4 \mathrm{~m}$
(c) $80.5 \mathrm{~m}$
(d) $79.2 \mathrm{~m}$
Answer:
(b) $78.4 \mathrm{~m}$
Question 8.
A ball is projected vertically upwards with a velocity $\mathrm{v}$. It comes back to ground in time $t$. Which $\mathrm{v}-1$ graph shows the motion correctly?

Answer:

Question 9.
If one object is dropped vertically downward and another object is thrown horizontally from the same height, then the ratio of vertical distance covered by both objects at any instant is
(a) 1
(b) 2
(c) 4
(d) 0.5
Answer:
(a) 1
Question 10.
A ball is dropped from some height towards the ground. Which one of the following represents the correct motion of the ball?

Answer:

Question 11.
If a particle executes uniform circular motion in the xy plane in clockwise direction, then the angular velocity is in
(a) +y direction
(b) $+z$ direction
(c) $-z$ direction
(d) - $x$ direction
Answer:
(c) -z direction
Question 12.
If a particle executes uniform circular motion, choose the correct statement [NEET 2016]
(a) The velocity and speed are constant.
(b) The acceleration and speed are constant.
(c) The velocity and acceleration are constant.
(d) The speed and magnitude of acceleration are constant.
Answer:
(d) The speed and magnitude of acceleration are constant.
Question 13.
If an object is thrown vertically up with the initial speed $u$ from the ground, then the time taken by the object to return back to ground is
(a) $\frac{u^2}{2 g}$
(b) $\frac{u^2}{g}$
(c) $\frac{u}{2 g}$
(d) $\frac{2 u}{g}$
Answer:
(d) $\frac{2 u}{g}$

Question 14.
Two objects are projected at angles $30^{\circ}$ and $60^{\circ}$ respectively with respect to the horizontal direction. The range of two objects are denoted as $\mathrm{R}_{30^{\circ}}$ and $\mathrm{R}_{60^{\circ}}$ - Choose the correct relation from the following:
(a) $\mathrm{R}_{30^{\circ}}=\mathrm{R}_{60^{\circ}}$
(b) $\mathrm{R}_{30^{\circ}}=4 \mathrm{R}_{60^{\circ}}$
(c) $\mathrm{R}_{30^{\circ}}=\frac{\mathrm{R}_{60^{\circ}}}{2}$
(d) $\mathrm{R}_{30^{\circ}}=2 \mathrm{R}_{60^{\circ}}$
Answer:
(a) $R_{30^{\circ}}=R_{60^{\circ}}$
Question 15.
An object is dropped in an unknown planet from height $50 \mathrm{~m}$, it reaches the ground in $2 \mathrm{~s}$. The acceleration due to gravity in this unknown planet is
(a) $\mathrm{g}=20 \mathrm{~m} \mathrm{~s}^{-2}$
(b) $\mathrm{g}=25 \mathrm{~m} \mathrm{~s}^{-2}$
(c) $\mathrm{g}=15 \mathrm{~m} \mathrm{~s}^{-2}$
(d) $\mathrm{g}=30 \mathrm{~m} \mathrm{~s}^{-2}$
Answer:
(a) $\mathrm{g}=25 \mathrm{~m} \mathrm{~s}^{-2}$
Short Answer Questions

