9th Class Physics New Book 2025 Chapter 3 Solved Exercise
9th Class Physics New Book 2025 Chapter 3 Solved Exercise
Class 9th Physics New Book 2025 Chapter3 MCQs
Tick the correct answer.
3.1 When we kick a stone, we get hurt. This is due to:
(a) inertia (b) velocity (c) momentum (d) reaction
Option ( d ) is correct.
Explanation:-
Newton’s Third Law of motion. Action and reaction are equal in magnitude but opposite in direction
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3.2 An object will continue its motion with constant acceleration until:
(a) the resultant force on it begins to decrease.
(b) the resultant force on it is zero.
(c) the resultant force on it begin to increase.
(d) the resultant force is at right angle to its tangential velocity.
Option (a) is correct.
Explanation: –
Newton’s First Law of motion. An object moving with constant speed in straight line continues it unless an external force tries to stop it.
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3.3 Which of the following is a non-contact force?
(a) Friction (b) Air resistance (c) Electrostatic force (d) Tension in the string
Option (c) is correct.
Explanation: –
Non contact force is a force that acts on an object without physical contact.
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3.5 A particle of mass m moving with a velocity v collides with another particle of the same mass at rest. The velocity of the first particle after collision is:
(a) v (b) -v (c) 0 (d) -1/2
Option (c) is correct.
Explanation: –
The moving particle comes to rest because collision is perfectly elastics and both the particles have equal masses
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3.6 Conservation of linear momentum is equivalent to:
(a) Newton’s fist law of motion
(b) Newton’s second law of motion
(c) Newton’s third law of motion
(d) None of these
Option (c) is correct
Explanation: –
Action and reaction are equal in magnitude but opposite in direction. Before collision is action and after the collision is reaction.
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3.9 A lubricant is usually introduced between two surfaces to decrease friction. The lubricant:
(a) decreases temperature
(b) acts as ball bearings
(c) prevents direct contact of the surfaces
(d) provides rolling friction
Option (c) is correct.
Explanation: –
A thin layer of lubricant reduces friction because their will be no direct contact between the surfaces.
Physics Class 9 Solved Exercise Chapter 3 Short Answer Questions
Q1 What kind of changes in motion may be produced by a force ?
Force can change the speed, direction, shape, size and state of an object.
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Q2 Give 5 examples of contact forces.
A force that is exerted by one object on the other at the point of contact is called contact force.
Examples of contact forces are friction, thrust, air resistance, drag force, normal force, tension force and elastic force.
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Q3 An object moves with a constant velocity in free space. How long will the object continue to move with this velocity ?
Due to Newton’s first law of motion the object will move forever unless the object experiences an external force.
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Q4 Define impulse of force ?
Impulse is the total change in momentum of an object.
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Q5 Why has not Newton’s first law been proved on the Earth ?
There is no proof of Newton’s first law on Earth. Because according to science, Earth is continuously rotating about the sun and spinning about its own axis.
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Q6 When sitting in a car which suddenly accelerates from rest, you are pushed back into the seat, why ?
It happens due to inertia. Body is initially at rest. When car suddenly accelerates, body resists due to inertia but seat pushes you forward. As a result, seat pushes you back.
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Q9 Define terminal velocity of an object.
Terminal velocity is the constant maximum velocity of a freely falling object when the force of gravity is balanced by the air resistance and no further acceleration occurs.
9th Class Physics New Book 2025 Chapter 3 Constructed Response Questions
3.1 Two ice skaters weighing 60 kg and 80 kg push off against each other on a frictionless ice track. The 60 kg skater gains a velocity of 4 metre per second. Considering all the relevant calculations involved, explain how Newton’s third law applies to this situation.
According to Newton’s third law of motion both the skaters will apply equal and opposite forces on each other.
Law of conservation of momentum
Initial momentum = Final momentum
0 = (60)(4) + (80)(v)
-80v = 240
v= -3 meter per second
The negative sign shows that the 80 kg skater will move in opposite direction with a speed of 3 meter per second.
Hence Newton’s third law tells us that the lighter skater accelerates more and will gain a greater velocity but the heavier skater accelerates less and will gain smaller velocity.
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3.2 Inflatable airbags are installed in the vehicles as safety equipment. In terms of momentum, what is the advantage of airbags over seatbelts?
When suddenly vehicle stops, the passenger’s momentum will change to zero. Airbags increase the time over which the momentum changes.
