Important Questions for CBSE Class 9 Science Chapter 9 - Force and Laws of Motion

Very Short Answer Questions                             (1 Mark)

1. A batsman hits a cricket ball which then rolls on level ground. After covering a short distance, the ball comes to rest. The ball slows to a stop because – 

1. The batsman did not hit the ball hard enough.

2. Velocity is proportional to the force exerted on the ball.

3. There is a force on the ball opposing the motion.

4. There is no unbalanced force on the ball, so the ball would want to come to rest.

Ans: (c) There is a force on the ball opposing the motion.

2. What is the momentum of an object of mass m, moving with a velocity v?

1. \[{{(mv)}^{2}}\]

2. \[m{{v}^{2}}\]

3. $\dfrac{1}{2}m{{v}^{2}}$

4. \[mv\]

Ans: (d) $mv$

3. Using a horizontal force of $200N$, we intend to move a wooden cabinet across a floor at a constant velocity. What is the friction force that will be exerted on the cabinet?

Ans: In order to move the cabinet across the floor at a constant velocity, the net force experienced by it must be zero. Thus a frictional force of $200N$ must be exerted on the cabinet to move it across the floor at a constant velocity, against the horizontal force of $200N$.

4. What is the S.I. unit of momentum?

1. $kgms$

2. $\dfrac{ms}{kg}$

3. $kgm{{s}^{-1}}$

4. $\dfrac{kg}{ms}$

Ans: (c) $kgm{{s}^{-1}}$

5. What is the numerical formula for force?

1. $\mathsf{F=ma}$

2. $\mathsf{F=}\dfrac{\mathsf{m}}{\mathsf{a}}$

3. $\mathsf{F=m}{{\mathsf{a}}^{\mathsf{2}}}$

4. $F={{m}^{2}}a$

Ans: (a) $\mathrm{F=ma}$

6. If the initial velocity is zero then the force acting is 

1. Retarding

2. Acceleration

3. Both

4. None

Ans: (a) Acceleration

7. What is the S.I. unit of force?

1. $\dfrac{kgm}{{{s}^{2}}}$

2. $\dfrac{kgm}{s}$

3. $\dfrac{kg{{m}^{2}}}{{{s}^{2}}}$

4. $kg{{m}^{2}}{{s}^{2}}$

Ans: (a) $\dfrac{\mathrm{kgm}}{{{\mathrm{s}}^{\mathrm{2}}}}$

8. Newton’s first law of motion is also called

1. Law of Inertia

2. Law of Momentum

3. Law of Action & Reaction

4. None of these

Ans: (a) Law of Inertia

9. Which law explains swimming?

1. Newton’s first law

2. Newton’s second law

3. Newton’s third law

4. All of these

Ans: (c) Newton’s third law

10. The S.I. unit of weight is:

1. $N$

2. $Nm$

3. $\dfrac{N}{s}$

4. $\dfrac{Nm}{s}$

Ans: (a) $\mathrm{N}$

11. Which equation defines Newton’s Second law of motion?

1. $\mathsf{F=ma=}\dfrac{\mathsf{dp}}{\mathsf{dt}}$

2. $\mathsf{F=m}\dfrac{\mathsf{da}}{\mathsf{dt}}\mathsf{=P}$

3. $\dfrac{\mathsf{dF}}{\mathsf{dt}}\mathsf{=ma=P}$

4. $\mathsf{F=ma=P}$

Ans: (d) $\mathrm{F=ma=P}$

12. The people in the bus are pushed backwards when the bus starts suddenly due to

1. Inertia due to Rest

2. Inertia due to Motion

3. Inertia due to Direction

4. Inertia

Ans: (a) Inertia due to Rest

13. If the force acting on the body is zero. Its momentum is

1. zero

2. constant

3. Both

4. None

Ans: (b) constant

14. The inability of the body to change its state of rest or motion is

1. Momentum

2. Force

3. Inertia

4. Acceleration.

Ans: (c) Inertia

Short Answer Questions                                          (2 Marks)

1. An object experiences a net zero external unbalanced force. Is it possible for the object to be traveling with a non-zero velocity? If yes, state the conditions that must be placed on the magnitude and direction of the velocity. If no, provide a reason.

Ans: 

• Yes, it is possible for an object to be traveling with a non-zero velocity if it experiences a net zero external unbalanced force. This is based on Newton’s First law of motion, which states that an object will change its state of motion if and only if the net force acting on it is non-zero.  

• Thus, in order to change its motion or to bring about motion, some external unbalanced force is required. 

• In this case, the object experiences a net zero unbalanced force, which will cause it to move with some non-zero velocity provided that the object was previously moving with a constant velocity.

2. When a carpet is beaten with a stick, dust comes out of it. Explain.

Ans: When a carpet is beaten with a stick, dust comes out of it because of Newton’s First Law of Motion, the law of Inertia. Initially, the dust particles and the carpet are in a state of rest. When the carpet is beaten with a stick, it causes the carpet to move, while the dust particles, due to inertia of rest, will resist the change in motion. Thus, the carpet’s forward motion will exert a backward force on the dust particles, which makes them move in the opposite direction. Therefore the dust comes out of the carpet.

3. Why is it advised to tie any luggage kept on the roof of a bus with a rope?

Ans: 

• It is advised to tie any luggage kept on the roof of a bus with a rope because of Newton’s First Law of Motion, the law of Inertia. 

• When the bus moves, the luggage will be in inertia of motion and say the bus suddenly stops, then the luggage tends to resist this change in motion, causing it to move forward and fall off, if not tied up by a rope. 

• Similarly, when the bus decelerates or changes its direction while turning, the inertia of motion of the luggage will try to resist this change in motion, causing the luggage to move oppositely and fall off, if not tied up by a rope.

4. A stone of $1kg$ is thrown with a velocity of $20m{{s}^{-1}}$ across the frozen surface of a lake and comes to rest after traveling a distance of $50m$. What is the force of friction between the stone and the ice?

Ans: Given:

Mass of stone: $\mathrm{m=1kg}$

Initial velocity of stone: $\mathrm{u=20m}{{\mathrm{s}}^{\mathrm{-1}}}$

Final velocity of stone:  $\mathrm{v=0m}{{\mathrm{s}}^{\mathrm{-1}}}$ (as it comes to rest)

Distance traveled on ice: $\mathrm{s=50m}$

To find: Force of friction between stone and ice.

First, we need to find the deceleration:

It is known that – ${{\mathrm{v}}^{\mathrm{2}}}\mathrm{=}{{\mathrm{u}}^{\mathrm{2}}}\mathrm{+2as}$

Thus, ${{\mathrm{0}}^{\mathrm{2}}}\mathrm{=(20}{{\mathrm{)}}^{\mathrm{2}}}\mathrm{+2a(50)}$

$\Rightarrow \mathrm{0=400+100a}$

$\Rightarrow -400=100a$

$\Rightarrow a=-4m{{s}^{-2}}$

The negative sign implies deceleration.

Next, finding the frictional force:

$\mathrm{F=ma}$

$\Rightarrow \mathrm{F=(1)(}-4)$

$\Rightarrow \mathrm{F=}-4N$

Thus, the force of friction between stone and ice is $-4N$.

5. An automobile vehicle has a mass of $1500kg$. What must be the force between the vehicle and road if the vehicle is to be stopped with a negative acceleration of $1.7m{{s}^{-2}}$?

Ans: Given:

Mass of vehicle:$\mathrm{m=1500kg}$

Negative acceleration:$\mathrm{a=}-\mathrm{1}\mathrm{.7m}{{\mathrm{s}}^{\mathrm{-2}}}$

To find: Force of friction between road and vehicle.

It is known that - $\mathrm{F=ma}$

$\Rightarrow \mathrm{F=(1500)(}-1.7)$

$\Rightarrow \mathrm{F=}-2550N$

Thus the force between the vehicle and road if the vehicle is to be stopped with a negative acceleration of $1.7m{{s}^{-2}}$is $-2550N$.

6. An object of mass $100kg$ is accelerated uniformly from a velocity of $5m{{s}^{-1}}$ to $8m{{s}^{-1}}$ in $6s$. Calculate the initial and final momentum of the object. Also, find the magnitude of the force exerted on the object.

Ans: Given:

Mass of object:$\mathrm{m=100kg}$

Initial velocity of object:$\mathrm{u=5m}{{\mathrm{s}}^{\mathrm{-1}}}$

Final velocity of object: $\mathrm{v=8m}{{\mathrm{s}}^{\mathrm{-1}}}$

Time duration of acceleration:$t=6s$

To find: 

• Initial momentum

• Final momentum

• Force exerted on the object

It is known that – $momentum=\operatorname{ma}ss\times velocity$

$Initial\_momentum=\operatorname{ma}ss\times initial\_velocity$

$\Rightarrow Initial\_momentum=100\times 5$

$\Rightarrow Initial\_momentum=500kgm{{s}^{-1}}$

$Final\_momentum=\operatorname{ma}ss\times final\_velocity$

$\Rightarrow Final\_momentum=100\times 8$

$\Rightarrow Final\_momentum=800kgm{{s}^{-1}}$

Now, the force – $\mathrm{F=ma}$

$\mathrm{F=m(}\dfrac{v-u}{t})$

$\Rightarrow \mathrm{F=100(}\dfrac{8-5}{6})$

$\Rightarrow \mathrm{F=100(}\dfrac{3}{6})$

$\Rightarrow \mathrm{F=50N}$

Thus,

• Initial momentum:$500kgm{{s}^{-1}}$

• Final momentum: $800kgm{{s}^{-1}}$

• Force exerted on object: $50N$

7. Akhtar, Kiran, and Rahul were riding in a motorcar that was moving with a high velocity on an expressway when an insect hit the windshield and got stuck on the windscreen. Akhtar and Kiran started pondering over the situation. Kiran suggested that the insect suffered a greater change in momentum as compared to the change in momentum of the motorcar (because the change in the velocity of the insect was much more than that of the motorcar). Akhtar said that since the motorcar was moving with a larger velocity, it exerted a larger force on the insect. And as a result, the insect died. Rahul while putting an entirely new explanation said that both the motorcar and the insect experienced the same force and a change in their momentum. Comment on these suggestions.

