Text Book Back Questions and Answers - Chapter 3 Laws of Motion 11th Science Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Laws of Motion
Textual QuestionsSolved
Multiple Choice Questions
Question 1.
When a car takes a sudden left turn in the curved road, passengers are pushed towards the right due to
(a) inertia of direction
(b) inertia of motion
(c) inertia of rest
(d) absence of inertia
Answer:
(a) inertia of direction
Question 2.
An object of mass $m$ held against a vertical wall by applying horizontal force $\mathrm{F}$ as shown in the figure. The minimum value of the force $F$ is 
(a) Less than $\mathrm{mg}$
(b) Equal to $\mathrm{mg}$
(c) Greater than $\mathrm{mg}$
(d) Cannot determine
Answer:
(c) Greater than $\mathrm{mg}$
Question 3.
A vehicle is moving along the positive $\mathrm{x}$ direction, if sudden brake is applied, then
(a) frictional force acting on the vehicle is along negative $\mathrm{x}$ direction
(b) frictional force acting on the vehicle is along positive $\mathrm{x}$ direction
(c) no frictional force acts on the vehicle
(d) frictional force acts in downward direction
Answer:
(a) frictional force acting on the vehicle is along negative $\mathrm{x}$ direction
Question 4.
A book is at rest on the table which exerts a normal force on the book. If this force is considered as reaction force, what is the action force according to Newton's third law?

(a) Gravitational force exerted by Earth on the book
(b) Gravitational force exerted by the book on Earth
(c) Normal force exerted by the book on the table
(d) None of the above
Answer:
(c) Normal force exerted by the book on the table
Question 5
. Two masses $\mathrm{m}$ and $\mathrm{m}_2$ are experiencing the same force where $\mathrm{m}_1<\mathrm{m}_2$ The ratio of their acceleration $\frac{a_1}{a_2}$ is -
(a) 1
(b) less than 1
(c) greater than 1
(d) all the three cases
Answer:
(c) greater than 1
Question 6.
Choose appropriate free body diagram for the particle experiencing net acceleration along negative y direction. (Each arrow mark represents the force acting on the system).

Answer:

Question 7.
A particle of mass $\mathrm{m}$ sliding on the smooth double inclined plane (shown in figure) will experience -
(a) greater acceleration along the path $\mathrm{AB}$
(b) greater acceleration along the path $\mathrm{AC}$
(c) same acceleration in both the paths
(d) no acceleration in both the paths

Answer:
(a) greater acceleration along the path $\mathrm{AC}$
Question 8.
Two blocks of masses $\mathrm{m}$ and $2 \mathrm{~m}$ are placed on a smooth horizontal surface as shown. In the first case only a force $F_1$ is applied from the left. Later only a force $F_2$ is applied from the right. If the force acting at the interface of the two blocks in the two cases is same, then $F_1$ : $\mathrm{F}_2$ is

(a) $1: 1$
(b) $1: 2$
(c) $2: 1$
(d) $1: 3$
Answer:
(c) $2: 1$
Question 9.
Force acting on the particle moving with constant speed is -
(a) always zero
(b) need not be zero
(c) always non zero
(d) cannot be concluded
Answer:
(b) need not be zero
Question 10.
An object of mass $m$ begins to move on the plane inclined at an angle 0 . The coefficient of static friction of inclined surface is lay. The maximum static friction experienced by the mass is -
(a) $\mathrm{mg}$
(b) $\mu_{\mathrm{s}} \mathrm{mg}$
(c) $\mu_{\mathrm{s}} \mathrm{mg} \sin \theta$

(d) $\mu_{\mathrm{s}} \mathrm{mg} \cos \theta$
Answer:
(d) $\mu_{\mathrm{s}} \mathrm{mg} \cos \theta$
Question 11.
When the object is moving at constant velocity on the rough surface -
(a) net force on the object is zero
(b) no force acts on the object
(c) only external force acts on the object
(d) only kinetic friction acts on the object
Answer:
(a) net force on the object is zero
Question 12.
When an object is at rest on the inclined rough surface -
(a) static and kinetic frictions acting on the object is zero
(b) static friction is zero but kinetic friction is not zero
(c) static friction is not zero and kinetic friction is zero
(d) static and kinetic frictions are not zero
Answer:
(c) static friction is not zero and kinetic friction is zero
Question 13.
The centrifugal force appears to exist -
(a) only in inertial frames
(b) only in rotating frames
(c) in any accelerated frame
(d) both in inertial and non-inertial frames
Answer:
(b) only in rotating frames
Question 14.
Choose the correct statement from the following -
(a) Centrifugal and centripetal forces are action reaction pairs
(b) Centripetal forces is a natural force
(c) Centrifugal force arises from gravitational force
(d) Centripetal force acts towards the center and centrifugal force appears to act away from the center in a circular motion
Answer:
(d) Centripetal force acts towards the center and centrifugal force appears to act away from the center in a circular motion

Question 15.
If a person moving from pole to equator, the centrifugal force acting on him -
(a) increases
(b) decreases
(c) remains the same
(d) increases and then decreases
Answer:
(a) increases
Short Answer Questions
Question 1.
Explain the concept of inertia. Write two examples each for inertia of motion, inertia of rest and inertia of direction.
Answer:
The inability of objects to move on its own or change its state of motion is called inertia. Inertia means resistance to change its state. There are three types of inertia:
1. Inertia of rest:
The inability of an object to change its state of rest is called inertia of rest.
Example:
- When a stationary bus starts to move, the passengers experience a sudden backward push.
- A book lying on the table will remain at rest until it is moved by some external agencies.
2. Inertia of motion:
The inability of an object to change its state of uniform speed (constant speed) on its own is called inertia of motion.
Example:
- When the bus is in motion, and if the brake is applied suddenly, passengers move forward and hit against the front seat.

