NCERT Solutions for Class 9 Science Chapter 9: Force and Laws Of Motion

NCERT Solutions for Class 9 Science Chapter 9 - Force and Laws of Motion

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Access NCERT Solutions For Class 9 Science Chapter 9 – Force and Laws of Motion part-1

Access NCERT Solutions For Class 9 Science Chapter 9 – Force and Laws of Motion

INTEXT EXERCISE 1

1. Which of the following has more inertia:

(a) a rubber ball and a stone of the same size?

Ans: The inertia of an object is measured by its mass. Heavier or more massive objects offer larger inertia.

Stone is heavier than the rubber ball of the same size. e. Hence, inertia of the stone is greater than that of a rubber ball.


(b) a bicycle and a train?

Ans: Train is heavier than bicycle. Hence, inertia of the train is greater than that of the

bicycle


(c) a five-rupees coin and a one-rupee coin?

Ans: A five rupee coin is heavier than a one rupee coin. . Hence, inertia of the five rupee coin is greater than that of the 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:  The ball's velocity changes four times.

First change: The ball's speed changes from 0 to a specific amount as the football player kicks it. value. As a result, the ball's velocity is altered. 

Second change:Another player is kicking the ball to the goal post in the second change. As a result of this, the  direction of the ball is changed. As a result, its speed varies. In this case, the player used force. to change the velocity of the ball.

Third change: The ball is being collected by the goalie in the third change. The ball finally comes to a halt. As a result, its speed is lowered to zero from a specific value The pace of the ball has changed. In this situation, the goalie utilised a counter-force to slow down or modify the pace of the ball.


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

Ans: Because of the inertia of rest, when the branch is quickly moved, the leaves attached to it tend to stay in their resting position. The leaves and branch junctions are put under a lot of stress as a result of this. This strain causes some leaves to detach off the branch.

Fourth change-The goalkeeper kicks the ball to his teammates. As a result, the ball's velocity increases from zero to a certain number. As a result, its velocity shifts once more. In this case, the goalkeeper used force to change the ball's velocity.


4. Why do you fall in the forward direction when a moving bus brakes to a stop and fall backwards when it accelerates from rest?
Ans: We move in the forward direction when a moving bus is braking because our upper portion of the body and the bus are both in motion when the bus is moving, and when the bus is breaking our body is trying to be in motion due to inertia of motion and thereby we experience a forward push. Similarly, when the bus accelerates from the rest, the passenger tends to fall backwards. This is because the passenger's inertia tends to oppose the bus's forward motion when the bus accelerates. Therefore, when the bus accelerates, the passenger tends to fall backwards.


INTEXT EXERCISE 2

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

Ans: With his foot, a horse pushes the earth in a rearward way. According to Newton's third law of motion, the Earth exerts a reaction force on the horse in the forward direction. As a result, the cart advances.


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

Ans: When a significant volume of water is discharged from a hose at a high velocity, Newton's Third Law of Motion states that the water pushes the hose backwards with the same force. As a result, gripping a hose that ejects a significant volume of water at a rapid rate is difficult for a firefighter.


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

Given:

Mass of the rifle, ${m_1} = 4$ kg

Mass of the bullet, ${m_2} = 50$ g $ = 0.05$ kg

Recoil velocity of the rifle $ = {v_1}$

Initial velocity of bullet, ${v_2} = 35$ m/s

Ans: As, the riffle is at rest, its initial velocity, $v = 0$

Total initial momentum of the rifle and bullet system $ = \left( {{m_1} + {m_2}} \right)v = 0$

Total momentum of the rifle and bullet system after firing:

$ = {m_1}{v_1} + {m_2}{v_2}$ 

$ = 4\left( {{v_1}} \right) + 0.05 \times 35$

$ = 4{v_1} + 1.75$

According to the law of conservation of momentum,

Total momentum after the firing = Total momentum before the firing

$4{v_1} + 1.75 = 0$

$4{v_1} =  - 1.75$

${v_1} = \dfrac{{ - 1.75}}{4}$

${v_1} =  - 0.4375$ m/s

The negative sign indicates that the rifle recoils backwards with a velocity ${v_1} =  - 0.4375$ m/s


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

Given:

Mass of one of the objects, ${m_1} = 100$ g $ = 0.1$ kg

Mass of the other object, ${m_2} = 200$g $ = 0.2$ kg

Velocity of m1 before collision, ${v_1} = 2$ m/s

Velocity of m2 before collision, ${v_2} = 1$ m/s

Velocity of m1 after collision, ${v_3} = 1.67$ m/s

Ans:

Velocity of m2 after collision $ = {v_4}$

According to the law of conservation of momentum:

Total momentum before collision $ = $ Total momentum after collision

${m_1}{v_1} + {m_2}{v_2} = {m_3}{v_3} + {m_4}{v_4}$

$\left( {0.1} \right)2 + \left( {0.2} \right)1 = \left( {0.1} \right)1.67 + \left( {0.2} \right){v_4}$

$0.2 + 0.2 = 0.167 + 0.2{v_4}$

$0.4 = 0.167 + 0.2{v_4}$

$0.4 - 0.167 = 0.2{v_4}$

$0.233 = 0.2{v_4}$

${v_4} = \dfrac{{0.233}}{{0.2}}$

${v_4} = 1.165$ m/s

Hence, the velocity of the second object becomes $1.165$  m/s after the collision.


NCERT EXERCISE

1. An object experiences a net zero external unbalanced force. Is it possible for the object to be travelling 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. An object can travel at a non-zero velocity even if it has a net zero external unbalanced force. This is only possible if the item moves at a consistent speed in a specified direction. As a result, the body is not subjected to any net imbalanced forces. The item will continue to travel at a velocity greater than zero. A net non-zero external unbalanced force must be supplied to the item to change its state of motion.