Ans: 

• Kiran’s statement – the insect suffered a greater change in momentum as compared to the change in momentum of the motorcar (because the change in the velocity of the insect was much more than that of the motorcar). 

Based on the law of conservation of momentum, the change of momentum experienced by both the insect and car should be equal. The change in velocity of the insect will be greater, due to its small mass, while the change in velocity of the car is insignificant, due to its larger mass. But the change in momentum before and after collision would be the same. Thus, Kiran’s statement is false.

• Akhtar’s statement – since the motorcar was moving with a larger velocity, it exerted a larger force on the insect. And as a result, the insect died. 

According to Newton’s third law of motion, for every action, there is an equal and opposite reaction. Thus both the car and insect would experience the same force. So, we can say that Akhtar’s statement is also false.

• Rahul’s statement – both the motorcar and the insect experienced the same force and a change in their momentum.

Inferring from the law of conservation of momentum and Newton’s third law of motion, we can say that Rahul’s statement is true.

8. State Newton’s second law of motion?

Ans: Newton’s Second law of motion states that when an unbalanced external force acts on an object, its velocity changes, that is, the object gets accelerated. It can also be stated as the time rate of change of the momentum of a body is equal in both magnitude and direction to the force applied on it.

Mathematically – $\mathrm{F=ma}$, where ‘F’ is the force, ‘m’ is the mass of the object, and ‘a’ is the acceleration.

9. What is the momentum of a body of mass $200g$ moving with a velocity of $15m{{s}^{-1}}$?

Ans: Given: 

Mass of body:$\mathrm{m=200g=0}\mathrm{.2kg}$

Velocity of body: $\mathrm{v=15m}{{\mathrm{s}}^{\mathrm{-1}}}$

To find: Momentum of the body.

It is known that – $momentum=\operatorname{ma}ss\times velocity$

$\Rightarrow momentum=0.2\times 15$

$\Rightarrow momentum=3kgm{{s}^{-1}}$

Thus, the momentum of the body is $3kgm{{s}^{-1}}$.

10. Define force and what are the various types of forces?

Ans: Force is defined as the push or pulls on an object that produces a change in the state or shape of the object. It can also cause a change in the speed and/or direction of motion of the object.

The various types of force are:

1. Mechanical force

2. Gravitational force

3. Frictional force

4. Electrostatic force

5. Electromagnetic force

6. Nuclear force

11. A force of $25N$ acts on a mass of $500g$ resting on a frictionless surface. What is the acceleration produced?

Ans: Given: 

Mass:$\mathrm{m=500g=0}\mathrm{.5kg}$

Force exerted: $\mathrm{F=25N}$

To find: Acceleration.

It is known that – $\mathrm{F=ma}$

$\Rightarrow a=\dfrac{F}{m}$

$\Rightarrow a=\dfrac{25}{0.5}$

$\Rightarrow a=50m{{s}^{-2}}$

Thus, the acceleration produced is \[50m{{s}^{-2}}\].

12. State Newton’s first law of Motion?

Ans: Newton’s first law of motion is also called the law of Inertia. It states that an object remains in a state of rest or of uniform motion in a straight line unless compelled to change that state by an applied force. Or, an object rest will continue to be at rest and an object in motion will continue to be in motion until and unless it is acted upon by an external force.

13. A body of mass $5kg$ starts and rolls down $32m$of an inclined plane in $4s$. Find the force acting on the body?

Ans: Given:

Mass of body:$\mathrm{m=5kg}$

Initial velocity of body: $\mathrm{u=0m}{{\mathrm{s}}^{\mathrm{-1}}}$ (as it is starting to roll)

Distance travelled on inclined plane: $\mathrm{s=32m}$

Time duration of rolling:$t=4s$

To find: Force acting on the body.

First we need to find the acceleration:

It is known that – $\mathrm{s=ut+}\dfrac{1}{2}\mathrm{a}{{\mathrm{t}}^{\mathrm{2}}}$

Thus, $\mathrm{32=(0}\times \mathrm{4)+}\dfrac{1}{2}(a\times {{4}^{2}})$

$\Rightarrow \mathrm{32=}\dfrac{1}{2}(a\times 16)$

$\Rightarrow \mathrm{32=}(a\times 8)$

$\Rightarrow a=4m{{s}^{-2}}$

Next, finding the force:

$\mathrm{F=ma}$

$\Rightarrow \mathrm{F=(5}\times 4)$

$\Rightarrow \mathrm{F=20}N$

Thus, the force acting on the body is $20N$.

14. On a certain planet, a small stone tossed up at $15m{{s}^{-1}}$ vertically upwards takes $7.5s$ to return to the ground. What is the acceleration due to gravity on the planet?

Ans: Given:

Initial velocity of stone:$\mathrm{u=15m}{{\mathrm{s}}^{\mathrm{-1}}}$

Final velocity of stone: $\mathrm{v=0m}{{\mathrm{s}}^{\mathrm{-1}}}$ (as it becomes zero at the highest point)

Total time duration of flight (tossed up and falling down to the ground):$t=7.5s$

To find: Acceleration due to gravity of the planet.

It is known that – $\mathrm{v=u+at}$

Thus, $\mathrm{0=15+(at)}$

$\mathrm{t=}\dfrac{-15}{a}$  this denotes the time for one-half of the entire flight.

Thus the total duration of the flight is twice this duration.

i.e. $7.5s=2t$

$\Rightarrow 7.5=2(\dfrac{-15}{a})$

Now, the acceleration due to gravity is –

$a=\dfrac{2\times (-15)}{7.5}$

\[\Rightarrow a=-4m{{s}^{-2}}\]

Thus, the acceleration due to gravity of the planet is $-4m{{s}^{-2}}$.

15. Why is the weight of the object more at the poles than at the equator?

Ans: 

• The weight of the object is more at the poles than at the equator because the acceleration due to gravity is slightly greater at the poles than at the equator. This is because - $g=\dfrac{GM}{{{r}^{2}}}$, meaning acceleration due to gravity is inversely proportional to the square of the radius. Since the radius of the earth at the equator is greater than at the poles, the acceleration due to gravity is slightly less at the equator, than at the poles.

• Also, we know that weight is directly proportional to acceleration due to gravity $(\because w=m\times g\Rightarrow w\alpha g)$.

• Using these two implications, we can say that at the equator, where the radius is larger, the acceleration due to gravity is smaller, the weight is lower. And at the poles, where the radius is smaller, the acceleration due to gravity is greater, the weight is higher.

16. Why does the passenger sitting in a moving bus are pushed in the forward direction when the bus stops suddenly?

Ans: The passengers sitting in the moving bus are pushed in the forward direction when the bus stops suddenly due to inertia because the passengers' upper body continues to be in a state of motion, while the lower part of the body that is in contact with the seat remains at rest. As a result, the passenger’s upper body is pushed in the forward direction, in the direction in which the bus was moving before coming to a halt.

17. Why does the boat move backward when the sailor jumps in the forward direction?

Ans: The boat moves backward when the sailor jumps in the forward direction because of Newton’s third law of motion which states that for every action, there is an equal and opposite reaction. Thus, when the sailor jumps in the forward direction he is causing an action force due to which the boat moves backward. In response, the boat exerts an equal and opposite force (the reaction force) on the sailor due to which he is pushed in the forward direction.

18. Derive the law of conservation of momentum from Newton’s third law?

Ans: Newton’s third law of motion states that for every action, there is an equal and opposite reaction, that acts on different bodies.

Say we have two objects A and B of masses ${{m}_{A}}$ and ${{m}_{B}}$ are traveling in the same direction along a straight line at different velocities ${{u}_{A}}$ and ${{u}_{B}}$, respectively. 

Consider that there are no other external unbalanced forces acting on them. 

Let ${{u}_{A}}>{{u}_{B}}$ and the two objects collide with each other.

During collision which lasts for a time $t$, A exerts a force ${{F}_{AB}}$ on B and B exerts a force ${{F}_{BA}}$ on  A. 

Say, ${{v}_{A}}$ and ${{v}_{B}}$ are the velocities of the two A and B after the collision, respectively.

It is known that – $momentum=\operatorname{ma}ss\times velocity$

Momentum of A before collision: \[{{m}_{A}}\times {{u}_{A}}\]

Momentum of A after collision: \[{{m}_{A}}\times {{v}_{A}}\]

Momentum of B before collision: \[{{m}_{B}}\times {{u}_{B}}\]

Momentum of B after collision: \[{{m}_{B}}\times {{v}_{B}}\]

It is also known that force can also be defined as the rate of change of momentum, i.e. $F=ma=m(\dfrac{v-u}{t})=\dfrac{mv-mu}{t}$

Now, the rate of change of momentum of A during collision is $\dfrac{{{m}_{A}}{{v}_{A}}-{{m}_{A}}{{u}_{A}}}{t}$, which is the force ${{F}_{AB}}$.

And the rate of change of momentum of B during collision is \[\dfrac{{{m}_{B}}{{v}_{B}}-{{m}_{B}}{{u}_{B}}}{t}\], which is the force ${{F}_{BA}}$.