- An athlete running is a race will continue to run even after reaching the finishing point.
3. Inertia of direction:
The inability of an object to change its direction of motion on its own is called inertia of direction.
Example:
- When a stone attached to a string is in whirling motion, and if the string is cut suddenly, the stone will not continue to move in circular motion but moves tangential to the circle.
- When a bus moving along a straight line takes a turn to the right. The passengers are thrown towards left.
Question 2.
State Newton's second law.
Answer:
The force acting on an object is equal to the rate of change of its momentum $\overline{\mathrm{F}}=\frac{d \bar{p}}{d t}$
Question 3.
Define one newton.
Answer:
One newton is defined as the force which acts on $1 \mathrm{~kg}$ of mass to give an acceleration $1 \mathrm{~ms}^{-2}$ in the direction of the force.
Question 4.
Show that impulse is the change of momentum.
Answer:
According to Newton's Second Law
$
\mathrm{F}=\frac{d p}{d t} \text { i.e. } \mathrm{dp}=\mathrm{Fdt}
$

Integrate it over a time interval from $t_i$ to $t_f$
$
\int_i^f d p=\int_{t_i}^{t_f} \mathrm{~F} d t
$
$
\mathrm{P}_f-\mathrm{P}_i=\int_{t_i}^{t_2} \mathrm{~F} \cdot d t
$
$\mathrm{P}_{\mathrm{i}} \rightarrow$ initial momentum of the object at $\mathrm{t}_{\mathrm{i}}$
$\mathrm{P}_{\mathrm{f}} \rightarrow$ Final momentum of the object at $\mathrm{t}_{\mathrm{f}}$
$\mathrm{P}_{\mathrm{f}}-\mathrm{P}_{\mathrm{i}}=\Delta \mathrm{p}=$ change in momentum during the time interval $\Delta \mathrm{t}$.
$\int_{t_i}^{t_f} \mathrm{~F} \cdot d t=\mathrm{J}$ is called the impulse.
If the force is constant over the time interval $\Delta \mathrm{t}$, then
$
\begin{aligned}
& \mathrm{J}=\int_{t_i}^{t_f} \mathrm{~F} \cdot d t=\mathrm{F} \int_{t_i}^{t_f} d t=\mathrm{F}\left(t_f-t_i\right)=\mathrm{F} \Delta t \\
& \mathrm{~J}=\mathrm{F} \cdot \Delta t=\Delta p
\end{aligned}
$
Hence the proof.
Question 5.
Using free body diagram, show that it is easy to pull an object than to push it.
Answer:
When a body is pushed at an arbitrary angle $\theta\left[0\right.$ to $\left.\frac{\pi}{2}\right]$, the applied force $F$ can be resolved into two components as $\mathrm{F} \sin 0$ parallel to the surface and $\mathrm{F} \cos 0$ perpendicular to the surface as shown in figure. The total downward force acting on the body is $\mathrm{mg}+\mathrm{F} \cos \theta$. It implies that the normal force acting on the body increases. Since there is no acceleration along the vertical direction the normal force $\mathrm{N}$ is equal to

$
\mathrm{N}_{\text {push }}=\mathrm{mg}+\mathrm{F} \cos \theta
$
As a result the maximal static friction also increases and is equal to $f_S^{\max }=\mu_r \mathrm{~N}_{\text {push }}=\mu_{\mathrm{s}}(\mathrm{mg}+\mathrm{F} \cos \theta) \ldots \ldots$ (2)
Equation (2) shows that a greater force needs to be applied to push the object into motion.

When an object is pulled at an angle $\theta$, the applied force is resolved into two components as shown in figure. The total downward force acting on the object is -
$
\mathrm{N}_{\text {pull }}=\mathrm{mg}-\mathrm{F} \cos \theta
$

Equation (3) shows that the normal force is less than $-\mathrm{N}_{\text {push. }}$. From equations (1) and (3), it is easier to pull an object than to push to make it move.
Question 6.
Explain various types of friction. Suggest a few methods to reduce friction.
Answer:
There are two types of Friction:
(1) Static Friction:
Static friction is the force which opposes the initiation of motion of an object on the surface. The magnitude of static frictional force $f_{\mathrm{s}}$ lies between
$
0 \leq f_s \leq \mu_s \mathrm{~N}
$
where, $\mu_{\mathrm{s}}-$ coefficient of static friction
$\mathrm{N}$ - Normal force

(2) Kinetic friction:
The frictional force exerted by the surface when an object slides is called as kinetic friction. Also called as sliding friction or dynamic friction, $\mathrm{f}_{\mathrm{k}}-\mu_{\mathrm{k}} \mathrm{N}$
where $\mu_{\mathrm{k}}$ - the coefficient of kinetic friction
N - Normal force exerted by the surface on the object
Methods to reduce friction:
Friction can be reduced
- By using lubricants
- By using Ball bearings
- By polishing
- By streamlining
Question 7.
What is the meaning by 'pseudo force'?
Answer:
Pseudo force is an apparent force which has no origin. It arises due to the non-inertial nature of the frame considered.
Question 8.
State the empirical laws of static and kinetic friction.
Answer:
The empirical laws of friction are:
- Friction is independent of surface of contact.
- Coefficient of kinetic friction is less than coefficient of static friction.
- The direction of frictional force is always opposite to the motion of one body over the other.
- Frictional force always acts on the object parallel to the surface on which the objet is placed,
- The magnitude of frictional force between any two bodies in contact is directly proportional to the normal reaction between them.
Question 9.
State Newton's third law.
Answer:
Newton's third law states that for every action there is an equal and opposite reaction.
Question 10.
What are inertial frames?
Answer:

Inertial frame is the one in which if there is no force on the object, the object moves at constant velocity.
Question 11.
Under what condition will a car skid on a leveled circular road?
Answer:
On a leveled circular road, if the static friction is not able to provide enough centripetal 'force to turn, the vehicle will start to skid
$\mu_s<\frac{v^2}{r g}$
Long Answer Questions
Question 1.
Prove the law of conservation of linear momentum. Use it to find the recoil velocity of a gun when a bullet is fired from it.
Answer:
In nature, conservation laws play a very important role. The dynamics of motion of bodies can be analysed very effectively using conservation laws. There are three conservation laws in mechanics. Conservation of total energy, conservation of total linear momentum, and conservation of angular momentum. By combining Newton's second and third laws, we can derive the law of conservation of total linear momentum. When two particles interact with each other, they exert equal and opposite forces on each other.
The particle 1 exerts force $\overrightarrow{\mathrm{F}}_{12}$ on particle 2 and particle 2 exerts an exactly equal and opposite force $\overrightarrow{\mathrm{F}}_{12}$ on particle 1 according to Newton's third law.
$\overrightarrow{\mathrm{F}}_{12}=-\overrightarrow{\mathrm{F}}_{12}$
In terms of momentum of particles, the force on each particle (Newton's second law) can be written as -
$\overrightarrow{\mathrm{F}}_{12}=\frac{d \vec{p}_1}{d t}$ and $\overrightarrow{\mathrm{F}}_{21}=\frac{d \vec{p}_2}{d t}$

Here $\vec{p}_1$ is the momentum of particle 1 which changes due to the force $\vec{F}_{12}$ exerted by particle 2. Further Po is the momentum of particle $\vec{p}_2$ This changes due to $\vec{F}_{21}$ exerted by particle 1.
Substitute equation (2) in equation (1)
$
\begin{aligned}
& \frac{d \vec{p}_1}{d t}=-\frac{d \vec{p}_2}{d t} \ldots \\
& \frac{d \vec{p}_1}{d t}+\frac{d \vec{p}_2}{d t}=0 \ldots \\
& \frac{d}{d t}\left(\vec{p}_1+\vec{p}_2\right)=0
\end{aligned}
$
It implies that $\vec{p}_1+\vec{p}_2=$ constant vector (always).
$\vec{p}_1+\vec{p}_2$ is the total linear momentum of the two particles $\left(\vec{p}_{t o t}=\vec{p}_1+\vec{p}_2\right)$.It is also called as total linear momentum of the system. Here, the two particles constitute the system. From this result, the law of conservation of linear momentum can be stated as follows.
If there are no external forces acting on the system, then the total linear momentum of the system $\left(\vec{p}_{t o t}\right)$ is always a constant vector. In other words, the total linear momentum of the
system is conserved in time. Here the word 'conserve' means that $\vec{p}_1$ and $\vec{p}_2$ can vary, in such a way that $\vec{p}_1+\vec{p}_2$ is a constant vector.

The forces $\vec{F}_{12}$ and $\vec{F}_{21}$ are called the internal forces of the system, because they act only between the two particles. There is no external force acting on the two particles from outside. In such a case the total linear momentum of the system is a constant vector or is conserved.
Meaning of law of conservation of momentum:
1. The Law of conservation of linear momentum is a vector law. It implies that both the magnitude and direction of total linear momentum are constant. In some cases, this total momentum can also be zero.
2. To analyse the motion of a particle, we can either use Newton's second law or the law of conservation of linear momentum. Newton's second law requires us to specify the forces involved in the process. This is difficult to specify in real situations. But conservation of linear momentum does not require any force involved in the process. It is convenient and hence important.

For example, when two particles collide, the forces exerted by these two particles on each other is difficult to specify. But it is easier to apply conservation of linear momentum during the collision process.
Examples:
Consider the firing of a gun. Here the system is Gun+bullet. Initially the gun and bullet are at rest, hence the total linear momentum of the system is zero. Let $\vec{p}_1$ be the momentum of the bullet and $\vec{p}_2$ the momentum of the gun before firing. Since initially both are at rest,

Total momentum before firing the gun is zero, $\vec{p}_1+\vec{p}_2=0$.
According to the law of conservation of linear momentum, total linear momentum has to be zero after the firing also.

When the gun is fired, a force is exerted by the gun on the bullet in forward direction. Now the momentum of the bullet changes from $\vec{p}_1$ to $\vec{p}_1$ To conserve the total linear momentum of the system, the momentum of the gun must also change from $\vec{p}_2$ to $\vec{p}_2^{\prime}$. Due to the conservation of linear momentum, $\vec{p}_1+\vec{p}_2^{\prime}=0$.
It implies that $\vec{p}_1^{\prime}=\vec{p}_2^{\prime}$, the momentum of the gun is exactly equal, but in the opposite direction to the momentum of the bullet. This is the reason after firing, the gun suddenly moves backward with the momentum $\left(-\vec{p}_2^{\prime}\right)$. It is called 'recoil momentum'. Th is is an example of conservation of total linear momentum.

Question 2.
What are concurrent forces? State Lami's theorem.
Answer:
Concurrent force:
A collection of forces is said to be concurrent, if the lines of forces act at a common point.
Lami's Theorem:
If a system of three concurrent and coplanar forces is in equilibrium, then Lami's theorem states that the magnitude of each force of the system is proportional to sine of the angle between the other two forces.
i.e. $\left|\overrightarrow{\mathrm{F}}_1\right| \propto \sin \alpha,\left|\overrightarrow{\mathrm{F}}_2\right| \propto \sin \beta,\left|\overrightarrow{\mathrm{F}}_2\right| \propto \sin \gamma$
Question 3.
Explain the motion of blocks connected by a string in
1. Vertical motion
2. Horizontal motion.
Answer:
When objects are connected by strings and When objects are connected by strings and a force $\mathrm{F}$ is applied either vertically or horizontally or along an inclined plane, it produces a tension $T$ in the string, which affects the acceleration to an extent. Let us discuss various cases for the same.
Case 1:
Vertical motion:
Consider two blocks of masses $\mathrm{m}_1$ and $\mathrm{m}_2\left(\mathrm{~m}_1>\mathrm{m}_2\right)$ connected by a light and in-extensible string that passes over a pulley as shown in Figure.