This increase in time will reduce the force because rate of change of momentum is equal to the applied force. Hence airbags reduce the force on the passenger and provide better protection.
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3.4 When a cricket ball hits high, a fielder tries to catch it. While holding the ball he/she draws hands backward. Why?
When the ball is caught, momentum changes to zero. By drawing the hands backward, the fielder increases the time taken to stop the ball. This increase in time will reduce the force because rate of change of momentum is equal to the applied force. This prevents injury.
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3.5 When someone jumps from a small boat onto the river bank, why does the jumper often fall into the water? Explain.
Newton’s third law of motion explains it.
Due to Newton’s third law of motion the boat exerts a backward force on the person. This makes the boat move and reduces stability, so the person may fall into the water.
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3.6 Imagine that if friction vanishes suddenly from everything, then what could be the scenario of daily life activities?
If friction vanishes suddenly from everything, daily life will be completely disturbed. Because holding, driving and walking will be impossible.
9th Class Physics New Book 2025 Chapter 3 Comprehensive Questions
Q1:- Explain the concept of force by practical examples.
Force is an agency which changes or tries to change the state of an object. Force transfers energy to the object.
Pushing a door, kicking a ball, pulling a door and squeezing a ball are examples of force. Friction, gravity, compression and tension are also forces.
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Q2:- Describe Newton’s laws of motion.
Newton’s first law of motion:-
A object does not change its state of rest or motion in a straight line unless an external force is applied on it.
Newton’s first law is also called law of inertia. Inertia is that property of matter due to which it resists to change its state.
Examples
A book on a table at rest will be at rest unless an external force is applied on it.
A car moving on a straight road will continue it unless an external force is applied on it
Newton’s second law of motion:-
When we apply a force on an object, it produces acceleration in it and this acceleration is directly proportional to the applied force and inversely proportional to the mass of that object.
$$a\;\propto F\\a\;\propto\frac1m\\After\;combining,\;we\;get\\a\propto\frac Fm\\a\;=\;k\;\frac Fm\\If\;\\a=\;1ms^{-2}\\F\;=\;1\;N\\m\;=\;1\;kg\\then\\K\;=\;1\\So\\a=\frac Fm\\of\\F\;=\;ma$$
Which is mathematical form of Newton’s second law of motion
Examples
If we apply the same force on two balls, small ball will accelerate faster than a heavy ball.
Hitting a ball harder makes it go faster.
Newton’s Third law of motion:-
For every action, there is always an equal and opposite reaction.
It can also be expressed as of one body exerts a force on the second body. the second body also exerts an equal and opposite force on the first body.
Examples
Lets the check the force on a book which is lying on the table. The force acting downward on the book is the weight. The book exerts a downward force on the table equal to its weight. The table also exerts a reaction force on the book.
When a bullet is fired from a gun, the bullet moves in the forward direction as an action and the gun recoils back as a reaction.
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Q3:- Define momentum and express Newton’s 2nd law of motion in terms of change in momentum.
The momentum of a moving body is the product of its mass and velocity.
P = m v
According to Newton’s Second law of motion.
$$F\;=\;m\;a\\F\;=\;m\;(\frac{\triangle v}{\triangle t})\\F\;=\;\frac{m\;(\;v_f\;-\;v_i)}{\triangle t}\\F\;=\frac{\;m\;v_f\;-\;m\;v_i}{\triangle t}\\F\;=\;\frac{p_f\;-\;p_i}{\triangle t}\\F\;=\;\frac{\triangle p}{\triangle t}$$
Equation shows that rate of change of momentum of a body is equal to the applied force.
Momentum is a vector quantity and its unit is Ns.
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Q4:- State and explain the principle of conservation of momentum.
Law of conservation of momentum states that for an isolated system momentum remains conserved.
Its mean that if no external force acts on an isolated system, the final momentum of the system is equal to the initial momentum of the system.
Let consider a system of two hard interacting balls.
$$Mass\;of\;1st\;ball\;=\;m_1\\Mass\;of\;2nd\;ball\;=\;m_2\\Velcoity\;of\;1st\;ball\\before\;collision=\;v_1\\Velcoity\;of\;2nd\;ball\\before\;collision\;=\;v_2\\Velcoity\;of\;1st\;ball\\after\;collision=\;v_1^\boldsymbol’\\Velcoity\;of\;2nd\;ball\\after\;collision=\;v_2^\boldsymbol’\\$$
When there is a collision of the two balls, there is a transfer of momentum from one ball to the other.