Question 1.
Explain what is meant by Cartesian coordinate system?
Answer:
At any given instant of time, the frame of reference with respect to which the position of the object is described in terms of position coordinates $(\mathrm{x}, \mathrm{y}, \mathrm{z})$ is called Cartesian coordinate system.
Question 2.
Define a vector. Give examples.
Answer:
Vector is a quantity which is described by the both magnitude and direction. Geometrically a vector is directed line segment.
Example - force, velocity, displacement.
Question 3.
Define a scalar. Give examples.
Answer:
Scalar is a property which can be described only by magnitude.
Example - mass, distance, speed.
Question 4.
Write a short note on the scalar product between two vectors.
Answer:
The scalar product (or dot product) of two vectors is defined as the product of the magnitudes of both the vectors and the cosine of the angle between them. Thus if there are two vectors $\overrightarrow{\mathrm{A}}$ and $\overrightarrow{\mathrm{B}}$ having an angle 0 between them, then their scalar product is defined as $\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}=\mathrm{AB}$ $\cos 0$. Here, $\mathrm{AB}$ and are magnitudes of $\vec{A}$ and $\overrightarrow{\mathrm{B}}$.
Question 5.
Write a short note on vector product between two vectors.
Answer:
The vector product or cross product of two vectors is defined as another vector having a magnitude equal to the product of the magnitudes of two vectors and the sine of the angle between them. The direction of the product vector is perpendicular to the plane containing the two vectors, in accordance with the right hand screw rule or right hand thumb rule. Thus, if $\overrightarrow{\mathrm{A}}$ and $\vec{B}$ are two vectors, then their vector product is written as $\vec{A} \times \vec{B}$ which is a vector $C$ defined by $\overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}}=(\mathrm{AB} \sin 0) \hat{n}$
The direction $\hat{n}$ of $\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}}$, i.e., $\overrightarrow{\mathrm{c}}$ is perpendicular to the plane containing the vectors $\overrightarrow{\mathrm{A}}$ and $\vec{B}$
Question 6.
How do you deduce that two vectors are perpendicular?
Answer:
If two vectors $\vec{A}$ and $\vec{B}$ are perpendicular to each other than their scalar product $\vec{A} \vec{B}=0$ because $\cos 90^{\circ}=0$. Then he vectors $\vec{A}$ and $\vec{B}$ are said to be mutually orthogonal.
Question 7.
Define displacement and distance.
Answer:
Distance is the actual path length traveled by an object in the given interval of time during the motion. It is a positive scalar quantity. Displacement is the difference between the final and initial positions of the object in a given interval of time. It can also be defined as the shortest distance between these two positions of the object. It is a vector quantity.
Question 8.
Define velocity and speed.
Answer:
Speed is defined as the ratio of total distance covered to the total time taken, it is a scalar quantity and always it is positive. Velocity is defined as the ratio of the displacement vector to the corresponding time interval. It is a vector quantity or it can also be defined as rate of change of displacement.
Question 9.
Define acceleration.
Answer:
Acceleration of a particle is defined as the rate of change of velocity or it can also be defined as the ratio of change in velocity to the given interval of time.

Question 10 .
What is the difference between velocity and average velocity.

Answer:

Question 11.
Define a radian?
One radian is the angle subtended at the center of a circle by an arc that is equal in length to
the radius of the circle.
$1 \mathrm{rad}=57.295^{\circ}$
Question 12 .
Define angular displacement and angular velocity.
Answer:
1. Angular displacement:
The angle described by the particle about the axis of rotation in a given time is called angular displacement.
2. Angular velocity:
The rate of change of angular displacement is called angular velocity.
Question 13.
What is non uniform circular motion?
If the speed of the object in circular motion is not constant, then we have non-uniform circular motion. For example, when the bob attached to a string moves in vertical circle, the speed of the bob is not the same at all time Whenever the speed is not same in circular motion, the particle will have both centripetal and tangential acceleration.
Question 14.
Write down the kinematic equations for angular motion.
Answer:
Kinematic equations for circular motion are -
1. $\omega=\omega_0+\alpha t$
2. $\theta=\omega_0 t+\frac{1}{2} \alpha t^2$
3. $\omega^2=\omega_o^2+2 \alpha \theta$
4. $\theta=\frac{\left(\omega_0+\omega\right)}{2} t$

Here,
$\omega_0=$ initial angular velocity
$\omega=$ final angular velocity
$\theta=$ angular displacement
$\alpha=$ angular acceleration
$\mathrm{t}=$ time.
Question 15.
Write down the expression for angle made by resultant acceleration and radius vector in the non uniform circular motion.
Answer:
The angle made by resultant acceleration and radius vector in the non uniform circular motion
is -
$
\tan \theta=\frac{a_t}{\left(\frac{V^2}{r}\right)} \text { or } \theta=\tan ^{-1}\left(\frac{a_t}{\left(\frac{V^2}{r}\right)}\right)
$

Long Answer Questions
Question 1.
Explain in detail the triangle law of addition.
Answer:
Let us consider two vectors $\vec{A}$ and $\overrightarrow{\mathrm{B}}$ as shown in figure. To find the resultant of the two vectors we apply the triangular.
Law of addition as follows: present the vectors $\mathrm{A}$ and by the two adjacent sides of a triangle taken in the same order. Then the resultant is given by the third side of the triangle as shown in figure.