Let the tension in the string be $\mathrm{T}$ and acceleration a. When the system is released, both the blocks start moving, $\mathrm{m}_2$ vertically upward and $\mathrm{m}_{\mathrm{k}}$, downward with same acceleration a. The gravitational force $\mathrm{m}_1 \mathrm{~g}$ on mass $\mathrm{m}_1$ is used in lifting the mass $\mathrm{m}_2$. The upward direction is
chosen as y direction. The free body diagrams of both masses are shown in Figure.

Applying Newton's second law for mass $\mathrm{m}_2 \mathrm{~T} \hat{j}-\mathrm{m}_2 \mathrm{~g} \hat{j}=\mathrm{m}_2$ a $\hat{j}$ The left hand side of the above equation is the total force that acts on $\mathrm{m}_2$ and the right hand side is the product of mass and acceleration of $\mathrm{m}_2$ in $\mathrm{y}$ direction.
By comparing the components on both sides, we get $\mathrm{T}=\mathrm{m}_2 \mathrm{~g}=\mathrm{m}_2 \mathrm{a} \ldots \ldots \ldots .(1)$
Similarly, applying Newton's second law for mass $\mathrm{m}_2$
$\mathrm{T} \hat{j}-\mathrm{m}_1 \mathrm{~g} \hat{j}=\mathrm{m}_1 \mathrm{a} \hat{j}$
As mass $\mathrm{m}_1$ moves downward $(-\hat{j})$, its acceleration is along $(-\hat{j})$
By comparing the components on both sides, we get
$\mathrm{T}=\mathrm{m}_1 \mathrm{~g}=-\mathrm{m}_1 \mathrm{a}$
$\mathrm{m}_1 \mathrm{~g}-\mathrm{T}=\mathrm{m}_1 \mathrm{a}$.........(2)
Adding equations (1) and (2), we get
$\mathrm{m}_1 \mathrm{~g}-\mathrm{m}_2 \mathrm{~g}=\mathrm{m}_1 \mathrm{a}+\mathrm{m}_2 \mathrm{a}$
$\left(\mathrm{m}_1-\mathrm{m}_2\right) \mathrm{g}=\left(\mathrm{m}_1+\mathrm{m}_2\right) \mathrm{a} \ldots \ldots \ldots . .(3)$
From equation (3), the acceleration of both the masses is -
$
\mathrm{a}=\left(\frac{m_1-m_2}{m_1+m_2}\right) \mathrm{g}
$
If both the masses are equal $\left(\mathrm{m}_1=\mathrm{m}_2\right)$, from equation (4)
$
a=0
$
This shows that if the masses are equal, there is no acceleration and the system as a whole will be at rest.
To find the tension acting on the string, substitute the acceleration from the equation (4) into the equation (1).
$\mathrm{T}=\mathrm{m}_2 \mathrm{~g}=\mathrm{m}_2\left(\frac{m_1-m_2}{m_1+m_2}\right)$
$\mathrm{T}=\mathrm{m}_2 \mathrm{~g}+\mathrm{m}_2\left(\frac{m_1-m_2}{m_1+m_2}\right) \mathrm{g} \ldots \ldots \ldots .(5)$
By taking $\mathrm{m}_2$ g common in the RHS of equation (5)

$
\begin{aligned}
\mathrm{T} & =m_2 g\left(1+\frac{m_1-m_2}{m_1+m_2}\right) \\
\mathrm{T} & =m_2 g\left(\frac{m_1+m_2+m_1-m_2}{m_1+m_2}\right) \\
\mathrm{T} & =\left(\frac{2 m_1 m_2}{m_1+m_2}\right) g
\end{aligned}
$
Equation (4) gives only magnitude of acceleration.
For mass $\mathrm{m}_1$, the acceleration vector is given by $\vec{a}=\frac{m_1-m_2}{m_1+m_2} \hat{j}$
For mass $\mathrm{m}_2$, the acceleration vector is given by $\vec{a}=\frac{m_1-m_2}{m_1+m_2} \hat{j}$
Case 2:
Horizontal motion:
In this case, mass $\mathrm{m}_2$ is kept on a horizontal table and mass $\mathrm{m}_1$, is hanging through a small pulley as shown in figure. Assume that there is no friction on the surface

As both the blocks are connected to the un stretchable string, if $\mathrm{m}_1$ moves with an acceleration a downward then $\mathrm{m}_2$ also moves with the same acceleration a horizontally. The forces acting on mass $\mathrm{m}_2$ are
- Downward gravitational force $\left(\mathrm{m}_2 \mathrm{~g}\right)$
- Upward normal force $(\mathrm{N})$ exerted by the surface
- Horizontal tension (T) exerted by the string
The forces acting on mass $m_1$ are
- Downward gravitational force $\left(\mathrm{m}_1 \mathrm{~g}\right)$
- Tension (T) acting upwards
The free body diagrams for both the masses is shown in figure.