Based on Newton’s third law of motion, the force ${{F}_{AB}}$  exerted by Aon B and force ${{F}_{BA}}$ exerted by B on A are equal in magnitude but opposite in direction.

i.e.  ${{F}_{AB}}=-{{F}_{BA}}$

Using the formulae – 

${{F}_{AB}}=-{{F}_{BA}}$

\[\Rightarrow \dfrac{{{m}_{A}}{{v}_{A}}-{{m}_{A}}{{u}_{A}}}{t}=-(\dfrac{{{m}_{B}}{{v}_{B}}-{{m}_{B}}{{u}_{B}}}{t})\]

Simplifying,

\[\Rightarrow {{m}_{A}}{{v}_{A}}-{{m}_{A}}{{u}_{A}}=-({{m}_{B}}{{v}_{B}}-{{m}_{B}}{{u}_{B}})\]

\[\Rightarrow {{m}_{A}}{{v}_{A}}-{{m}_{A}}{{u}_{A}}=-{{m}_{B}}{{v}_{B}}+{{m}_{B}}{{u}_{B}}\]

Rearranging,

\[{{m}_{A}}{{v}_{A}}+{{m}_{B}}{{v}_{B}}={{m}_{A}}{{u}_{A}}+{{m}_{B}}{{u}_{B}}\]

Here, \[{{m}_{A}}{{u}_{A}}+{{m}_{B}}{{u}_{B}}\]is the total sum of momentum of A and B before collision and \[{{m}_{A}}{{v}_{A}}+{{m}_{B}}{{v}_{B}}\] is the total sum of momentum of A and B after collision.

This equation implies that the final momentum of the two objects after the collision is equal to the initial momentum of the two objects before the collision.

Thus the law of conservation of momentum, for an isolated system, is that the total initial momentum for an event is equal to total initial momentum.

19. An astronaut has $80kg$ mass on earth.

i) What is his weight on earth?

Ans: Given:

Mass of astronaut: $\mathrm{m=80kg}$

To find his weight on earth.

It is known that,

Acceleration due to gravity on earth: ${{g}_{e}}=10m{{s}^{-2}}$

Acceleration due to gravity on mars:${{g}_{m}}=3.7m{{s}^{-2}}$

Weight: $w=m\times g$

Weight on earth: ${{w}_{e}}=m\times {{g}_{e}}$

$\Rightarrow {{w}_{e}}=80\times 10$

${{w}_{e}}=800N$

ii) What will be his mass and weight on mars with ${{g}_{m}}=3.7m{{s}^{-2}}$?

Ans: Given:

Mass of astronaut:$\mathrm{m=80kg}$

To find his mass and weight on mars.

It is known that,

Acceleration due to gravity on earth: ${{g}_{e}}=10m{{s}^{-2}}$

Acceleration due to gravity on mars:${{g}_{m}}=3.7m{{s}^{-2}}$

Weight: $w=m\times g$

Weight on mars: ${{w}_{m}}=m\times {{g}_{m}}$

$\Rightarrow {{w}_{m}}=80\times 3.7$

${{w}_{m}}=296N$

The mass of astronauts remains the same on mars because it is a constant value. 

Thus, mass on mars is $\mathrm{m=80kg}$.

Short Answer Questions                                     (3 Marks)

1. Which of the following has more inertia:

a. A rubber ball and a stone of the same size?

Ans: Inertia depends on the mass of the object. The larger the mass, the greater will be its inertia and vice-versa. In the case of a rubber ball and a stone of the same size, it is clear that the stone will have greater inertia than the ball. It is because, despite being the same size, the stone weighs more than the rubber ball.

b. A bicycle and a train?

Ans: Inertia depends on the mass of the object. The larger the mass, the greater will be its inertia and vice-versa. In the case of a bicycle and a train, it is clear that the train will have greater inertia than the bicycle because the train weighs more than the bicycle.

c. A five rupees coin and a one-rupee coin?

Ans: Inertia depends on the mass of the object. The larger the mass, the greater will be its inertia and vice-versa. In the case of a five rupees coin and a one-rupee coin, the five rupees coin will have greater inertia than the one-rupee coin because five rupees coin weighs more than a one-rupee coin.

2. In the following example, try to identify the number of times the velocity of the ball changes:

“A football player kicks a football to another player of his team who kicks the football towards the goal. The goalkeeper of the opposite team collects the football and kicks it towards a player of his own team”.

Also, identify the agent supplying the force in each case.

Ans: In the example, the number of times the velocity of football changes is four.

(i) The velocity of the football changes first when a player kicks the ball towards another player on his team. Here, the agent supplying the force is the foot of the football player who is kicking the ball.

(ii) The velocity of the football changes second when that another player kicks the ball towards the goal. Here, the agent supplying the force is the foot of the other player who is now kicking the ball towards the goal.

(iii) The velocity of the football changes for the third time when the goalkeeper of the opposite team stops the football by collecting it. Here, the agent supplying the force are the hands of the goalkeeper who collects the ball.

(iv) The velocity of the football changes for the fourth time when the goalkeeper kicks it towards a player of his team. Here, the agent supplying the force is the foot of the goalkeeper who is now kicking the ball towards his teammate.

3. Explain why some of the leaves may get detached from a tree if we vigorously shake its branch.

Ans: Some of the leaves may get detached from a tree if we vigorously shake its branch because the branches of the tree will come into motion while the leaves tend to continue in their state of rest. This is due to the inertia of rest of the leaves. The force of shaking will act on the leaves with the change in direction rapidly, which results in the leaves detaching and falling off from the tree.

4. Why do you fall in the forward direction when a moving bus breaks to a stop and fall back when it accelerates from rest?

Ans: The passengers sitting in the moving bus are pushed in the forward direction when the bus stops suddenly due to inertia because the passengers' upper body continues to be in a state of motion, while the lower part of the body that is in contact with the seat remains at rest. As a result, the passenger’s upper body is pushed in the forward direction, in the direction in which the bus was moving before coming to a halt.

Similarly, the passengers sitting in the bus are pushed in the backward direction when the bus accelerates from rest due to inertia, because the passengers’ upper body continues to be in a state of rest, while the lower part of the body that is in contact with the seat is set in motion. As a result, the passenger’s upper body is pushed in the backward direction, in the opposite to which the bus starts to move.

5. If action is always equal to the reaction, explain how a horse can pull a cart.

Ans: 

• In this case, we are dealing with unbalanced forces. It is true that the horse exerts an action force on the cart and experiences a reaction force from the cart. But also, the horse creates an action force on the ground over which it is walking, and experiences a reaction force from the ground.

• In pulling the cart, the action force of the horse pulling the cart is greater than the reaction force of the cart, resisting the pull. Thus the cart moves in the direction of the pull of the horse.

• In stepping on the ground, the horse creates an action force on the ground in the backward direction during each step, while the ground creates a reaction force pushing the horse forward.

• In this was a horse can pull a cart.

6. Explain why is it difficult for a fireman to hold a hose, which ejects large amounts of water at a high velocity.

Ans: It is difficult for a fireman to hold a hose, which ejects large amounts of water at a high velocity because of Newton’s third law of motion. In this case, the water being ejected out in the forward direction with great force (action) will create a backward force that results in the backward movement (reaction) of the hose pipe. As a result of this backward force and movement, it will be difficult for the fireman to hold the hose properly with stability.

7. From a rifle of mass $4kg$, a bullet of mass $50g$ is fired with an initial velocity of $35m{{s}^{-1}}$. Calculate the initial recoil velocity of the rifle.

Ans: Given:

Mass of rifle: ${{m}_{1}}=4kg$

Mass of bullet: ${{m}_{2}}=50g=0.05kg$ 

Initial velocity of rifle: ${{u}_{1}}=0m{{s}^{-1}}$ (it is stationary during firing)

Initial velocity of bullet:${{u}_{2}}=0m{{s}^{-1}}$(it starts from rest, inside the barrel of the rifle)

Fired velocity of bullet:${{v}_{2}}=35m{{s}^{-1}}$

To find: Recoil velocity of rifle:${{v}_{1}}$

By using the law of conservation of momentum, for an isolated system, is that the total initial momentum for an event is equal to total initial momentum, we get – 

\[{{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}}={{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}\]

Here, \[{{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}\]is the total sum of momentum of rifle and bullet before firing and \[{{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}}\] is the total sum of momentum of rifle and bullet after firing.

Substituting the values in – \[{{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}}={{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}\]

\[\Rightarrow (4\times {{v}_{1}})+(0.05\times 35)=(4\times 0)+(0.05\times 0)\]

\[\Rightarrow (4\times {{v}_{1}})+(17.5)=0\]

\[\Rightarrow (4\times {{v}_{1}})=-17.5\]

\[\Rightarrow {{v}_{1}}=-4.375m{{s}^{-1}}\](The negative sign indicates the backward direction in which the rifle moves when it recoils)

Thus, the recoil velocity of the rifle is \[4.375m{{s}^{-1}}\].

8. An $8000kg$ engine pulls a train of $5$ wagons, each of $2000kg$, along a horizontal track. If the engine exerts a force of $40000N$ and the track offers a friction force of $5000N$, then calculate:

(a) The net accelerating force

Ans: Given:

Force exerted by engine on wagons: $F=40000N$

Frictional exerted on wagons:$f=5000N$

Mass of engine:${{m}_{e}}=8000kg$

Mass of each wagon:${{m}_{w}}=2000kg$

Mass of all five wagons:${{m}_{W}}=5\times {{m}_{w}}=5\times 2000=10000kg$

Mass of entire train: ${{m}_{T}}={{m}_{e}}+{{m}_{W}}=8000+10000=18000kg$

To find accelerating force.

Net accelerating force can be found by subtracting the frictional force from the force exerted by the engine on the wagons.

Thus, $NetAcceleratingForce=ForceOfEngine-FrictionalForce$

$\Rightarrow NetAcceleratingForce=F-f$

$\Rightarrow NetAcceleratingForce=40000-5000$

$\Rightarrow NetAcceleratingForce=35000N$

(b) The acceleration of the train

Ans:  Given:

Force exerted by engine on wagons: $F=40000N$

Frictional exerted on wagons:$f=5000N$

Mass of engine:${{m}_{e}}=8000kg$

Mass of each wagon:${{m}_{w}}=2000kg$

Mass of all five wagons:${{m}_{W}}=5\times {{m}_{w}}=5\times 2000=10000kg$

Mass of entire train: ${{m}_{T}}={{m}_{e}}+{{m}_{W}}=8000+10000=18000kg$

To find the acceleration of the train.