To explain further, the head of the first vector $\overrightarrow{\mathrm{A}}$ is connected to the tail of the second vect $\overrightarrow{\mathrm{B}}$ Let $O$ he the angle between $\vec{A}$ and $\vec{B}$. Then $\vec{R}$ is the resultant vector connecting the tail of the first vector $\vec{A}$ to the head of the second vector $\vec{B}$ The magnitude of $\vec{R}$. (resultant) given geometrically by the length of $(\mathrm{OQ})$ and the direction of the resultant vector is the angle between $\overrightarrow{\mathrm{R}}$. and $\overrightarrow{\mathrm{A}}$. Thus we write
$
\overrightarrow{\mathrm{R}}=\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}} \overrightarrow{\mathrm{OQ}}=\overrightarrow{\mathrm{OP}}+\overrightarrow{\mathrm{PQ}}
$
1. Magnitude of resultant vector:
The magnitude and angle of the resultant vector ar determined by using triangle law of vectors as follows.From figure, consider the triangle $\mathrm{ABN}$, which is obtained by extending the side $\mathrm{OA}$ to $\mathrm{ON} . \mathrm{ABN}$ is a right angled triangle.

From figure, let $R$ is the magnitude of the resultant of $\overrightarrow{\mathrm{A}}$ and $\overrightarrow{\mathrm{B}}$. $\cos \theta=\frac{A N}{B} \therefore \mathrm{AN}=\mathrm{B} \cos \theta$ and $\sin \theta=\frac{B N}{B} \therefore \mathrm{BN}=\mathrm{B} \sin \theta$
For $\Delta \mathrm{OBN}$, we have $\mathrm{OB}^2=\mathrm{ON}^2+\mathrm{BN}^2$
$
\begin{aligned}
& \Rightarrow \mathrm{R}^2=(\mathrm{A}+\mathrm{B} \cos \theta)^2+(\mathrm{B} \sin \theta)^2 \\
& \Rightarrow R^2=A^2+B^2 \cos ^2 \theta+2 A B \cos \theta B^2 \sin ^2 \theta \\
& \Rightarrow \mathrm{R}^2=\mathrm{A}^2+\mathrm{B}^2\left(\cos ^2 \theta+\sin ^2 \theta\right)+2 \mathrm{AB} \cos \theta \\
& \Rightarrow \mathrm{R}^2=\sqrt{A^2+B^2+2 A B \cos \theta} \\
&
\end{aligned}
$
2. Direction of resultant vectors:
If 0 is the angle between $\vec{A}$ and $\vec{B}$ then,
$
|\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}|=\sqrt{\mathrm{A}^2+\mathrm{B}^2+2 \mathrm{AB} \cos \theta}
$
If $\mathrm{R}$ makes an angle $\alpha$ with $\overrightarrow{\mathrm{A}}$, then in $\mathrm{AOBN}$, $\tan \alpha=\frac{B N}{O N}=\frac{B N}{O A+A N}$
$
\tan \alpha=\frac{B \sin \theta}{A+B \cos \theta} \Rightarrow \alpha=\tan ^{-1}\left(\frac{B \sin \theta}{A+B \cos \theta}\right)
$
Question 2.
Discuss the properties of scalar and vector products.
Answer:
Properties of scalar product of two vectors are:
(1) The product quantity $\vec{A} \cdot \vec{B}$ is always a scalar. It is positive if the angle between the vectors is acute (i.e., $<90^{\circ}$ ) and negative if the angle between them is obtuse (i.e. $90^{\circ}<0<$ $\left.180^{\circ}\right)$
(2) The scalar product is commutative, i.e. $\vec{A} \vec{B} \neq \vec{B} \cdot \vec{A}$