Applying Newton's second law for $\mathrm{m}_1$
$\mathrm{T} \hat{i}-\mathrm{m}_1 \mathrm{~g} \hat{j}=-\mathrm{m}_1 \mathrm{a} \hat{j}$ (alongy direction)
By comparing the components on both sides of the above equation, $\mathrm{T}-\mathrm{m}_1 \mathrm{~g}=-\mathrm{m}_1 \mathrm{a} \ldots \ldots \ldots \ldots .(1)$
Applying Newton's second law for $\mathrm{m}_2$
$\mathrm{Ti}=\mathrm{m}_1$ ai (along $\mathrm{x}$ direction)
By comparing the components on both sides of above equation, $\mathrm{T}=\mathrm{m}_2 \mathrm{a} \ldots \ldots \ldots \ldots . .(2)$
There is no acceleration along $\mathrm{y}$ direction for $\mathrm{m}_2$.
$
\mathrm{N} \hat{j}-\mathrm{m}_2 \mathrm{~g} \hat{j}=0
$
By comparing the components on both sides of the above equation
$
\begin{aligned}
& \mathrm{N}-\mathrm{m}_2 \mathrm{~g}=0 \\
& \mathrm{~N}=\mathrm{m}_2 \mathrm{~g} \ldots \ldots \ldots .(3)
\end{aligned}
$
By substituting equation (2) in equation (1), we can find the tension $T$
$
\begin{aligned}
& \mathrm{m}_2 \mathrm{a}-\mathrm{m}_1 \mathrm{~g}=-\mathrm{m}_1 \mathrm{a} \\
& \mathrm{m}_2 \mathrm{a}+\mathrm{m}_1 \mathrm{a}=\mathrm{m}_1 \mathrm{~g} \\
& \mathrm{a}=\frac{m_1}{m_1+m_2} \mathrm{~g} \ldots \ldots \ldots .
\end{aligned}
$
Tension in the string can be obtained by substituting equation (4) in equation (2)
$
\mathrm{T}=\frac{m_1 m_2}{m_1+m_2} \mathrm{~g}
$
Comparing motion in both cases, it is clear that the tension in the string for horizontal motion is half of the tension for vertical motion for same set of masses and strings. This result has an
important application in industries. The ropes used in conveyor belts (horizontal motion) work for longer duration than those of cranes and lifts (vertical motion).

Question 4 .
Briefly explain the origin of friction. Show that in an inclined plane, angle of friction is pqual to angle of repose.
Answer:
If a very gentle force in the horizontal direction is given to an object at rest on the table it does not move. It is because of the opposing force exerted by the surface on the object which resists its motion. This force is called the frictional force. During the time of Newton and Galileo, frictional force was considered as one of the natural forces like gravitational force. But in the twentieth century, the understanding on atoms, electron and protons has changed the perspective.
The frictional force is actually the electromagnetic force between the atoms on the two surfaces. Even well polished surfaces have irregularities on the surface at the microscopic level. The component of force parallel to the inclined plane $(\mathrm{mg} \sin \theta)$ tries to move the object down. The component of force perpendicular to the inclined plane $(\mathrm{mg} \cos \theta)$ is balanced by the Normal force $(\mathrm{N})$. $\mathrm{N}=\mathrm{mg} \cos \theta$
When the object just begins to move, the static friction attains its maximum value $\mathrm{f}_{\mathrm{s}}=f_s^{\max }=\mu_{\mathrm{s}} \mathrm{N}$
This friction also satisfies the relation $f_s^{\max }=\mu_{\mathrm{s}} \mathrm{mg} \sin \theta$
Equating the right hand side of equations (1) and (2), $\left(f_s^{\max }\right) / \mathrm{N}=\sin \theta / \cos \theta$
From the definition of angle of friction, we also know that $\tan \theta=\mu_{\mathrm{s}} \ldots \ldots \ldots .(3)$
in which $\theta$ is the angle of friction.
Thus the angle of repose is the same as angle of friction. But the difference is that the angle of repose refers to inclined surfaces and the angle of friction is applicable to any type of surface.
Question 5 .
State Newton's three laws and discuss their significance.
Answer:
First Law:
Every object continues to be in the state of rest or of uniform motion (constant velocity) unless there is external force acting on it.
Second Law:
The force acting on an object is equal to the rate of change of its momentum
Third Law:
For every action there is an equal and opposite reaction.

Significance of Newton's Laws:
1. Newton's laws are vector laws. The equation $\overline{\mathrm{F}}=$ mà is a vector equation and essentially it is equal to three scalar equations. In Cartesian coordinates, this equation can be written as $F_x \hat{i}$ $+\mathrm{F}_{\mathrm{y}} \hat{j}+\mathrm{F}_{\mathrm{z}} \hat{j}=\mathrm{ma}_{\mathrm{x}} \hat{i}+\mathrm{ma}_{\mathrm{y}} \hat{j}+\mathrm{ma}_{\mathrm{z}} \hat{j}$
By comparing both sides, the three scalar equations are
$\mathrm{F}_{\mathrm{x}}=m \mathrm{ma}_{\mathrm{x}}$ The acceleration along the $\mathrm{x}$-direction depends only on the component of force acting along the $\mathrm{x}$ - direction.
$\mathrm{F}_{\mathrm{y}}=\mathrm{ma}_{\mathrm{y}}$ The acceleration along the $\mathrm{y}$ direction depends only on the component of force acting along the $y$-direction.
$\mathrm{F}_{\mathrm{z}}=\mathrm{ma}_{\mathrm{z}}$ The acceleration along the $\mathrm{z}$ direction depends only on the component of force acting along the $z$ - direction.
From the above equations, we can infer that the force acting along $y$ direction cannot alter the acceleration along $\mathrm{x}$ direction. In the same way, $\mathrm{F}_{\mathrm{z}}$ cannot affect $\mathrm{a}_{\mathrm{y}}$ and $\mathrm{a}_{\mathrm{x}}$. This understanding is essential for solving problems.
2. The acceleration experienced by the body at time $t$ depends on the force which acts on the body at that instant of time. It does not depend on the force which acted on the body before the time t. This can be expressed as $\overline{\mathrm{F}}(\mathrm{t})=\mathrm{ma}(\mathrm{t})$
Acceleration of the object does not depend on the previous history of the force. For example, when a spin bowler or a fast bowler throws the ball to the batsman, once the ball leaves the hand of the bowler, it experiences only gravitational force and air frictional force. The acceleration of the ball is independent of how the ball was bowled (with a lower or a higher speed).
3. In general, the direction of a force may be different from the direction of motion. Though in some cases, the object may move in the same direction as the direction of the force, it is not always true. A few examples are given below.
Case 1:
Force and motion in the same direction:
When an apple falls towards the Earth, the direction of motion (direction of velocity) of the apple and that of force are in the same downward direction as shown in the Figure. 