It is known that – $\mathrm{F=ma}$

$\Rightarrow a=\dfrac{F}{m}$

$\Rightarrow a=\dfrac{NetAcceleratingForce}{MassOfTrain}$

$\Rightarrow a=\dfrac{35000}{18000}$

$\Rightarrow a=1.944m{{s}^{-2}}$

(c) The force of wagon $1$ on wagon $2$.

Ans: Given:

Force exerted by engine on wagons: $F=40000N$

Frictional exerted on wagons:$f=5000N$

Mass of engine:${{m}_{e}}=8000kg$

Mass of each wagon:${{m}_{w}}=2000kg$

Mass of all five wagons:${{m}_{W}}=5\times {{m}_{w}}=5\times 2000=10000kg$

Mass of entire train: ${{m}_{T}}={{m}_{e}}+{{m}_{W}}=8000+10000=18000kg$

To find the force exerted by wagon $1$ on wagon $2$

Here, wagon $1$ exerts a pulling force on the remaining $4$ wagons

$\mathrm{F=ma}$

$\Rightarrow {{\mathrm{F}}_{21}}\mathrm{=(4}{{\mathrm{m}}_{w}}\mathrm{)}\times \mathrm{a}$

$\Rightarrow {{\mathrm{F}}_{21}}\mathrm{=(4}\times 2000\mathrm{)}\times 1.944$

$\Rightarrow {{\mathrm{F}}_{21}}\mathrm{=15552}N$

9. Two objects, each of mass $1.5kg$, are moving in the same straight line but in opposite directions. The velocity of each object is $2.5m{{s}^{-1}}$ before the collision during which they stick together. What will be the velocity of the combined object after collision?

Ans: Given:

Mass of object 1: ${{m}_{1}}=1.5kg$

Mass of object 2: ${{m}_{2}}=1.5kg$ 

Initial velocity of object 1: ${{u}_{1}}=2.5m{{s}^{-1}}$

Initial velocity of object 2:${{u}_{2}}=-2.5m{{s}^{-1}}$(negative sign because it is moving in the opposite direction)

Mass of combined object after collision: $m={{m}_{1}}+{{m}_{2}}=1.5+1.5=3kg$

To find: Final velocity of the combined object after collision:$v$

By using the law of conservation of momentum, for an isolated system, is that the total initial momentum for an event is equal to total initial momentum, we get – 

\[mv={{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}\]

Here, \[{{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}\]is the total sum of the momentum of objects before the collision and \[mv\] is the total momentum of the combined objects after the collision.

Substituting the values in – \[mv={{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}\]

\[\Rightarrow (3\times v)=(1.5\times 2.5)+(1.5\times -2.5)\]

\[\Rightarrow (3\times v)=(3.75)+(-3.75s)\]

\[\Rightarrow (3\times v)=0\]

$\Rightarrow v=0m{{s}^{-1}}$

Thus, the velocity of the combined object after collision is \[0m{{s}^{-1}}\].

10. According to the third law of motion when we push on an object, the object pushes back on us with an equal and opposite force. If the object is a massive truck parked along the roadside, it will probably not move. A student justifies this by answering that the two opposite and equal forces cancel each other. Comment on this logic and explain why the truck does not move.

Ans: 

• According to the third law of motion when we push on an object, the object pushes back on us with an equal and opposite force. This pair of forces is called the action-reaction pair. 

• In the case of the massive truck parked alongside the road, the action-reaction pair is the weight of the truck exerting a force on the road in the downward direction (action), and the static friction of the road in the upward direction (reaction), which keeps the truck at rest. These two equal and opposite forces cancel out each other, which is why the truck will not move.

• For it to move, we need to apply additional external force to overcome the static friction of the road. 

• Thus, as the student explained, the truck does not move because the two opposite and equal forces of the truck and road cancel out each other is valid.

11. A hockey ball of mass $200g$ traveling at $10m{{s}^{-1}}$ is struck by a hockey stick so as to return it along its original path with a velocity at $5m{{s}^{-1}}$. Calculate the change of momentum that occurred in the motion of the hockey ball by the force applied by the hockey stick.

Ans: Given:

Mass of hockey ball: $m=200g=0.2kg$

Initial velocity of hockey ball: $u=10m{{s}^{-1}}$

Final velocity of hockey ball:$v=-5m{{s}^{-1}}$ (because it moves back in its original direction)

To find: Change in momentum of hockey ball due to the force of hockey stick

\[ChangeOfMomentum=mv-mu\]

Here, \[mu\]is the initial momentum of the hockey ball and \[mv\] is the final momentum of the hockey ball.

Substituting the values in –\[ChangeOfMomentum=mv-mu\]

\[\Rightarrow ChangeOfMomentum=(0.2\times -5)-(0.2\times 10)\]

\[\Rightarrow ChangeOfMomentum=(-1)-(2)\]

\[\Rightarrow ChangeOfMomentum=-3kgm{{s}^{-1}}\]

Thus, the change in momentum of hockey ball due to the force of hockey stick is \[-3kgm{{s}^{-1}}\].

12. A bullet of mass $10g$ traveling horizontally with a velocity of $150m{{s}^{-1}}$ strikes a stationary wooden block and comes to rest in $0.03s$. Calculate the distance of penetration of the bullet into the block. Also, calculate the magnitude of the force exerted by the wooden block on the bullet.

Ans: Given:

Mass of bullet: $m=10g=0.01kg$

Initial velocity of bullet: $u=150m{{s}^{-1}}$

Final velocity of bullet: $\mathrm{v=0m}{{\mathrm{s}}^{\mathrm{-1}}}$ (as it comes to rest after penetration)

Time duration of bullet travel:$t=0.03s$

To find:

• Distance of penetration of bullet into the block

• Force exerted by the block on the bullet

1. Distance of penetration:

It is known that – $\mathrm{v=u+at}$

Thus, $\mathrm{0=150+(a}\times 0.03\mathrm{)}$

$\Rightarrow \mathrm{(a}\times 0.03\mathrm{)=}-\mathrm{150}$

$\Rightarrow \mathrm{a=}-5000m{{s}^{-2}}$

Now,

$\mathrm{s=ut+}\dfrac{1}{2}\mathrm{a}{{\mathrm{t}}^{\mathrm{2}}}$

$\Rightarrow \mathrm{s=(150}\times 0.03\mathrm{)+}\dfrac{1}{2}(-5000\times {{0.03}^{2}})$

$\Rightarrow \mathrm{s=(}4.5\mathrm{)+}(-2.25)$

$\Rightarrow \mathrm{s=}2.25m$

2. Next, finding the force:

$\mathrm{F=ma}$

$\Rightarrow \mathrm{F=(0}\mathrm{.01}\times -5000)$

$\Rightarrow \mathrm{F=}-5\mathrm{0}N$

Thus,

• Distance of penetration of bullet into the block is $2.25m$

• Force exerted by the block on the bullet is $-50N$

13. An object of mass $1kg$ traveling in a straight line with a velocity of $10m{{s}^{-1}}$ collides with and sticks to, a stationary wooden block of mass $5kg$. Then they both move off together in the same straight line. Calculate the total momentum just before the impact and just after the impact. Also, calculate the velocity of the combined object.

Ans: Given:

Mass of object 1: ${{m}_{1}}=1kg$

Mass of object 2: ${{m}_{2}}=5kg$ 

Initial velocity of object 1: ${{u}_{1}}=10m{{s}^{-1}}$

Initial velocity of object 2:${{u}_{2}}=0m{{s}^{-1}}$(because it is stationary)

Mass of combined object after collision: $m={{m}_{1}}+{{m}_{2}}=1+5=6kg$

To find: 

• Momentum before impact

• Momentum after impact

• The final velocity of the combined object after collision:$v$

1. Momentum before impact is the Initial momentum:

\[InitialMomentum={{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}\]

\[\Rightarrow InitialMomentum=(1\times 10)+(5\times 0)\]

\[\Rightarrow InitialMomentum=10kgm{{s}^{-1}}\]

2. Momentum after impact is the Final momentum:

\[FinalMomentum=mv\]

By using the law of conservation of momentum, for an isolated system, is that the total initial momentum for an event is equal to total initial momentum, we get – 

\[mv={{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}\]

Here, \[{{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}\]is the total sum of the momentum of objects before the collision and \[mv\] is the total momentum of the combined objects after the collision.

Thus we get – \[FinalMomentum=mv={{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}=10kgm{{s}^{-1}}\]

\[\Rightarrow FinalMomentum=10kgm{{s}^{-1}}\]

3. Finding the final velocity of the combined object:

\[FinalMomentum=10kgm{{s}^{-1}}\]

\[\Rightarrow mv=10\]

\[\Rightarrow (6\times v)=10\]

\[\Rightarrow v=1.67m{{s}^{-1}}\]

Thus,

• Momentum before impact is \[10kgm{{s}^{-1}}\]

• Momentum after impact is \[10kgm{{s}^{-1}}\]

• Final velocity of the combined object after collision is \[1.67m{{s}^{-1}}\]

14. How much momentum will a dumb-bell of mass $10kg$ transfer to the floor if it falls from a height of $80cm$? Take its downward acceleration to be $10m{{s}^{-1}}$.

Ans: Given:

Mass of dumbbell: $m=10kg$

Initial velocity of dumbbell: $u=0m{{s}^{-1}}$(as it starts from rest)

Height of fall of dumbbell: $\mathrm{h=80cm=0}\mathrm{.8m}$ 

Acceleration due to gravity: $g=10m{{s}^{-2}}$

To find: Momentum transferred to the ground by dumbbell.