(3) The vectors obey distributive law i.e. $\vec{A}(\vec{B}+\vec{C})=\vec{A} \cdot \vec{B}+\vec{A} \cdot \vec{C}$
(4) The angle between the vectors $\theta=\cos ^{-1}\left[\frac{\vec{A} \cdot \vec{B}}{A B}\right]$
(5) The scalar product of two vectors will be maximum when $\cos \theta=1$, i.e. $\theta=0^{\circ}$, i.e., when the vectors are parallel;
$
(\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}})_{\max }=\mathrm{AB}
$
(6) The scalar product of two vectors will be minimum, when $\cos \theta=-1$, i.e. $\theta=180^{\circ}$. $(\vec{A} \cdot \vec{B})_{\min }=-\mathrm{AB}$ when the vectors are anti-parallel.
(7) If two vectors $\vec{A}$ and $\vec{B}$ are perpendicular to each other than their scalar product $\vec{A} \cdot \vec{B}=$ 0 , because $\cos 90^{\circ} 0$. Then the vectors $\vec{A}$ and $\vec{B}$ are said to be mutually orthogonal.
(8) The scalar product of a vector with itself is termed as self-dot product and is given by ( $\overrightarrow{\mathrm{A}})^2=\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{A}}=\mathrm{AA} \cos 0=\mathrm{A}^2$. Here angle $0=0^{\circ}$.
The magnitude or norm of the vector $\vec{A}$ is $|\vec{A}|=A=\sqrt{\vec{A} \cdot \vec{A}}$.
(9) In case of a unit vector $\hat{n}$
$\hat{n} \cdot \hat{n}=1 \times 1 \times \cos 0=1$. For example, $\hat{i}-\hat{i}=\hat{j} \cdot \hat{j}=\hat{k} \cdot \hat{k}=1$.
(10) In the case of orthogonal unit vectors, $\hat{i}, \hat{j}$ and $\hat{k}$
$\hat{i} \cdot \hat{j}=\hat{j} \cdot \hat{k}=\hat{k} \cdot \hat{i}=1.1 \cos 90^{\circ}=0$
(11) In terms of components the scalar product of $\vec{A}$ and $\vec{B}$ can be written as $\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}=\left(\mathrm{A}_{\mathrm{x}} \hat{i}+\mathrm{A}_{\mathrm{y}} \hat{j}+\mathrm{A}_{\mathrm{z}} \hat{k}\right) \cdot\left(\mathrm{B}_{\mathrm{x}} \hat{i}+\mathrm{B}_{\mathrm{y}} \hat{j}+\mathrm{B}_{\mathrm{z}} \hat{k}\right)$
$=A_x B_x+A_y B_y+A_z B_z$, with all other terms zero.
The magnitude of vector $|\vec{A}|$ is given by
$
|\overrightarrow{\mathrm{A}}|=\mathrm{A}=\sqrt{\mathrm{A}_x^2+\mathrm{A}_y^2+\mathrm{A}_z^2}
$

Properties of vector product of two vectors are:
(1) The vector product of any two vectors is always another vector whose direction is perpendicular to the plane containing these two vectors, i.e., orthogonal to both the vectors $\overrightarrow{\mathrm{A}}$ and $\vec{B}$, even though the vectors $\vec{A}$ and $\vec{B}$ may or may not be mutually orthogonal.
(2) The vector product of two vectors is not commutative, i.e., $\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}} \neq \overrightarrow{\mathrm{B}} \times \overrightarrow{\mathrm{A}}$. But, $\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}}=-\overrightarrow{\mathrm{B}} \times \overrightarrow{\mathrm{A}}$
Here it is worthwhile to note that $|\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}}|=$
$|\overrightarrow{\mathrm{B}} \times \overrightarrow{\mathrm{A}}|=\mathrm{AB} \sin 0$ i.e., in the case of the product vectors $\overrightarrow{\mathrm{B}}=-\overrightarrow{\mathrm{B}}$ and $\overrightarrow{\mathrm{B}} \times \overrightarrow{\mathrm{A}}$, the magnitudes are equal but directions are opposite to each other.
(3) The vector product of two vectors will have maximum magnitude when $\sin 0=1$, i.e., $0=$ $90^{\circ}$ i.e., when the vectors $\overrightarrow{\mathrm{A}}$ and $\overrightarrow{\mathrm{B}}$ are orthogonal to each other.
$
(\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}})_{\max }=\mathrm{AB} \hat{n}=\mathrm{AB} \hat{n}
$
(4) The vector product of two non-zero vectors will be $\operatorname{minimum}$ when $\sin \theta=0$, i.e $\theta=0^{\circ}$ or $180^{\circ}$
$
\overrightarrow{(\mathrm{A}} \times \overrightarrow{\mathrm{B}})_{\min }=0
$
i. e., the vector product of two non - zero vectors vanishes, if the vectors are either parallel or anti parallel.
(5) The self-cross product, i.e., product of a vector with itself is the null vector $\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{A}}=\mathrm{AA} \sin 0^{\circ} \hat{n}=\overrightarrow{0}$ In physics the null vector 0 is simply denoted as zero.
(6) The self-vector products of unit vectors are thus zero.