Case 2:
Force and motion not in the same direction:
The Moon experiences a force towards the Earth. But it actually moves in elliptical orbit. In this case, the direction of the force is different from the direction of motion as shown in Figure.

Case 3 :
Force and motion in opposite direction: If an object is thrown vertically upward, the direction of motion is upward, but gravitational force is downward as shown in the Figure.

Case 4:
Zero net force, but there is motion:
When a raindrop gets detached from the cloud it experiences both downward gravitational force and upward air drag force. As it descends towards the Earth, the upward air drag force increases and after a certain time, the upward air drag force cancels the downward gravity. From then on the raindrop moves at constant velocity till it touches the surface of the Earth. Hence the raindrop comes with zero net force, therefore with zero acceleration but with nonzero terminal velocity. It is shown in the Figure

4. If multiple forces $\overrightarrow{\mathrm{F}}_1, \overrightarrow{\mathrm{F}}_2, \overrightarrow{\mathrm{F}}_3, \ldots \ldots . \overrightarrow{\mathrm{F}}_n$ act on the same body, then the total force $\left(\overrightarrow{\mathrm{F}}_{n e t}\right)$ is equivalent to the vectorial sum of the individual forces. Their net force provides the acceleration.
$
\overrightarrow{\mathrm{F}}_{n e t}=\overrightarrow{\mathrm{F}}_1+\overrightarrow{\mathrm{F}}_2+\overrightarrow{\mathrm{F}}_3+\ldots \ldots \ldots+\overrightarrow{\mathrm{F}}_n
$

Newton's second law for this case is -
$
\overrightarrow{\mathrm{F}}_{n e t}=\mathrm{ma}
$
In this case the direction of acceleration is in the direction of net force.
Example:
Bow and arrow

5. Newton's second law can also be written in the following form.
Since the acceleration is the second derivative of position vector of the body $\left(\vec{a}=\frac{d^2 \vec{r}}{d t^2}\right)$ the force on the body is -
$
\overline{\mathrm{F}}=\mathrm{m} \frac{d^2 \vec{r}}{d t^2}
$
From this expression, we can infer that Newton's second law is basically a second order ordinary differential equation and whenever the second derivative of position vector is not zero, there must be a force acting on the body.
6. If no force acts on the body then Newton's second law, $\mathrm{m}=\frac{d \vec{v}}{d t}=0$
It implies that $\overline{\mathrm{v}}=$ constant. It is essentially Newton's first law. It implies that the second law is consistent with the first law. However, it should not be thought of as the reduction of second law to the first when no force acts on the object. Newton's first and second laws are independent laws. They can internally be consistent with each other but cannot be derived from each other.
7. Newton's second law is cause and effect relation. Force is the cause and acceleration is the effect. Conventionally, the effect should be written on the left and cause on the right hand side of the equation. So the correct way of writing Newton's second law is $\mathrm{m} \overline{\mathrm{a}}=\overline{\mathrm{F}}$ or $\frac{d \vec{p}}{d t}=\overline{\mathrm{F}}$
Question 6.
Explain the similarities and differences of centripetal and centrifugal forces.
Answer:
Salient features of centripetal and centrifugal forces.
Centripetal Force:
- It is a real force which is exerted on the body by the external agencies like gravitational force, tension in the string, normal force etc.
- Acts in both inertial and non-inertial frames

- It acts towards the axis of rotation or center of the circle in circular motion $\left|\overrightarrow{\mathrm{F}}_{\mathrm{C}_{\mathrm{P}}}\right|=m \omega^2 \mathbf{r}=\frac{m v^2}{r}$
- Real force and has real effects
- Origin of centripetal force is interaction between two objects.
- In inertial frames centripetal force has to be included when free body diagrams are drawn.
Centrifugal Force:
- It is a pseudo force or fictitious force which cannot arise from gravitational force, tension force, normal force etc.
- Acts only in rotating frames (non-inertial frame)
- It acts outwards from the axis of rotation or radially outwards from the center of the circular motion
- Pseudo force but has real effects
- Origin of centrifugal force is inertia. It does not arise from interaction. In an inertial frame the object's inertial motion appears as centrifugal force in the rotating frame.
- In inertial frames there is no centrifugal force. In rotating frames, both centripetal and centrifugal force have to be included when free body diagrams are drawn.
Question 7.
Briefly explain 'centrifugal force' with suitable examples.
Answer:
To use Newton's first and second laws in the rotational frame of reference, we need to include a Pseudo force called centrifugal force. This centrifugal force appears to act on the object with respect to rotating frames.
Circular motion can be analysed from two different frames of reference. One is the inertial frame (which is either at rest or in uniform motion) where Newton's laws are obeyed. The other is the rotating frame of reference which is a non - inertial frame of reference as it is accelerating.

When we examine the circular motion from these frames of reference the situations are entirely different. To use Newton's first and second laws in the rotational frame of reference, we need to include a pseudo force called 'centrifugal force'. This 'centrifugal force' appears to act on the object with respect to rotating frames. To understand the concept of centrifugal force, we can take a specific case and discuss as done below.

Free body diagram of a particle including the centrifugal force Consider the case of a whirling motion of a stone tied to a string. Assume that the stone has angular velocity $\omega$ in the inertial frame (at rest). If the motion of the stone is observed from a frame which is also rotating along with the stone with same angular velocity $\omega$ then, the stone appears to be at rest.
This implies that in addition to the inward centripetal force $-\mathrm{m} \omega^2 \mathrm{r}$ there must be an equal and opposite force that acts on the stone outward with value $+\mathrm{m} \omega^2 \mathrm{r}$. So the total force acting on the stone in a rotating frame is equal to zero $\left(-m \omega^2 r+m \omega^2 r=0\right)$. This outward force $+m \omega^2 r$ is called the centrifugal force. The word 'centrifugal' means 'flee from center'.
Note that the 'centrifugal force' appears to act on the particle, only when we analyse the motion from a rotating frame. With respect to an inertial frame there is only centripetal force which is given by the tension in the rstring. For this reason centrifugal force is called as a 'pseudo force'. A pseudo force has no origin. It arises due to the non inertial nature of the frame considered. When circular motion problems are solved from a rotating frame of reference, while drawing free body diagram of a particle, the centrifugal force should necessarily be included as shown in the figure.