It is known that – ${{v}^{2}}={{u}^{2}}+2gh$

$\Rightarrow {{v}^{2}}={{(0)}^{2}}+(2\times 10\times 0.8)$

$\Rightarrow {{v}^{2}}=16$

$\Rightarrow v=4m{{s}^{-1}}$

Now, \[Momentum=mv\]

\[\Rightarrow Momentum=(10\times 4)\]

\[\Rightarrow Momentum=40kgm{{s}^{-1}}\]

Thus, the momentum transferred to the ground by dumbbell is \[40kgm{{s}^{-1}}\]

15. A force of $15N$acts for $5s$on a body of mass $5kg$ which is initially at rest. Calculate.

(a) Final velocity of the body

Ans: Given:

Mass of body: $m=5kg$

Initial velocity of body: $u=0m{{s}^{-1}}$(as it starts from rest)

Force acting on the body: $F=15N$

Time: $t=5s$

To find the final velocity of the body.

First we need to find the acceleration produced.

$\mathrm{F=ma}$

$\Rightarrow \mathrm{a=}\dfrac{F}{m}$

$\Rightarrow \mathrm{a=}\dfrac{15}{5}$

$\Rightarrow a=3m{{s}^{-2}}$

It is known that – $\mathrm{v=u+at}$

Thus, $\mathrm{v=0+(3}\times 5\mathrm{)}$

$\Rightarrow v\mathrm{=15}m{{s}^{-1}}$

(b) The displacement of the body

Ans: Given:

Mass of body: $m=5kg$

Initial velocity of body: $u=0m{{s}^{-1}}$(as it starts from rest)

Force acting on the body: $F=15N$

Time: $t=5s$

To find the displacement of the body.

Next, the distance of penetration:

$\mathrm{s=ut+}\dfrac{1}{2}\mathrm{a}{{\mathrm{t}}^{\mathrm{2}}}$

$\Rightarrow \mathrm{s=(0}\times 5)\mathrm{+}\dfrac{1}{2}(3\times {{5}^{2}})$

$\Rightarrow \mathrm{s=(0)+}(37.5)$

$\Rightarrow \mathrm{s=37}.5m$

16. Differentiate between mass and weight?

Ans: The difference between mass and weight is given below,

17. A scooter is moving with a velocity of $20m{{s}^{-1}}$when brakes are applied. The mass of the scooter and the rider is $180kg$. The constant force applied by the brakes is $500N$.

(a) How long should the brakes be applied to make the scooter comes to a halt?

Ans:  Given:

Mass of scooter and rider: $m=180kg$

Initial velocity of scooter: $u=20m{{s}^{-1}}$

Final velocity of scooter: $v=0m{{s}^{-1}}$(as it halts after applying the brake)

Force of the brake: $F=-500N$(as it opposes the motion)

To find the time duration over which brake should be applied to stop the scooter.

Now, the force – $\mathrm{F=ma}$

$\mathrm{F=m(}\dfrac{v-u}{t})$

$\Rightarrow -500\mathrm{=180(}\dfrac{0-20}{t})$

Rearranging,

$\Rightarrow t\mathrm{=(}\dfrac{180\times (-20)}{-500})$

$\Rightarrow t\mathrm{=(}\dfrac{-3600}{-500})$

$\Rightarrow t\mathrm{=}7.2s$

(b) How far does the scooter travel before it comes to rest?

Ans: Given:

Mass of scooter and rider: $m=180kg$

Initial velocity of scooter: $u=20m{{s}^{-1}}$

Final velocity of scooter: $v=0m{{s}^{-1}}$(as it halts after applying the brake)

Force of the brake: $F=-500N$(as it opposes the motion)

To find distance travelled by scooter before coming to halt.

Acceleration – $\mathrm{a=(}\dfrac{v-u}{t})$

$\Rightarrow \mathrm{a=(}\dfrac{0-20}{7.2})$

$\Rightarrow \mathrm{a=}-2.78m{{s}^{-2}}$

Acceleration is negative because it is retarding the motion of the scooter.

$\mathrm{s=ut+}\dfrac{1}{2}\mathrm{a}{{\mathrm{t}}^{\mathrm{2}}}$

$\Rightarrow \mathrm{s=(20}\times 7.2)\mathrm{+}\dfrac{1}{2}(-2.78\times {{7.2}^{2}})$

$\Rightarrow \mathrm{s=(144)+}(-72.1)$

$\Rightarrow \mathrm{s=}71.9m$

18. State Newton’s third law of motion and how does it explain the walking of man on the ground?

Ans: Newton’s third law of motion states that for every action, there is an equal and opposite reaction acting on different bodies. This implies the existence of the action-reaction force pair. That is, for every action force created an equal and opposite reaction force will be created.

The walking of a man on the ground can be explained with Newton’s third law of motion. During walking on the ground, the man creates an active force on the ground in the backward direction during each step, while the ground creates a reaction force pushing the man forward enabling him to walk.

19. With what speed must a ball be thrown vertically up in order to rise to a maximum height of $45m$? And for how long will it be in the air?

Ans: Given:

Final velocity of stone: $v=0m{{s}^{-1}}$(as it attains zero velocity at maximum height)

Height to which stone is to be thrown: $h=45m$

Acceleration due to gravity: $g=-10m{{s}^{-2}}$(it is negative because the stone is thrown against gravity upwards)

To find:

• The initial velocity with which the stone is to be thrown: $u$

• Time duration over which the stone stays in the air

1. Initial velocity:

It is known that – ${{v}^{2}}={{u}^{2}}+2gh$

$\Rightarrow {{0}^{2}}={{u}^{2}}+(2\times -10\times 45)$

\[\Rightarrow 0={{u}^{2}}+(-900)\]

\[\Rightarrow {{u}^{2}}=900\]

$\Rightarrow u=30m{{s}^{-1}}$

2. Time:

It is known that – $\mathrm{v=u+gt}$

Thus, $\mathrm{0=30+(}-10\times t\mathrm{)}$

$-10t=-30$

$\Rightarrow t=3s$

It takes $3s$ to go up and another $3s$ to come down. So we can say that the total time the stone is air bound is $3s+3s=6s$

Thus,

• The initial velocity with which the stone is to be thrown is $30m{{s}^{-1}}$

• Time duration over which the stone stays in the air is $6s$

20. State Newton’s second law of motion and derive it mathematically?

Ans: Newton’s second law of motion states that the rate of change of momentum of an object is proportional to the applied unbalanced force in the direction of the force.

Mathematical derivation:

Say we have an object of mass $m$that is moving along a straight line with an initial velocity, $u$. 

It is then uniformly accelerated to velocity, $v$ in time, $t$ by the application of a constant force, $F$ throughout the time, $t$. 

Initial Momentum of object: \[{{p}_{1}}=m\times u\]

Final Momentum of object: \[{{p}_{2}}=m\times v\]

Now, the change of momentum is the Final momentum subtracted by the Initial momentum

Thus, $\Delta p={{p}_{2}}-{{p}_{1}}=mv-mu=m(v-u)$

$\Rightarrow \Delta p=m(v-u)$

The rate of change of momentum is $\dfrac{\Delta p}{t}$

i.e. $\dfrac{\Delta p}{t}=\dfrac{m(v-u)}{t}$

We know that the applied force is proportional to the rate of change of momentum of the object.

$F=\dfrac{\Delta p}{t}$

$\Rightarrow F=\dfrac{m(v-u)}{t}$

But, acceleration $a=\dfrac{v-u}{t}$

Using these, we get

$\Rightarrow F=ma$

The SI unit of force is Newton ($Kgm{{s}^{-2}}$)

The second law of motion gives a method to measure the force acting on an object as a product of its mass and acceleration.

21. A bullet traveling at $360m{{s}^{-1}}$ strikes a block of softwood. The mass of the bullet is $2.0g$. Does the bullet come to rest after penetrating $10cm$ into the wood?

1. Find the average deceleration force exerted by the wood.

Ans: Given:

Mass of bullet: $m=2.0g=0.002kg$

Initial velocity of bullet: $u=360m{{s}^{-1}}$

Final velocity of bullet: $\mathrm{v=0m}{{\mathrm{s}}^{\mathrm{-1}}}$ (as it comes to rest after penetration)

Distance travelled by the bullet into the block:$s=10cm=0.1m$

To find the average deceleration force exerted by the wood block.

It is known that – ${{v}^{2}}={{u}^{2}}+2as$

$\Rightarrow {{0}^{2}}={{360}^{2}}+(2\times a\times 0.1)$

\[\Rightarrow 0=129600+(0.2a)\]

\[\Rightarrow 0.2a=-129600\]

$\Rightarrow a=-648000m{{s}^{-2}}$

The acceleration is negative because it opposes the motion of bullet.

Next, force

$\mathrm{F=ma}$

$\Rightarrow \mathrm{F=(0}\mathrm{.002}\times -648000)$

$\Rightarrow \mathrm{F=}-1296N$

(b) Find the time taken by the bullet to come to rest.

Ans: Given:

Mass of bullet: $m=2.0g=0.002kg$

Initial velocity of bullet: $u=360m{{s}^{-1}}$

Final velocity of bullet: $\mathrm{v=0m}{{\mathrm{s}}^{\mathrm{-1}}}$ (as it comes to rest after penetration)

Distance travelled by the bullet into the block: $s=10cm=0.1m$

To find the time taken by the bullet to come to rest.

It is known that – $\mathrm{v=u+at}$

Thus, $\mathrm{0=360+(}-648000\times t\mathrm{)}$

$\Rightarrow -648000t\mathrm{=}-36\mathrm{0}$

$\Rightarrow t=5.56\times {{10}^{-4}}s$

22. Two objects A and B are dropped from a height. The object B being dropped $1s$ after A was dropped. How long after A was dropped will A and B be $10m$apart?

Ans: Given: Object B is dropped one second after object A.

To find: Time at which A and B will be $10m$ apart

We can say that the initial velocity of both A and B as ${{\mathrm{u}}_{A}}={{\mathrm{u}}_{B}}\mathrm{=0m}{{\mathrm{s}}^{\mathrm{-1}}}$, since they are dropped from rest.