$
\hat{i} \times \hat{i}=\hat{j} \times \hat{j}=\hat{k} \times \hat{k}=0
$
(7) In the case of orthogonal unit vectors, $\hat{i}, \hat{j} . \hat{k}$, in accordance with the right hand screw rule:
$
\hat{i} \times \hat{j}=\hat{k}, \hat{j} \times \hat{k}=\hat{i} \text { and } \hat{k} \times \hat{i}=\hat{j}
$

Also, since the cross product is not commutative, $\hat{j} \mathrm{x} \hat{i}=-\hat{k}, \hat{k} \times \hat{j}=-\hat{i}$ and $\hat{i} \times \hat{k}=\hat{j}$
(8) In terms of components, the vector product of two vectors $\vec{A}$ and $\overrightarrow{\mathrm{B}}$ is -
$
\begin{aligned}
\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}} & =\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
\mathrm{~A}_x & \mathrm{~A}_y & \mathrm{~A}_z \\
\mathrm{~B}_x & \mathrm{~B}_y & \mathrm{~B}_z
\end{array}\right| \\
& =\hat{i}\left(\mathrm{~A}_y \mathrm{~B}_z-\mathrm{A}_z \mathrm{~B}_y\right)+\hat{j}\left(\mathrm{~A}_z \mathrm{~B}_x-\mathrm{A}_x \mathrm{~B}_z\right)+\hat{k}\left(\mathrm{~A}_x \mathrm{~B}_y-\mathrm{A}_y \mathrm{~B}_x\right)
\end{aligned}
$
Note that in the $\hat{j}^{\text {th }}$ component the order of multiplication is different than $\hat{i}^{\text {th }}$ and $\hat{k}^{\text {th }}$ components.
(9) If two vectors $\vec{A}$ and $\vec{B}$ form adjacent sides in a parallelogram, then the magnitude of $\mid \vec{A}$ $x \vec{B} \mid$ will give the area of the parallelogram as represented graphically in figure.
$|\vec{A} \times \vec{B}|=|\vec{A}||\vec{B}| \sin \theta$

(10) Since we can divide a parallelogram into two equal triangles as shown in the figure, the area of a triangle with $\vec{A}$ and $\vec{B}$ as sides is $\frac{1}{2}|\vec{A} \times \vec{B}|$. This is shown in the Figure. A number of quantities used in Physics are defined through vector products. Particularly physical quantities representing rotational effects like torque, angular momentum, are defined through vector products.

Question 3 .
Derive the kinematic equations of motion for constant acceleration.
Answer:
Consider an object moving in a straight line with uniform or constant acceleration ' $a$ '. Let $u$ be the velocity of the object at time $t=0$, and $\mathrm{v}$ be velocity of the body at a later time $t$.
Velocity - time relation:
(1) The acceleration of the body at any instant is given by the first derivative of the velocity with respect to time, $\mathrm{a}=\frac{d v}{d t}$ or $\mathrm{dv}=\mathrm{a} \mathrm{dt}$
Integrating both sides with the condition that as time changes from 0 to $t$, the velocity changes from $u$ to v. For the constant acceleration,
$
\begin{aligned}
& \int_u^v d v=\int_0^t a d t=a \int_0^t d t \Rightarrow[v]_u^v=a[t]_0^t \\
& v-u=a t \text { or } v=u+a t \quad \ldots(i)
\end{aligned}
$
(2) The velocity of the body is given by the first derivative of the displacement with respect to time.
$\mathrm{v}=\frac{d s}{d t}$ or $\mathrm{ds}=\mathrm{vdt}$
and since $\mathrm{v}=\mathrm{u}+$ at,
we get $\mathrm{ds}=(\mathrm{u}+$ at $) \mathrm{dt}$
Assume that initially at time $t=0$, the particle started from the origin. At a later time $t$, the particle displacement is s. Further assuming that acceleration is time-independent, we have
$
\begin{aligned}
& \int_0^s d s=\int_0^t u d t+\int_9^t a t d t \text { or } s=u t+\frac{1}{-} a t^2 \\
& \text { Velocity - displacem }
\end{aligned}
$

(3) The acceleration is given by the first derivative of velocity with respect to time. $\mathrm{a}=\frac{d v}{d t}=\frac{d v}{d s}=\frac{d s}{d t}=\frac{d v}{d s} \mathrm{v}[$ since $\mathrm{ds} / \mathrm{dt}=\mathrm{v}]$ where $\mathrm{s}$ is displacement traverse This is rewritten as a $=\frac{1}{2} \frac{d v^2}{d s}$ or $\mathrm{ds}=\frac{1}{2 a} d\left(v^2\right)$ Integrating the above equation, using the fact when the velocity changes from $\mathrm{u}^2$ to $\mathrm{v}^2$, displacement changes from 0 to $\mathrm{s}$, we get
$
\begin{aligned}
\int_0^s d s & =\int_u^v \frac{1}{2 a} d\left(v^2\right) \\
s & =\frac{1}{2 a}\left(v^2-u^2\right) \\
v^2 & =u^2+2 a s
\end{aligned}
$
We can also derive the displacement 5 in terms of initial velocity $u$ and final velocity $v$. From equation we can write,
at $=\mathrm{v}-\mathrm{u}$
Substitute this in equation, we get
$
\begin{aligned}
& s=u t+\frac{1}{2}(v-u) t \\
& s=\frac{(u+v) t}{2}
\end{aligned}
$