Question 8.
Briefly explain 'rolling friction'.
Answer:
The invention of the wheel plays a crucial role in human civilization. One of the important applications is suitcases with rolling on coasters. Rolling wheels makes it easier than carrying luggage. When an object moves on a surface, essentially it is sliding on it. But wheels move on the surface through rolling motion. In rolling motion when a wheel moves on a surface, the point of contact with surface is always at rest. Since Rolling and kinetic friction the point of
contact is at rest, there is no relative motion between the wheel and surface.

Hence the frictional fore is very less. At the same time if an object moves without a wheel, there is a relative motion between the object and the surface. As a result frictional force is larger. This makes it difficult to move the object. The figure shows the difference between rolling and kinetic friction. Ideally in pure rolling, motion of the point of contact with the surface should be at rest, but in practice it is not so.
Due to the elastic nature of the surface at the point of contact there will be some deformation on the object at this point on the wheel or surface as shown in figure. Due to this deformation, there will be minimal friction between wheel and surface. It is called 'rolling friction. In fact, rolling friction' is much smaller than kinetic friction.
Question 9.
Describe the method of measuring angle of repose.
Answer:
When objects are connected by strings and When objects are connected by strings and a force $\mathrm{F}$ is applied either vertically or horizontally or along an inclined plane, it produces a tension $T$ in the string, which affects the acceleration to an extent. Let us discuss various cases for the  same.

Case 1:
Vertical motion:
Consider two blocks of masses $\mathrm{m}_1$ and $\mathrm{m}_2\left(\mathrm{~m}_1>\mathrm{m}_2\right)$ connected by a light and in extensible string that passes over a pulley as shown in Figure.

Let the tension in the string be $\mathrm{T}$ and acceleration a. When the system is released, both the blocks start moving, $\mathrm{m}_2$ vertically upward and $\mathrm{m}_{\mathrm{k}}$, downward with same acceleration a. The gravitational force $\mathrm{m}_1$ g on mass $\mathrm{m}_1$ is used in lifting the mass $\mathrm{m}_2$. The upward direction is chosen as y direction. The free body diagrams of both masses are shown in Figure.

Applying Newton's second law for mass $m_2$
$\mathrm{T} \hat{j}-\mathrm{m}_2 \mathrm{~g} \hat{j}=\mathrm{m}_2 \mathrm{a} \hat{j}$ The left hand side of the above equation is the total force that acts on $\mathrm{m} 2$ and the right hand side is the product of mass and acceleration of $\mathrm{m}_2$ in $\mathrm{y}$ direction.
By comparing the components on both sides, we get $\mathrm{T}=\mathrm{m}_2 \mathrm{~g}=\mathrm{m}_2 \mathrm{a}$

Similarly, applying Newton's second law for mass $\mathrm{m}_2$
$\mathrm{T} \hat{j}-\mathrm{m}_1 \mathrm{~g} \hat{j}=\mathrm{m}_1 \mathrm{a} \hat{j}$
As mass mj moves downward $(-\hat{j})$, its acceleration is along $(-\hat{j})$
By comparing the components on both sides, we get
$\mathrm{T}=\mathrm{m}_1 \mathrm{~g}=-\mathrm{m}_1 \mathrm{a}$
$\mathrm{m}_1 \mathrm{~g}-\mathrm{T}=\mathrm{m}_1 \mathrm{a}$
Adding equations (1) and (2), we get
$\mathrm{m}_1 \mathrm{~g}-\mathrm{m}_2 \mathrm{~g}=\mathrm{m}_1 \mathrm{a}+\mathrm{m}_2 \mathrm{a}$
$\left(\mathrm{m}_1-\mathrm{m}_2\right) \mathrm{g}=\left(\mathrm{m}_1+\mathrm{m}_2\right) \mathrm{a} \ldots \ldots \ldots \ldots .(3)$
From equation (3), the acceleration of both the masses is $\mathrm{a}=\left(\frac{m_1-m_2}{m_1+m_2}\right) \mathrm{g} \ldots \ldots \ldots \ldots(4)$
If both the masses are equal $\left(\mathrm{m}_1=\mathrm{m}_2\right)$, from equation (4) $\mathrm{a}=0$
This shows that if the masses are equal, there is no acceleration and the system as a whole will be at rest.
To find the tension acting on the string, substitute the acceleration from the equation (4) into the equation (1).
$\mathrm{T}=\mathrm{m}_2 \mathrm{~g}=\mathrm{m}_2\left(\frac{m_1-m_2}{m_1+m_2}\right)$
$\mathrm{T}=\mathrm{m}_2 \mathrm{~g}+\mathrm{m}_2\left(\frac{m_1-m_2}{m_1+m_2}\right) \mathrm{g} \ldots \ldots \ldots .(5)$
By taking $\mathrm{m} 2 \mathrm{~g}$ common in the RHS of equation (5)
$
\begin{aligned}
\mathrm{T} & =m_2 g\left(1+\frac{m_1-m_2}{m_1+m_2}\right) \\
\mathrm{T} & =m_2 g\left(\frac{m_1+m_2+m_1-m_2}{m_1+m_2}\right) \\
\mathrm{T} & =\left(\frac{2 m_1 m_2}{m_1+m_2}\right) g
\end{aligned}
$