It is known that – $\mathrm{s=ut+}\dfrac{1}{2}\mathrm{a}{{\mathrm{t}}^{\mathrm{2}}}$

For object A – ${{\mathrm{s}}_{A}}\mathrm{=}{{\mathrm{u}}_{A}}\mathrm{t+}\dfrac{1}{2}\mathrm{g}{{\mathrm{t}}_{A}}^{\mathrm{2}}$

For object B –${{\mathrm{s}}_{B}}\mathrm{=}{{\mathrm{u}}_{B}}\mathrm{t+}\dfrac{1}{2}\mathrm{g}{{\mathrm{t}}_{B}}^{\mathrm{2}}$

We take acceleration to be acceleration due to gravity because it is being dropped from a height downwards to the earth.

Acceleration due to gravity: $g=10m{{s}^{-2}}$

Since we need to find the time at which A and B will be $10m$ apart

Let’s say - ${{\mathrm{s}}_{A}}-{{\mathrm{s}}_{B}}\mathrm{=10m}$

Also, since object B is dropped one second after object A, we can say that ${{t}_{B}}={{t}_{A}}-1$

Substituting in – ${{\mathrm{s}}_{A}}-{{\mathrm{s}}_{B}}\mathrm{=10m}$

$\Rightarrow {{\mathrm{s}}_{A}}-{{\mathrm{s}}_{B}}\mathrm{=}{{\mathrm{u}}_{A}}\mathrm{t+}\dfrac{1}{2}\mathrm{g}{{\mathrm{t}}_{A}}^{\mathrm{2}}-{{\mathrm{u}}_{B}}\mathrm{t+}\dfrac{1}{2}\mathrm{g}{{\mathrm{t}}_{B}}^{\mathrm{2}}$

$\Rightarrow 10\mathrm{=}{{\mathrm{u}}_{A}}\mathrm{t+}\dfrac{1}{2}\mathrm{g}{{\mathrm{t}}_{A}}^{\mathrm{2}}-\left[ {{\mathrm{u}}_{B}}\mathrm{t+}\dfrac{1}{2}\mathrm{g(}{{\mathrm{t}}_{A}}-1{{)}^{\mathrm{2}}} \right]$

$\Rightarrow 10\mathrm{=(0}\times \mathrm{t)+}\dfrac{1}{2}\mathrm{(10}\times {{\mathrm{t}}_{A}}^{\mathrm{2}})-\left[ \mathrm{(0}\times \mathrm{t)+}\dfrac{1}{2}\mathrm{(10}\times {{\mathrm{(}{{\mathrm{t}}_{A}}-1)}^{\mathrm{2}}}) \right]$

$\Rightarrow 10\mathrm{=(5}\times {{\mathrm{t}}_{A}}^{\mathrm{2}})-\left[ 5\times \mathrm{(}{{\mathrm{t}}_{A}}^{\mathrm{2}}-2{{\mathrm{t}}_{A}}+1) \right]$

$\Rightarrow 10\mathrm{=5}{{\mathrm{t}}_{A}}^{\mathrm{2}}-\mathrm{5}{{\mathrm{t}}_{A}}^{\mathrm{2}}+10{{\mathrm{t}}_{A}}-5$

$\Rightarrow 10\mathrm{=}10{{\mathrm{t}}_{A}}-5$

\[\Rightarrow 10{{\mathrm{t}}_{A}}=15\]

$\Rightarrow {{\mathrm{t}}_{A}}=1.5s$

Thus, the time at which A and B will be $10m$ apart is $1.5s$

23. A boy throws a stone up with a velocity of $60m{{s}^{-1}}$.

(a) How long will it take to reach the maximum height? $(g=-10m{{s}^{-2}})$

Ans: Given:

Initial velocity of stone: $\mathrm{u=60m}{{\mathrm{s}}^{\mathrm{-1}}}$ 

Final velocity of stone: $v=0m{{s}^{-1}}$(as it attains zero velocity at maximum height)

Acceleration due to gravity: $g=-10m{{s}^{-2}}$(it is negative because the stone is thrown against gravity upwards)

To find time to reach maximum height.

It is known that – $\mathrm{v=u+at}$

$\Rightarrow \mathrm{v=u+gt}$

$\Rightarrow 0\mathrm{=60+(}-10\mathrm{t)}$

$\Rightarrow -10\mathrm{t=}-60$

$\Rightarrow t=6s$

(b) What will be the maximum height reached by the stone?

Ans: Given:

Initial velocity of stone: $\mathrm{u=60m}{{\mathrm{s}}^{\mathrm{-1}}}$ 

Final velocity of stone: $v=0m{{s}^{-1}}$(as it attains zero velocity at maximum height)

Acceleration due to gravity: $g=-10m{{s}^{-2}}$(it is negative because the stone is thrown against gravity upwards)

To find maximum height.

It is known that – $\mathrm{s=ut+}\dfrac{1}{2}\mathrm{a}{{\mathrm{t}}^{\mathrm{2}}}$

$\Rightarrow \mathrm{h=ut+}\dfrac{1}{2}\mathrm{g}{{\mathrm{t}}^{\mathrm{2}}}$

$\Rightarrow \mathrm{h=(60}\times \mathrm{6)+}\dfrac{1}{2}\mathrm{(}-10\times {{6}^{\mathrm{2}}})$

$\Rightarrow \mathrm{h=(360)+(}-180)$

$\Rightarrow \mathrm{h=180m}$

(c) What will be its velocity when it reaches the ground?

Ans: Given:

Initial velocity of stone: $\mathrm{u=60m}{{\mathrm{s}}^{\mathrm{-1}}}$ 

Final velocity of stone: $v=0m{{s}^{-1}}$(as it attains zero velocity at maximum height)

Acceleration due to gravity: $g=-10m{{s}^{-2}}$(it is negative because the stone is thrown against gravity upwards)

To find velocity when it reaches the ground.

Velocity when reaching the ground:

For this, we consider the initial velocity (from its maximum attained height) is zero. And the acceleration due to gravity becomes positive because it is falling down.

i.e. Initial velocity of stone: $\mathrm{u=0m}{{\mathrm{s}}^{\mathrm{-1}}}$ 

Acceleration due to gravity: $g=10m{{s}^{-2}}$

Now,  ${{v}^{2}}={{u}^{2}}+2as$

$\Rightarrow {{v}^{2}}={{u}^{2}}+2gs$

$\Rightarrow {{v}^{2}}={{0}^{2}}+(2\times 10\times 180)$

\[\Rightarrow {{v}^{2}}=3600\]

\[\Rightarrow v=\sqrt{3600}\]

$\Rightarrow v=60m{{s}^{-1}}$

24. A certain particle has a weight of $30N$at a place where the acceleration due to gravity is $9.8m{{s}^{-2}}$.

(a) What are its mass and weight at a place where the acceleration due to gravity is $3.5m{{s}^{-2}}$?

Ans: Given:

Weight of particle:$\mathrm{w=30N}$

Acceleration due to gravity on that planet: ${{g}_{1}}=9.8m{{s}^{-2}}$

Mass of particle: $m$

To find mass and weight of particle on planet with${{g}_{2}}=3.5m{{s}^{-2}}$.

Weight: $w=m\times g$

Mass of particle: $m=\dfrac{w}{g}$

$\Rightarrow m=\dfrac{30}{9.8}$

$\Rightarrow m=3.06kg$

Mass and Weight on planet with${{g}_{2}}=3.5m{{s}^{-2}}$

Mass is a constant quantity irrespective of place. 

So, $\Rightarrow m=3.06kg$

Weight: $w=m\times {{g}_{2}}$

$\Rightarrow w=3.06\times 3.5$

$w=10.71N$

(b) What will be its mass and weight at a place where the acceleration due to gravity is zero?

Ans: Given:

Weight of particle:$\mathrm{w=30N}$

Acceleration due to gravity on that planet: ${{g}_{1}}=9.8m{{s}^{-2}}$

Mass of particle: $m$

To find mass and weight of particle on planet with${{g}_{3}}=0m{{s}^{-2}}$ s.

Weight: $w=m\times g$

Mass of particle: $m=\dfrac{w}{g}$

$\Rightarrow m=\dfrac{30}{9.8}$

$\Rightarrow m=3.06kg$

Mass and Weight on planet with${{g}_{3}}=0m{{s}^{-2}}$

Mass is a constant quantity irrespective of place. 

So, $\Rightarrow m=3.06kg$

Weight: $w=m\times {{g}_{3}}$

$\Rightarrow w=3.06\times 0$

$w=0N$

25. Why does a person while firing a bullet holds the gun tightly to his shoulders?

Ans: A person while firing a bullet holds the gun tightly to his shoulder because of the recoil of the gun when the bullet is fired. This is in accordance with Newton’s third law of motion, which states that for every action, there is an equal and opposite reaction, that acts on different bodies. 

When a bullet is fired, the forward motion of the bullet (action) creates a recoil or backward motion of the gun (reaction). The action force being much greater will create an equivalent recoil force in the backward direction. 

If the person who holds the gun does not hold it properly to his shoulders that can result in injury. This is because the shoulder absorbs most of the force during recoil that enables the shooter to take a steady shot.

Thus, if not held tightly to his shoulders, the shot will not be precise and this can also cause the gun to fly away from his hands.

26. A car is moving with a velocity of $16m{{s}^{-1}}$ when brakes are applied. The force applied by the brakes is $1000N$. The mass of the car its passengers is $1200kg$.

1. How long should the brakes be applied to make the car come to a halt?

Ans: Given:

Mass of car and passengers: $m=1200kg$

Initial velocity of car: $u=16m{{s}^{-1}}$

Final velocity of car: $v=0m{{s}^{-1}}$(as it halts after applying the brake)

Force of the brake: $F=-1000N$(as it opposes the motion)

To find time duration over which brake should be applied to stop the car.