Question 4.
Derive the equations of motion for a particle (a) falling vertically (b) projected vertically.
Answer:
'Equations of motion for a particle falling vertically downward from certain height. Consider an object of mass $\mathrm{m}$ falling from a height $h$. Assume there is no air resistance. For convenience, let us choose the downward direction as positive $y$ - axis as shown in the figure. The object experiences acceleration ' $g$ ' due to gravity which is constant near the surface of the Earth. We can use kinematic equations to explain its motion. We have The acceleration $\vec{a}$ $=\mathrm{g} \hat{i}$
By comparing the components, we get, Equations of motion for a particle thrown vertically upwards, $\mathrm{a}_{\mathrm{x}}=0, \mathrm{a}_{\mathrm{x}}=0, \mathrm{a}_{\mathrm{y}}=\mathrm{g}$ Let us take for simplicity, $\mathrm{a}_{\mathrm{y}}=\mathrm{a}=\mathrm{g}$ in free fall

If the particle is thrown with initial velocity ' $u$ ' downward which is in negative $y-a x i$, then velocity and position at of the particle any time $t$ is given by
$
\begin{aligned}
& \mathrm{v}=\mathrm{u}+\mathrm{gt} \\
& \mathrm{v}=\mathrm{ut}+\frac{1}{2}-\mathrm{gt}^2
\end{aligned}
$
The square of the speed of the particle when it is at a distance $y$ from the hill -top, is $v^2=u^2$

$+2 \mathrm{gy}$
Suppose the particle starts from rest.
Then $\mathrm{u}=0$
Then the velocity $\mathrm{v}$, the position of the particle and $\mathrm{v}^2$ at any time $t$ are given by (for a point $y$ from the hill - top)
$
\begin{aligned}
& \mathrm{v}=\mathrm{gt} \ldots \ldots \ldots \ldots \text {.(i) } \\
& \mathrm{y}=\frac{1}{2}-\mathrm{gt}^2 \ldots \ldots \ldots \ldots \text { (ii) } \\
& \mathrm{v}^2=2 \mathrm{gy} \ldots \ldots \ldots \ldots \text { (iii) }
\end{aligned}
$
The time $(\mathrm{t}=\mathrm{T})$ taken by the particle to reach the ground (for which $\mathrm{y}=\mathrm{h}$ ), is given by using equation (ii),
$\mathrm{h}=\frac{1}{2}-\mathrm{gT}^2$
$\mathrm{T}=\sqrt{\frac{2 h}{g}}$.
The equation (iv) implies that greater the height (h), particle takes more time (T) to reach the ground. For lesser height (h), it takes lesser time to reach the ground. The speed of the particle when it reaches the ground $(\mathrm{y}=\mathrm{h})$ can be found using equation (iii), we get,
$v_{\text {ground }}=\sqrt{2 g h}$
(vi)
The above equation implies that the body falling from greater height (h) will have higher velocity when it reaches the ground. The motion of a body falling towards the Earth from a small altitude $(\mathrm{h}<<\mathrm{R})$, purely under the force of gravity is called free fall. (Here $\mathrm{R}$ is radius of the Earth).
case (ii):
A body thrown vertically upwards:
Consider an object of mass $m$ thrown vertically upwards with an initial velocity $u$. Let us neglect the air friction. In this case we choose the vertical direction as positive $y$ axis as shown in the figure, then the acceleration $\mathrm{a}=-\mathrm{g}$ (neglect air friction) and $\mathrm{g}$ points towards the negative $y$ axis. The kinematic equations for this motion are,
The velocity and position of the object at any time $t$ are,

An object thrown vertically
$\mathrm{v}=\mathrm{u}-\mathrm{gt}$
$\mathrm{s}=\mathrm{ut}-\frac{1}{2}-\mathrm{gt}^2$
The velocity of the object at any position y (from the point where the object is thrown) is $v^2=u^2-2 g y$
Question 5.
Derive the equation of motion, range and maximum height reached by the particle thrown at an oblique angle 9 with respect to the horizontal direction.
Answer:
This projectile motion takes place when the initial velocity is not horizontal, but at some angle with the vertical, as shown in Figure.
(Oblique projectile)
Examples:
- Water ejected out of a hose pipe held obliquely.
- Cannon fired in a battle ground.