According to law of conservation of momentum
Momentum before the collision = momentum after the collision
$$\;m_1v_1+m_2v_2=m_1v_1^\boldsymbol’+m_2v_2^\boldsymbol’$$
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Q5:- Describe the motion of a block on a table taking into account the friction between the two surfaces. What is the static friction and kinetic friction ?
When block is placed on a table, two main forces act on it, weight in the downward direction and normal reaction in the upward direction. if we push it slightly, block will not move due to static friction between the block and the table. As we increase the applied force, the static friction also increases and attains its maximum limit.
When we apply a greater force then the maximum static friction, the block starts moving. When block moves, friction still exists. It is known as kinetic friction.
The force between two solid surfaces is called sliding friction. Sliding friction has two main categories.
Static Friction:-
The force that opposes the start of motion of a body at rest is called static friction. It has maximum value, after which the object will start moving.
$$static\;friction\;is\;denoted\\by\;F_s\;or\;f_s$$
Kinetic Friction:-
The friction which acts when object is moving is called kinetic friction. Its magnitude is generally less than maximum static friction.
$$kinetic\;friction\;is\;denoted\\by\;F_k\;or\;f_k$$
Class 9 Physics New Book 2025 Chapter 3 Numerical
$$\boldsymbol3\boldsymbol{\mathit.}\boldsymbol1\;\;\;A\;10\;kg\;block\;is\;placed\\on\;a\;smooth\;horiznotal\\surface.\;A\;horizontal\;force\\of\;5N\;is\;applied\;to\;the\;block.\\Find:\\(a)\;the\;acceleration\;produced\\in\;the\;block.\\(b)\;the\;velocity\;of\;block\;after\\5\;seconds.$$
$$\boldsymbol D\boldsymbol a\boldsymbol t\boldsymbol a\boldsymbol:\boldsymbol-\\\\Mass\;=\;m\;=\;10\;kg\\Force\;=\;F\;=\;5\;N\\Time\;=\;t\;=\;5\;s\\Initial\;velocity\;=\;v_i\;=\;0\\Acceleration\;=\;a\;=\;?\\Final\;velocity\;=\;v_f\;=\;?\\\\\boldsymbol F\boldsymbol o\boldsymbol r\boldsymbol m\boldsymbol u\boldsymbol l\boldsymbol a\boldsymbol:\boldsymbol-\\\boldsymbol F\boldsymbol\;\boldsymbol=\boldsymbol\;\boldsymbol m\boldsymbol\;\boldsymbol a\\\\5\;N\;=\;(\;10\;kg\;)\;a\\\frac{5\;N}{10\;kg}\;=\;a\\\frac{1\;kg\;m\;s^{-2}}{2\;kg}=\;a\\a\;=\;0.5\;m\;s^{-2}\\\\\boldsymbol F\boldsymbol o\boldsymbol r\boldsymbol m\boldsymbol u\boldsymbol l\boldsymbol a\boldsymbol:\boldsymbol-\\{\boldsymbol v}_{\mathbf f}\boldsymbol\;\boldsymbol=\boldsymbol\;{\boldsymbol v}_{\mathbf i}\boldsymbol\;\boldsymbol+\boldsymbol\;\boldsymbol a\boldsymbol\;\boldsymbol t\\\\v_f\;=\;0\;+\;(\;0.5\;m\;s^{-2}\;)\;(\;5\;s\;)\\v_f\;=\;2.5\;m\;s^{-2}\;\\$$