Now, the force – $\mathrm{F=ma}$

$\mathrm{F=m(}\dfrac{v-u}{t})$

$\Rightarrow -1000\mathrm{=1200(}\dfrac{0-16}{t})$

Rearranging,

$\Rightarrow t\mathrm{=(}\dfrac{1200\times (-16)}{-1000})$

$\Rightarrow t\mathrm{=(}\dfrac{-19200}{-1000})$

$\Rightarrow t\mathrm{=19}\mathrm{.2}s$

2. How far does the car travel before it comes to rest?

Ans: Given:

Mass of car and passengers: $m=1200kg$

Initial velocity of car: $u=16m{{s}^{-1}}$

Final velocity of car: $v=0m{{s}^{-1}}$(as it halts after applying the brake)

Force of the brake: $F=-1000N$(as it opposes the motion)

To find distance travelled by car before coming to halt.

Acceleration – $\mathrm{a=(}\dfrac{v-u}{t})$

$\Rightarrow \mathrm{a=(}\dfrac{0-16}{19.2})$

$\Rightarrow \mathrm{a=}-0.83m{{s}^{-2}}$

Acceleration is negative because it is retarding the motion of the scooter.

$\mathrm{s=ut+}\dfrac{1}{2}\mathrm{a}{{\mathrm{t}}^{\mathrm{2}}}$

$\Rightarrow \mathrm{s=(16}\times 19.2)\mathrm{+}\dfrac{1}{2}(-0.83\times {{19.2}^{2}})$

$\Rightarrow \mathrm{s=(307}\mathrm{.2)+}(-152.98)$

$\Rightarrow \mathrm{s=154}\mathrm{.2}m$

Long Answer Questions                       (5 Marks)

1. Two objects of masses $100g$ and $200g$ are moving along the same line and direction with velocities of $2m{{s}^{-1}}$ and $1m{{s}^{-1}}$ respectively. They collide and after the collision, the first object moves at a velocity of $1.67m{{s}^{-1}}$. Determine the velocity of the second object.

Ans: Given:

Mass of object 1: ${{m}_{1}}=100g=0.1kg$

Mass of object 2: ${{m}_{2}}=200g=0.2kg$ 

Initial velocity of object 1: ${{u}_{1}}=2m{{s}^{-1}}$

Initial velocity of object 2:${{u}_{2}}=1m{{s}^{-1}}$

Final velocity of object 1:${{v}_{1}}=1.67m{{s}^{-1}}$

To find: Final velocity of object 2:${{v}_{2}}$

By using the law of conservation of momentum, for an isolated system, is that the total initial momentum for an event is equal to total initial momentum, we get – 

\[{{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}}={{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}\]

Here, \[{{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}\]is the total sum of momentum of objects before collision and \[{{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}}\] is the total sum of momentum of objects after collision.

Substituting the values in – \[{{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}}={{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}\]

\[\Rightarrow (0.1\times 1.67)+(0.2\times {{v}_{2}})=(0.1\times 2)+(0.2\times 1)\]

\[\Rightarrow (0.167)+(0.2\times {{v}_{2}})=(0.2)+(0.2)\]

\[\Rightarrow (0.167)+(0.2\times {{v}_{2}})=0.4\]

\[\Rightarrow 0.2\times {{v}_{2}}=0.4-0.167\]

\[\Rightarrow 0.2\times {{v}_{2}}=0.233\]

$\Rightarrow {{v}_{2}}=1.165m{{s}^{-1}}$

Thus, the velocity of the second object is \[1.165m{{s}^{-1}}\].

2. A truck starts from rest and rolls down a hill with a constant acceleration. It travels a distance of $400m$ in $20s$. Find its acceleration. Find the force acting on it, if its mass is $7metricTonnes$.(Hint: $1metricTonne=1000kg$)

Ans: Given:

Mass of truck:$\mathrm{m=7metricTonne=7000kg}$

Initial velocity of truck: $\mathrm{u=0m}{{\mathrm{s}}^{\mathrm{-1}}}$ (as it is stars from rest)

Distance travelled: $\mathrm{s=400m}$

Time duration of travel:$t=20s$

To find: 

• Acceleration of the truck

• Force acting on the truck

1. First we need to find the acceleration:

It is known that – $\mathrm{s=ut+}\dfrac{1}{2}\mathrm{a}{{\mathrm{t}}^{\mathrm{2}}}$

Thus, $\mathrm{400=(0}\times 20\mathrm{)+}\dfrac{1}{2}(a\times {{20}^{2}})$

$\Rightarrow 400\mathrm{=}\dfrac{1}{2}(a\times 400)$

$\Rightarrow 800\mathrm{=}(a\times 400)$

$\Rightarrow a=2m{{s}^{-2}}$

2. Next, finding the force:

$\mathrm{F=ma}$

$\Rightarrow \mathrm{F=(7000}\times 2)$

$\Rightarrow \mathrm{F=14000}N$

Thus, the acceleration of the truck is $2m{{s}^{-2}}$, and the force acting on the truck is $14000N$.

3. A stone is dropped from a $100m$high tower. How long does it take to fall?

1. The first $50m$

Ans: Given:

Initial velocity of stone: $\mathrm{u=0m}{{\mathrm{s}}^{\mathrm{-1}}}$ (as it is starts from rest, before being dropped)

Height of the tower: $\mathrm{s=100m}$ 

Distance travelled in case A: ${{\mathrm{s}}_{1}}\mathrm{=50m}$ (first half distance)

Distance travelled in case B: ${{\mathrm{s}}_{2}}\mathrm{=50m}$ (next half distance)

To find: Time duration of travel:$t$ during the first $50m$.

It is known that – $\mathrm{s=ut+}\dfrac{1}{2}\mathrm{a}{{\mathrm{t}}^{\mathrm{2}}}$

Since the stone is dropped from a height, we can consider its acceleration to be equal to the acceleration due to gravity.

Acceleration of stone: Acceleration due to gravity  $\Rightarrow a=g=10m{{s}^{-2}}$

${{\mathrm{s}}_{1}}\mathrm{=ut+}\dfrac{1}{2}\mathrm{a}{{\mathrm{t}}^{\mathrm{2}}}$

$\Rightarrow \mathrm{50=(0}\times t\mathrm{)+}\dfrac{1}{2}(10\times {{t}^{2}})$

$\Rightarrow \mathrm{50=5}\times {{t}^{2}}$

$\Rightarrow {{t}^{2}}=10$

$\Rightarrow t=\sqrt{10}=3.16s$

$\Rightarrow {{t}_{1}}=3.16s$

(b) The second $50m$

Ans: Given:

Initial velocity of stone: $\mathrm{u=0m}{{\mathrm{s}}^{\mathrm{-1}}}$ (as it is starts from rest, before being dropped)

Height of the tower: $\mathrm{s=100m}$ 

Distance travelled in case A: ${{\mathrm{s}}_{1}}\mathrm{=50m}$ (first half distance)

Distance travelled in case B: ${{\mathrm{s}}_{2}}\mathrm{=50m}$ (next half distance)

To find: Time duration of travel:$t$ during the second $50m$.

It is known that – $\mathrm{s=ut+}\dfrac{1}{2}\mathrm{a}{{\mathrm{t}}^{\mathrm{2}}}$

Since the stone is dropped from a height, we can consider its acceleration to be equal to the acceleration due to gravity.

Acceleration of stone: Acceleration due to gravity  $\Rightarrow a=g=10m{{s}^{-2}}$

Time duration for the next $50m$can be found by subtracting time for the first half distance from the time for the total distance of travel.

$\mathrm{s=ut+}\dfrac{1}{2}\mathrm{a}{{\mathrm{t}}^{\mathrm{2}}}$

$\Rightarrow 10\mathrm{0=(0}\times t\mathrm{)+}\dfrac{1}{2}(10\times {{t}^{2}})$

$\Rightarrow 100\mathrm{=5}\times {{t}^{2}}$

$\Rightarrow {{t}^{2}}=20$

$\Rightarrow t=\sqrt{20}=4.47s$

$\Rightarrow t=4.47s$

Thus the time for the second half is – ${{t}_{2}}=t-{{t}_{1}}$

$\Rightarrow {{t}_{2}}=4.47-3.16$

$\Rightarrow {{t}_{2}}=1.31s$

4. A body of mass $10kg$ starts from rest and rolls down an inclined plane. It rolls down $10m$ in $2s$.

1. What is the acceleration attained by the body?

Ans: Given:

Mass of body:$\mathrm{m=10kg}$

Initial velocity of body: $\mathrm{u=0m}{{\mathrm{s}}^{\mathrm{-1}}}$ (as it is starting to roll)

Distance travelled on inclined plane: $\mathrm{s=10m}$

Time duration of rolling:$t=2s$

To find acceleration attained by the body.

It is known that – $\mathrm{s=ut+}\dfrac{1}{2}\mathrm{a}{{\mathrm{t}}^{\mathrm{2}}}$

Thus, $\mathrm{10=(0}\times 2\mathrm{)+}\dfrac{1}{2}(a\times {{2}^{2}})$

$\Rightarrow 10\mathrm{=}\dfrac{1}{2}(a\times 4)$

$\Rightarrow 10\mathrm{=}(a\times 2)$

$\Rightarrow a=5m{{s}^{-2}}$

2. What is the velocity of the body at $2s$?

Ans: Given:

Mass of body:$\mathrm{m=10kg}$

Initial velocity of body: $\mathrm{u=0m}{{\mathrm{s}}^{\mathrm{-1}}}$ (as it is starting to roll)

Distance travelled on inclined plane: $\mathrm{s=10m}$

Time duration of rolling:$t=2s$

To find velocity of body at $t=2s$.

It is known that – $\mathrm{v=u+at}$

Thus, $\mathrm{v=0+(5}\times 2\mathrm{)}$

$\Rightarrow v=10m{{s}^{-1}}$

3. What is the force acting on the body?

Ans: Given:

Mass of body:$\mathrm{m=10kg}$

Initial velocity of body: $\mathrm{u=0m}{{\mathrm{s}}^{\mathrm{-1}}}$ (as it is starting to roll)

Distance travelled on inclined plane: $\mathrm{s=10m}$

Time duration of rolling:$t=2s$

To find force acting on the body.