Consider an object thrown with initial velocity at an angle $\theta$ with the horizontal.
Then, $\overrightarrow{\mathrm{u}}=\mathrm{u}_{\mathrm{x}} \hat{\mathrm{i}}+\mathrm{u}_{\mathrm{y}} \hat{j}$
where $\mathrm{u}_{\mathrm{x}}=\mathrm{u} \cos \theta$ is the horizontal component and $\mathrm{u}_{\mathrm{y}}=\mathrm{u} \sin \theta$ the vertical component of velocity. Since the acceleration due to gravity is in the direction opposite to the direction of vertical component $\mathrm{u}_{\mathrm{y}}$, this component will gradually reduce to zero at the maximum height of the projectile. At this maximum height, the same gravitational force will push the projectile to move downward and fall to the ground. There is no acceleration along the $\mathrm{x}$ direction throughout the motion. So, the horizontal component of the velocity $\left(\mathrm{u}_{\mathrm{x}}=\mathrm{u} \cos \theta\right)$ remains the same till the object reaches the ground. Hence after the time t, the velocity along horizontal motion $\mathrm{v}_{\mathrm{x}}=\mathrm{u}_{\mathrm{x}}+\mathrm{a}_{\mathrm{x}} \mathrm{t}=\mathrm{u}_{\mathrm{x}}=\mathrm{u} \cos \theta$. The horizontal distance travelled by projectile $\mathrm{m}$ time $\mathrm{t}$ is $\mathrm{s}_{\mathrm{x}}=u_x t+\frac{1}{2} a_x t^2$
Here, $s_x=x, u_x=u \cos \theta, a_x=0$

Thus, $\mathrm{x}=\mathrm{u} \cos \theta$ or $\mathrm{t}=\frac{x}{u \cos \theta} \ldots \ldots$..(i)
Next, for the vertical motion $\mathrm{v}_{\mathrm{y}}=\mathrm{u}_{\mathrm{y}}+\mathrm{a}_{\mathrm{y}} \mathrm{t}$
Here $\mathrm{u}_{\mathrm{y}}=\mathrm{u} \sin \theta, \mathrm{a}_{\mathrm{y}}=-\mathrm{g}$ (acceleration due to gravity acts opposite to the motion).
Thus, $\mathrm{v}_{\mathrm{y}}=\mathrm{u} \sin \theta-\mathrm{gt}$
The vertical distance traveled by the projectile in the same time $t$ is
Here, $\mathrm{s}_{\mathrm{y}}=\mathrm{y}, \mathrm{u}_{\mathrm{y}}=\mathrm{u} \sin \theta, \mathrm{a}_{\mathrm{x}}=-\mathrm{g}$. Then
$\mathrm{y}=\mathrm{u} \sin \theta \mathrm{t}-\frac{1}{2}-\mathrm{gt}^2 \ldots \ldots \ldots$..(ii)
Substitute the value of $t$ from equation (i) in equation (ii), we have .
$
\begin{aligned}
& y=u \sin \theta \frac{x}{u \cos \theta}-\frac{1}{2} g \frac{x^2}{u^2 \cos ^2 \theta} \\
& y=x \tan \theta-\frac{1}{2} g \frac{x^2}{u^2 \cos ^2 \theta}
\end{aligned}
$
Thus the path followed by the projectile is an inverted parabola Maximum height $\left(\mathrm{h}_{\max }\right)$ : The maximum vertical distance travelled by the projectile during the journey is called maximum height. This is determined as follows:
For the vertical part of the motion.
$
v_y^2=u_y^2+2 a_y s
$
Here, $\mathrm{u}_{\mathrm{y}}=\mathrm{u} \sin \theta, \mathrm{a}=-\mathrm{g}, \mathrm{s}=\mathrm{h}_{\max }$, and at the maximum height $\mathrm{v}_{\mathrm{y}}=0$
Hence, $(0)^2=\mathrm{u}^2 \sin ^2 \theta=2 \mathrm{gh}_{\max }$ or $h_{\max }=\frac{u^2 \sin ^2 \theta}{2 g}$
Time of flight $\left(\mathrm{T}_{\mathrm{f}}\right)$ :
The total time taken by the projectile from the point of projection till it hits the horizontal plane is called time of flight. This time of flight is the time taken by the projectile to go from point $\mathrm{O}$ to $\mathrm{B}$ via point $\mathrm{A}$ as shown
we know that $\mathrm{s}_{\mathrm{y}}=\mathrm{y}=0$ (net displacement in $\mathrm{y}$-direction is zero),
$\mathrm{u}_{\mathrm{y}}=\mathrm{u} \sin \theta, \mathrm{a}_{\mathrm{y}}=-\mathrm{g}, \mathrm{t}=\mathrm{T}_{\mathrm{f}}$ Then