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$$\boldsymbol3\boldsymbol{\mathit.}\boldsymbol2\;\;\;\;The\;mass\;of\;person\;is\\80\;kg\;what\;will\;be\;his\;weight\\on\;the\;earth\;?\;what\;will\;be\;his\\weight\;on\;the\;moon.\;The\\value\;of\;acceleration\;due\;to\\gravity\;of\;moon\;is\;1.6\;m\;s^{-2}.$$
$$Mass\;=\;m\;=\;80\;kg\\Weight\;on\;Earth=\;w_e=?\\Gravitional\;acceleration\;on\\Earth=g=10\;m\;s^{-2}\\Gravitional\;acceleration\;on\\Moon=g_m=1.6\;m\;s^{-2}\\Weight\;on\;Moon=w_m\;=?\\\\\boldsymbol F\boldsymbol\;\boldsymbol=\boldsymbol\;\boldsymbol w\boldsymbol\;\boldsymbol=\boldsymbol\;\boldsymbol m\boldsymbol\;\boldsymbol g\\\\w_e\;=\;m\;g\\w_e=(80\;kg)\;(10\;ms^{-2})\;\\w_{\mathrm e}\;=\;800\;N\\\\w_m\;=\;m\;g_m\\w_m=(80\;kg)\;(1.6\;ms^{-2})\;\\w_{\mathrm m}\;=\;128\;N$$
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$$\boldsymbol3\boldsymbol{\mathit.}\boldsymbol3\;\;\;\;\;What\;force\;is\;required\\to\;increase\;the\;velocity\boldsymbol\;of\\800\;kg\;\;car\;from\;\;10\;m\;s^{-1}\\to\;\;30\;m\;s^{-1}\;\;in\;\;10\;seconds.$$
$$\boldsymbol D\boldsymbol a\boldsymbol t\boldsymbol a\boldsymbol{\mathit:}\boldsymbol{\mathit-}\\\\Mass\;=\;m\;=\;800\;kg\\Initial\;velocity\;=\;v_i\;=\;10\;m\;s^{-1}\\Final\;velocity\;=\;v_f\;=\;30\;m\;s^{-1}\\Time\;=\;t\;=\;10\;s\\Force\;=\;F\;=\;?\\\\\boldsymbol F\boldsymbol o\boldsymbol r\boldsymbol m\boldsymbol u\boldsymbol l\boldsymbol a\boldsymbol{\mathit:}\boldsymbol{\mathit-}\\{\boldsymbol v}_{\boldsymbol f}\boldsymbol{\mathit\;}\boldsymbol{\mathit=}\boldsymbol{\mathit\;}{\boldsymbol v}_{\boldsymbol i}\boldsymbol{\mathit\;}\boldsymbol{\mathit+}\boldsymbol{\mathit\;}\boldsymbol a\boldsymbol{\mathit\;}\boldsymbol t\\\\v_f\;-\;v_i\;=\;a\;t\;\\\\\frac{v_f\;-\;v_i}t\;=\;a\\\\a\;=\frac{v_f\;-\;v_i}t\\\\a\;=\;\frac{30\;m\;s^{-1}\;-\;10\;m\;s^{-1}}{10\;s}\\a\;=\;\frac{20\;m\;s^{-1}}{10\;s}\\\\a\;=\;2\;m\;s^{-2}\;\\\\\boldsymbol F\boldsymbol o\boldsymbol r\boldsymbol m\boldsymbol u\boldsymbol l\boldsymbol a\boldsymbol{\mathit:}\boldsymbol{\mathit-}\\\boldsymbol F\boldsymbol{\mathit\;}\boldsymbol{\mathit=}\boldsymbol{\mathit\;}\boldsymbol m\boldsymbol{\mathit\;}\boldsymbol a\\\\F\;=\;(\;800\;kg\;)\;(\;2\;m\;s^{-2}\;)\\\\\boldsymbol F\boldsymbol=\boldsymbol\;\mathbf{1600}\boldsymbol\;\boldsymbol N\\$$
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$$\boldsymbol3\boldsymbol{\mathit.}\boldsymbol4\;\;\;A\;5\;g\;bullet\;is\;fired\;from\\a\;gun.\;The\;bullet\;moves\;with\\a\;velocity\;of\;300\;m\;s^{-1}.\;\;If\\the\;mass\;of\;gun\;is\;\;10\;kg,\\find\;recoil\;speed\;of\;the\;gun.\\$$