$\mathrm{F=ma}$

$\Rightarrow \mathrm{F=(10}\times 5)$

$\Rightarrow \mathrm{F=50}N$

5. A body of mass $2kg$ is at rest at the origin of a frame of reference. A force of $5N$acts on it at $t=0s$. The force acts for $4s$ and then stops.

(a) What is the acceleration produced by the force on the body?

Ans:  Given:

Mass of body:$\mathrm{m=2kg}$

Initial velocity of body: $\mathrm{u=0m}{{\mathrm{s}}^{\mathrm{-1}}}$ (as it is starting from rest)

Force acting on the body: $F=5N$

Time duration for which force is exerted:$t=4s$

To find acceleration produced by the force on the body.

It is known that – $\mathrm{F=ma}$

$\Rightarrow a=\dfrac{F}{m}$

$\Rightarrow a=\dfrac{5}{2}$

$\Rightarrow a=2.5m{{s}^{-2}}$

(b) What is the velocity at $t=4s$?

Ans:  Given:

Mass of body:$\mathrm{m=2kg}$

Initial velocity of body: $\mathrm{u=0m}{{\mathrm{s}}^{\mathrm{-1}}}$ (as it is starting from rest)

Force acting on the body: $F=5N$

Time duration for which force is exerted:$t=4s$

To find velocity of body at $t=4s$.

It is known that – $\mathrm{v=u+at}$

Thus, $\mathrm{v=0+(2}\mathrm{.5}\times 4\mathrm{)}$

$\Rightarrow v=10m{{s}^{-1}}$

(c) Draw the v-t graph for the period $t=0s$ to $t=6s$.

Ans:  Given:

Mass of body:$\mathrm{m=2kg}$

Initial velocity of body: $\mathrm{u=0m}{{\mathrm{s}}^{\mathrm{-1}}}$ (as it is starting from rest)

Force acting on the body: $F=5N$

Time duration for which force is exerted:$t=4s$

To plot the v-t graph.

Plotting the v-t graph

Using the formula – $\mathrm{v=u+at}$

At $t=0s\Rightarrow v=0+(2.5\times 0)=0m{{s}^{-1}}$

At $t=1s\Rightarrow v=0+(2.5\times 1)=2.5m{{s}^{-1}}$

At $t=2s\Rightarrow v=0+(2.5\times 2)=5m{{s}^{-1}}$

At $t=3s\Rightarrow v=0+(2.5\times 3)=7.5m{{s}^{-1}}$

At $t=4s\Rightarrow v=0+(2.5\times 4)=10m{{s}^{-1}}$

At $t=5s\Rightarrow v=0+(2.5\times 5)=12.5m{{s}^{-1}}$

At $t=6s\Rightarrow v=0+(2.5\times 6)=15m{{s}^{-1}}$

The graph:

(Image will be Uploaded Soon)

(d) Find the distance traveled in $6s$.

Ans: Given:

Mass of body:$\mathrm{m=2kg}$

Initial velocity of body: $\mathrm{u=0m}{{\mathrm{s}}^{\mathrm{-1}}}$ (as it is starting from rest)

Force acting on the body: $F=5N$

Time duration for which force is exerted:$t=4s$

To find distance travelled in $t=6s$.

This can be found by calculating the area under the v-t graph.

(Image will be Uploaded Soon)

This is a triangle with base as $6$ and height as $15$

Area of triangle: $\dfrac{1}{2}\times b\times h=\dfrac{1}{2}\times 6\times 15=45$

Thus, the distance travelled in $t=6s$ is $45m$.

CBSE Class 9 Science Chapter-9 Important Questions - Free PDF Download

Why is Chapter 9 the Forces and Laws of Motion Important for Students?

Forces and Laws of Motion is a crucial chapter. In this chapter, students learn about the Laws of Motion as given by Sir Issac Newton, the various terminologies used for denoting forces of nature and motion of objects, the relation between objects, force, motion, etc. It gives students valuable knowledge about the working of the universe. 

Force and Laws of Motion Class 9 Important Questions with Answers 

Force and Laws of Motion important questions will ensure that students are prepared for all types of questions that might appear in exams. It will help them understand how well they have prepared, how good their time management is, and how they can improve. 

We suggest that students keep practicing the class 9 science chapter 9 important questions to familiarize themselves with the various types of questions, save them precious time in exams, and score good marks. 

The Relevance of Chapter 9 Forces and Laws of Motion

Chapter 9 of Class 9 Physics deals with Forces and Laws of Motion. The chapter talks about the multiple forces in nature, how they act upon objects, how things react, and most importantly, the fundamental Laws of Motion as given by Sir Issac Newton. 

Students need to study this chapter thoroughly and from a young age, so remember them later.  

How to Study for Chapter 9 Forces and Laws of Motion

Given below is the proforma we suggest our students follow for the best possible preparation- 

• Study the given class and study notes carefully, and make their notes for better understanding.

• Consult online classes on our website.

• Watch videos of inertia, forces of motion, and simple examples to understand the concept better.

• Practice the Force and Laws of Motion Class 9 Important Questions so that they get to know all the possible questions that might appear.

• Practice the questions repeatedly so that they get the best preparation possible and attain good marks in exams. 

Introduction to Forces and Laws of Motion

A force refers to the effort to change an object's state at rest or even at motion. It might also change the object's velocity and direction. The shape of an object can also be changed by force. 

Forces can be Two Types- 

• Balanced Forces: Balanced forces do not result in any changes in motion. When it is applied to an object, there will be no such force acting upon the object.

• Unbalanced Forces: Unbalanced Forces move in the direction of the force in the highest magnitude. It acts upon an object and can change its speed and direction of motion. 

Types of Forces 

There are different types of forces acting around us. 

• Gravitational Force: One of the most commonly known forces, gravitational force, refers to the force that exists due to the attraction between two bodies due to their masses. It is denoted by 

F= G (m1m2/r2) 

Here, G refers to the universal constant while m1 and m2 are the masses of the bodies and r is the distance between them. 

• Electromagnetic Force- Electromagnetic Force refers to the force exerted by two charged particles at each other. Two common examples of electromagnetic forces are Friction and Tension. 

• Nuclear Force- There are protons and neutrons in every atom. The nuclear force helps to bind neutrons and protons and holds them together in an atom. It is also known as Strong Force or Nuclear Interactions. This force is larger in magnitude than any other force. However, it has a concise range of influence, so in that respect, the other forces are more dominating. 

• Weak Force- Weak Forces are responsible for phenomenons known as beta decay. Sometimes a neutron changes itself into a proton, emits an electron, and a particle known as antineutrino. This process is known as Beta Decay. 

The weak force is a force of attraction that works at a concise range of 0.1% of a proton's diameter. These forces differ from gravitational, electromagnetic, and nuclear forces and are known as Weak Forces.

Three Main Laws of Motion

The three primary Laws of Motion are Newton's Laws of Motion. They are as follows-

• Newton's First Law of Motion states that an object remains at rest unless an external force acts upon it. Similarly, an object in motion stays in motion in the same direction unless acted upon by an external force. 

• Newton's Second Law of Motion states that the force acting on a body will be directly proportional to the rate of change in its momentum.

• Newton's Third Law of Motion states that "For every action, there is an equal and opposite reaction." 

The Terminology Used in this Chapter

• Net Force: Net force acting on a body refers to the fact that when multiple forces work on one body, it tends to change into one component. 

• Frictional Force: The force present between two surfaces in contact and opposes relative motion is known as Frictional Force. 

• Inertia: Inertia refers to all bodies' tendency to resist changes in a state of rest or motion. However, all bodies do not have the same inertia. The inertia of a body is directly proportional to its mass.

• Momentum: The momentum of an object refers to the product of its mass and velocity. p=mv The impacts that an object or body produces depend on its mass and speed, and this vector quantity has direction and magnitude. 

• Inertial and Non-Inertial Frames: Inertial Frame refers to the frame where Newton's Laws hold. On the other hand, Non-Inertial Frames refer to the reference frame where Newton's Law of Motion does not fit. A non-inertial frame undergoes acceleration with respect to an inertial frame. In a non-inertial frame, an accelerometer will detect a non-zero acceleration. 

Significance of Key Questions in CBSE Class 9 Science Chapter 9 - Force and Laws of Motion

The significance of Key Questions for CBSE Class 9 Science Chapter 9 - "Force and Laws of Motion" provided by Vedantu is substantial and cannot be emphasized enough. These curated questions play a crucial role in a student's exam preparation journey. They serve as a targeted resource for revising key concepts and assessing one's understanding of the subject matter. Vedantu's focus on quality education is evident in these questions, which are designed to align with the CBSE curriculum and examination pattern. By practicing these questions, students gain confidence in their problem-solving abilities and enhance their grasp of the fundamental principles of force and motion. These questions are not just aids for academic excellence but also tools for fostering a deeper understanding and appreciation of the laws that govern the physical world.

Why Choose Vedantu to Study?

Vedantu is a one-stop platform for all students who require class notes and solved questions to study. After careful assessment and research, we have provided the best possible class notes, revision notes, and class 9 science chapter 9 important questions for our students. These notes and questions are written by our experts who have immense knowledge of the respective subjects. They carefully go through all the syllabus, notes, and guidelines given by the board and the previous years' question papers before writing the letters and questions available on our website. 

Vedantu also conducts online classes on its website, which can help students understand the study material better. These sessions are also recorded and available on our website if anyone misses the classes and wishes to go through the classes later.

Conclusion 

To conclude, we can say that motion and motion laws are a significant chapter for students in Class 9. Not only does it hold importance in school exams, but it also has considerable weightage in board exams and competitive exams later on. So, we at Vedantu make sure that our students get all the study notes, revision notes, and essential questions of force and motion laws in one place. They are also available in downloadable PDF format so that students can study anytime and anywhere they want. We suggest students practice the critical questions repeatedly for the best preparation possible and fetch good marks. 

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