Equation (4) gives only magnitude of acceleration
For mass $\mathrm{m}_1$, the acceleration vector is given by $\vec{a}=\frac{m_1-m_2}{m_1+m_2} \hat{j}$
For mass $\mathrm{m}_2$, the acceleration vector is given by $\vec{a}=\frac{m_1-m_2}{m_1+m_2} \hat{j}$
Case 2 :
Horizontal motion:
In this case, mass $\mathrm{m}_2$ is kept on a horizontal table and mass $\mathrm{m}_1$, is hanging through a small pulley as shown in figure. Assume that there is no friction on the surface

As both the blocks are connected to the un stretchable string, if $\mathrm{m}_1$ moves with an acceleration a downward then $\mathrm{m}_2$ also moves with the same acceleration a horizontally. The forces acting on mass $m_2$ are
1. Downward gravitational force $\left(\mathrm{m}_2 \mathrm{~g}\right)$
2. Upward normal force $(N)$ exerted by the surface
3. Horizontal tension $(\mathrm{T})$ exerted by the string
The forces acting on mass $m_1$ are
1. Downward gravitational force $\left(\mathrm{m}_1 \mathrm{~g}\right)$
2. Tension (T) acting upwards

The free body diagrams for both the masses is shown in figure  2

Applying Newton's second law for $\mathrm{m}_1$
$\mathrm{T} \hat{i}-\mathrm{m}_1 \mathrm{~g} \hat{j}=-\mathrm{m}_1 \mathrm{a} \hat{j}$ (alongy direction)
By comparing the components on both sides of the above equation, $\mathrm{T}-\mathrm{m}_1 \mathrm{~g}=-\mathrm{m}_1 \mathrm{a} \ldots \ldots \ldots \ldots .(1)$
Applying Newton's second law for $\mathrm{m}_2$
$\mathrm{T} \hat{i}=\mathrm{m}_1 \mathrm{a} \hat{i}$ (along $\mathrm{x}$ direction)
By comparing the components on both sides of above equation,
$
\mathrm{T}=\mathrm{m}_2 \mathrm{a} \text {. }
$
There is no acceleration along y direction for $\mathrm{m}_2$.
$
\mathrm{N} \hat{j}-\mathrm{m}_2 \mathrm{~g} \hat{j}=0
$
By comparing the components on both sides of the above equation
$
\mathrm{N}-\mathrm{m}_2 \mathrm{~g}=0
$
$
\mathrm{N}=\mathrm{m}_2 \mathrm{~g}
$
By substituting equation (2) in equation (1), we can find the tension $\mathrm{T}$
$
\begin{aligned}
& \mathrm{m}_2 \mathrm{a}-\mathrm{m}_1 \mathrm{~g}=-\mathrm{m}_1 \mathrm{a} \\
& \mathrm{m}_2 \mathrm{a}+\mathrm{m}_1 \mathrm{a}=\mathrm{m}_1 \mathrm{~g} \\
& \mathrm{a}=\frac{m_1}{m_1+m_2} \mathrm{~g} \ldots \ldots \ldots
\end{aligned}
$
Tension in the string can be obtained by substituting equation (4) in equation (2)
$
\mathrm{T}=\frac{m_1 m_2}{m_1+m_2} \mathrm{~g}
$
Comparing motion in both cases, it is clear that the tension in the string for horizontal motion is half of the tension for vertical motion for same set of masses and strings. This result has an important application in industries. The ropes used in conveyor belts (horizontal motion) work for longer duration than those of cranes and lifts (vertical motion).
Question 10.
Explain the need for banking of tracks.
Answer:
In a leveled circular road, skidding mainly depends on the coefficient of static friction py. The coefficient of static friction depends on the nature of the surface which has a maximum limiting value. To avoid this problem, usually the outer edge of the road is slightly raised compared to inner edge. This is called banking of roads or tracks. This introduces an inclination, and the angle is called banking angle. "Let the surface of the road make angle 9 with horizontal surface. Then the normal force makes the same angle 9 with the vertical. When the car takes a turn.

there are two forces acting on the car:
(a) Gravitational force $\mathrm{mg}$ (downwards)
(b) Normal force $N$ (perpendicular to surface)
We can resolve the normal force into two components $\mathrm{N} \cos \theta$ and $\mathrm{N} \sin \theta$. The component balances the downward gravitational force ' $\mathrm{mg}$ ' and component will provide the necessary centripetal acceleration. By using Newton second law.

The banking angle 0 and radius of curvature of the road or track determines the safe speed of the car at the turning. If the speed of car exceeds this safe speed, then it starts to skid outward but frictional force comes into effect and provides an additional centripetal force to prevent the outward skidding.At the same time, if the speed of the car is little lesser than safe speed, it starts to skid inward and frictional force comes into effect, which reduces centripetal force to prevent inward skidding. However if the speed of the vehicle is sufficiently greater than the correct speed, then frictional force cannot stop the car from skidding.
Question 11.
Calculate the centripetal acceleration of Moon towards the Earth.
Answer:
The centripetal acceleration is given by $\mathrm{a}=\frac{v^2}{r}$ This expression explicitly depends on Moon's speed which is nontrivial. We can work with the formula $\omega^2 \mathrm{R}_{\mathrm{m}}=a_m$
$a_m$ is centripetal acceleration of the Moon due to Earth's gravity, $\omega$ is angular velocity $\mathrm{R}_{\mathrm{m}}$ is the distance between Earth and the Moon, which is 60 times the radius of the Earth. $\mathrm{R}_{\mathrm{m}}=60 \mathrm{R}=60 \times 6.4 \times 10^6=384 \times 10^6 \mathrm{~m}$
As we know the angular velocity $\omega=\frac{2 \pi}{T}$ and $T=27.3$ days $=27.3 \times 24 \times 60 \times 60$ second $=$ $2.358 \times 10^6 \mathrm{sec}$.
By substituting these values in the formula for acceleration $\mathrm{a}_6=\frac{\left(4 \pi^2\right)\left(384 \times 10^6\right)}{\left(2.358 \times 10^8\right)^2}=0.00272 \mathrm{~ms}^{-2}$