$$Law\mathit\;of\mathit\;conservation\mathit\;of\\momentum\mathit\;states\\that\mathit\;momentum\mathit\;before\mathit\;the\\collision\mathit\;is\;equal\mathit\;to\mathit\;the\\momentum\mathit\;after\mathit\;the\mathit\;collision\mathit.\\m_{\mathit1}u_{\mathit1}\mathit+m_{\mathit2}u_{\mathit2}\mathit=m_{\mathit1}v_{\mathit1}\mathit+m_{\mathit2}v_{\mathit2}\\\\\\Here,\;momentum\;before\;the\\collision\;is\;zero\\\\\;So\\m_{\mathit1}u_{\mathit1}\mathit\;\mathit+\mathit\;m_{\mathit2}u_{\mathit2}\;=\;0\\\\m_1u_1\mathit\;\mathit+\mathit\;m_2u_2\mathit\;\\\mathit=\mathit\;\mathit\;m_1v_1\mathit\;\mathit+\mathit\;m_2v_2\\After\;putting\;the\;values,\\we\;get\\\\0=(0.005)\;(300)+(10)\;v_2\;\\\\0\;=\;1.5\;+\;(\;10\;)\;v_2\\-1.5=\;(10\;)\;v_2\\\\\frac{-1.5\;}{10\;}=\;v_2\\\\v_2\;=-\;0.15\;m\;s^{-1}\\\\Negative\;\mathrm{si}gn\;appears\\due\;to\;the\;recoil\;speed\\of\;the\;gun.\\$$
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$$\boldsymbol3\boldsymbol{\mathit.}\boldsymbol7\;\;\;\;\;A\;cyclist\;weighing\\55\;kg\;rides\;a\;bicycle\;of\\mass\;5\;kg.\;He\;starts\;from\\rest\;and\;applies\;a\;force\;of\\90\;N\;for\;8\;seconds.\;Then\\he\;continues\;at\;a\;cons\tan t\\speed\;for\;another\;8\;seconds.\\Calculate\;the\;total\;dis\tan ce\\travelled\;by\;the\;cyclist.$$
$$Total\;mass\;=m\\=55kg\;+\;5kg=60\;kg\\\\Initial\;velocity=v_i=0\\\\Force\;=\;F\;=\;90\;N\\\\Initial\;time=t_1=8\;s\\\\Further\;time=t_2=8\;s\\\\Total\;dis\tan ce=S=?$$
$$\boldsymbol F\boldsymbol o\boldsymbol r\boldsymbol m\boldsymbol u\boldsymbol l\boldsymbol a\boldsymbol:\boldsymbol-\\\boldsymbol F\boldsymbol\;\boldsymbol=\boldsymbol\;\boldsymbol m\boldsymbol\;\boldsymbol a\;\\\\\frac Fm\;=\;a\\\\\frac{90\;N}{60\;kg}\;=\;a\\\\a\;=\;1.5\;m\;s^{-2}\\\\$$
$$Initial\;dis\tan ce\;=\;S_1\\\\\boldsymbol F\boldsymbol o\boldsymbol r\boldsymbol m\boldsymbol u\boldsymbol l\boldsymbol a\boldsymbol:\boldsymbol-\\\boldsymbol S\boldsymbol\;\boldsymbol=\boldsymbol\;{\boldsymbol v}_{\mathbf i}\boldsymbol\;\boldsymbol t\boldsymbol\;\boldsymbol+\boldsymbol\;\frac{\mathbf1}{\mathbf2}\boldsymbol\;\boldsymbol a\boldsymbol\;\boldsymbol t^{\mathbf2}\;\\\\S_1\;=\;0\;+\;(0.5)\;(\;1.5\;m\;s^{-2})\;{(8s)}^2\\\\S_1\;=\;48\;m\\\\\\$$
$$\boldsymbol F\boldsymbol o\boldsymbol r\boldsymbol m\boldsymbol u\boldsymbol l\boldsymbol a\boldsymbol:\boldsymbol-\\{\boldsymbol v}_{\mathbf f}\boldsymbol\;\boldsymbol=\boldsymbol\;{\boldsymbol v}_{\mathbf i}\boldsymbol\;\boldsymbol+\boldsymbol\;\boldsymbol a\boldsymbol\;\boldsymbol t\boldsymbol\;\\\\v_f\;=\;0\;+\;(\;1.5\;m\;s^{-2}\;)\;{(\;8\;s\;)}\\\\v_f\;=\;12\;m\;s^{-1}\\\\\\\\\\$$
$$\boldsymbol F\boldsymbol o\boldsymbol r\boldsymbol m\boldsymbol u\boldsymbol l\boldsymbol a\boldsymbol:\boldsymbol-\\\boldsymbol S\boldsymbol\;\boldsymbol=\boldsymbol\;{\boldsymbol v}_{\mathbf a\mathbf v}\boldsymbol\;\boldsymbol t\\\\Here\;t\;=\;t_{2\;}\;and\;\;v_{av}\;=\;v_f\;\;\\s_2\;=\;s\;(\;say\;)\\So\\\\S_2\;=\;(\;12\;m\;s^{-1}\;)\;(\;8\;s\;)\\\\S_2\;=\;96\;m\\\\\\\\\\\\$$
$$S\;=\;S_1\;+\;S_2\;\\\\S\;=\;48\;m\;+\;96\;m\\\\S\;=\;144\;m\\\\\\\\\\\\$$
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