$
\begin{aligned}
0 & =u \sin \theta \mathrm{T}_f-\frac{1}{2} g \mathrm{~T}_f^2 \\
\mathrm{~T}_f & =2 u \frac{\sin \theta}{g} \quad \ldots(v)
\end{aligned}
$
Horizontal range $(\mathrm{R})$ :
The maximum horizontal distance between the point of projection and the point on the horizontal plane where the projectile hits the ground is called horizontal range (R). This is found easily since the horizontal component of initial velocity remains the same. We can write Range $\mathrm{R}=$ Horizontal component of velocity $\mathrm{x}$ time of flight $=\mathrm{u} \cos \theta \times \mathrm{T}_f=\frac{u^2 \sin 2 \theta}{g}$ The horizontal range directly depends on the initial speed (u) and the sine of angle of projection $(\theta)$. It inversely depends on acceleration due to gravity ' $\mathrm{g}$ '.
For a given initial speed $\mathrm{u}$, the maximum possible range is reached when $\sin 2 \theta$ is maximum, $\sin 2 \theta=1$. This implies $2 \theta=\pi / 2$ or $\theta=\frac{\pi}{4}$ This means that if the particle is projected at 45 degrees with respect to horizontal, it attains maximum range, given by.
$
\mathrm{R}_{\max }=\frac{u^2}{g}
$
Question 6.
Derive the expression for centripetal acceleration.
Answer:
In uniform circular motion the velocity vector turns continuously without changing its magnitude (speed), as shown in figure.

Note that the length of the velocity vector is not changed during the motion, implying that the speed remains constant. Even though the velocity is tangential at every point in the circle, the acceleration is acting towards the center of the circle. This is called centripetal acceleration. It always points towards the center of the circle. This is shown in the figure.

The centripetal acceleration is derived from a simple geometrical relationship between position and velocity vectors.

Let the directions of position and velocity vectors shift through the same angle $\theta$ in a small interval of time $\Delta \mathrm{t}$, as shown in figure. For uniform circular motion, $\mathrm{r}=\left|\vec{r}_1\right|=\left|\vec{r}_2\right|$ and $\mathrm{v}=$ $\left|\vec{v}_1\right|=\left|\vec{v}_2\right|$. If the particle moves from position vector $\vec{r}_1$ to $\vec{r}_2$, the displacement is given by $\Delta$ $\overrightarrow{\mathrm{r}}=\vec{r}_2-\vec{r}_1$ and the change in velocity from $\vec{v}_1$ to $\vec{v}_2$ is given by $\Delta \overrightarrow{\mathrm{v}}=\vec{v}_2-\vec{v}_1$. The magnitudes of the displacement $\Delta \mathrm{r}$ and of $\Delta \mathrm{v}$ satisfy the following relation. $\frac{\Delta r}{r}=\frac{-\Delta v}{v}=\theta$ Here the negative sign implies that $\Delta \mathrm{v}$ points radially inward, towards the center of the circle.
$
\begin{aligned}
\Delta v & =-v\left(\frac{\Delta r}{r}\right) \\
a & =\frac{\Delta v}{\Delta t}=\frac{v}{r}\left(\frac{\Delta r}{\Delta t}\right)=-\frac{v^2}{r}
\end{aligned}
$
For uniform circular motion $\mathrm{v}=$ cor, where co is the angular velocity of the particle about center. Then the centripetal acceleration can be written as.
$
a=-\omega^2 r
$
Question 7.
Derive the expression for total acceleration in the non uniform circular motion.
Answer:
If the speed of the object in circular motion is not constant, then we have non-uniform circular motion. For example, when the bob attached to a string moves in vertical circle, the speed of the bob is not the same at all time. Whenever the speed is not same in circular motion, the particle will have both centripetal and tangential acceleration as shown in the figure.

The resultant acceleration is obtained by vector sum of centripetal and tangential acceleration Since centripetal acceleration is $\frac{v^2}{r}$, the magnitude of this resultant acceleration is given by $\dot{a}_{\mathrm{R}}=\sqrt{a_t^2+\left(\frac{v^2}{r}\right)^2}$
This resultant acceleration makes an angle 0 with the radius vector as shown in figure. This angle is given by $\tan \theta=\frac{a_t}{\left(v^2 / r\right)}$