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

Ans: Using a stick to beat a carpet; causing the carpet to move quickly, while dust particles trapped in the carpet's pores prefer to stay still, since inertia of an item resists any change in its state of rest or motion. The dust particles, according to Newton's first rule of motion, remain at rest as the carpet moves. As a result, dust particles emerge from the carpet.


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

Ans: According to Newton's First Law of Motion, luggage on a bus' roof tends to maintain its condition of rest when the bus is at rest and retain its state of motion when the bus is in motion. When the bus starts moving again after a period of rest, luggage on the roof may fall down to maintain the resting spot. Similarly, owing to inertia of motion, luggage on the roof top of a moving bus will tumble forward when it arrives in the rest state. To avoid this, any luggage kept on a bus's roof should be tied with a rope.


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

(a) The batsman did not hit the ball hard enough.

(b) Velocity is proportional to the force exerted on the ball.

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

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

Ans: Option(c). When the ball moves on the ground, the force of friction opposes its movement and after some time ball comes to a state of rest.


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

Given: 

Initial velocity of the truck , $u = 0$ (since the truck is initially at rest)

Distance travelled, s $ = 400$ m

Time taken, t $ = 20$ s

Ans:

According to the second equation of motion:

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

$400 = 0 + \dfrac{1}{2}a{\left( {20} \right)^2}$

$400 = \dfrac{1}{2}a\left( {400} \right)$

$400 = a\left( {200} \right)$

\[a = \dfrac{{400}}{{200}}\]

\[a = 2\] m/s2

\[1\] metric tonne \[ = 1000\] kg

\[\therefore 7\] metric tonnes \[ = 7000\] kg

Mass of truck, \[m = 7000\] kg

From Newton’s second law of motion:

Force, F = Mass × Acceleration

F = ma 

F= \[ = 7000 \times 2\]

F \[ = 14000\] N

Hence, the acceleration of the truck is \[2\] m/s2 and the force acting on the truck F \[ = 14000\] N


6. A stone of \[1\] kg is thrown with a velocity of \[20\]m s \[ - 1\] across the frozen surface of a lake and comes to rest after travelling a distance of \[50\] m. What is the force of friction between the stone and the ice?

Given:

Initial velocity of the stone, u \[ = 20\] m/s

Final velocity of the stone, v \[ = 0\] (finally the stone comes to rest)

Distance covered by the stone, s \[ = 50\] m

Ans:

According to the third equation of motion:\[\]

${v^2} = {u^2} + 2as$

${0^2} = {\left( {20} \right)^2} + 2 \times a \times 50$

$0 = 400 + 100a$

$100a =  - 400$

$a =  - \dfrac{{400}}{{100}}$

$a =  - 4$

a = −4 $\dfrac{m}{{{s^2}}}$

The negative sign indicates that acceleration is acting against the motion of the stone.

Mass of the stone, m $ = 1$ kg

From Newton’s second law of motion:

Force, F = Mass × Acceleration

F = ma

F $ = 1 \times  - 4$ 

F $ =  - 4$ N

Hence, the force of friction between the stone and the ice F $ =  - 4$ N .


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

(a) the net accelerating force;

Given:

Force exerted by the engine, F $ = 40000$ N

Frictional force offered by the track, ${F_{fraction}} = 5000$ N

Ans:

Net accelerating force,

${F_{net}} = F - {F_{friction}}$

\[{F_{net}} = 40000 - 5000\]

\[{F_{net}} = 35000\] N

Hence, the net accelerating force \[{F_{net}} = 35000\] N


(b) the acceleration of the train; and

Given:

The engine exerts a force of \[40000\]N on all the five wagons.

Net accelerating force on the wagons, \[{F_{net}} = 35000\] N

Mass of a wagon \[ = 2000\] kg

Number of wagons \[ = 5\]

Formula:

Total Mass of the wagons,

m = Mass of a wagon × Number of wagons

Ans:

Total Mass of the wagons,

m \[ = 2000 \times 5\]

m \[ = 10000\] kg

Mass of the engine, m′ \[ = 8000\] kg

Total mass, M = m + m′ 

\[ = 10000 + 8000\]

\[ = 18000\] kg

From Newton’s second law of motion:

\[Fa = Ma\]

\[a = \dfrac{{Fa}}{m}\]

\[a = \dfrac{{35000}}{{18000}}\]

\[a = 1.944\]m/s2

Hence, the acceleration of the wagons and the train \[a = 1.944\]m/s2


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

Ans: 

The force of wagon 1 on wagon 2 = mass of four wagons x acceleration

Mass of 4 wagons 

\[ = 4 \times 2000\]

\[ = 8000\] kg

F \[ = 8000\] kg \[ \times 1.944\]m/s2

F \[ = 1552\] N


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

Given:

Mass of the automobile vehicle, m \[ = 1500\]kg

Final velocity, \[v = 0\]

Acceleration of the automobile, a \[ =  - 1.7\] \[m{s^{ - 2}}\]

Ans:

From Newton’s second law of motion, 

Force = Mass × Acceleration 

\[ = 1500 \times \left( { - 1.7} \right)\]

\[ =  - 2550\] N

Hence, the force between the automobile and the road \[ =  - 2550\] N.

Negative sign shows that the force is acting in the opposite direction of the vehicle.


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

(a) (mv)2

(b) mv2

(c)1/2 mv2

(d) mv

Ans: 

(d) mv

Mass of the object \[ = m\]

Velocity \[ = v\]

Momentum = Mass × Velocity

Momentum \[ = mv\]


10. Using a horizontal force of \[200\] N, 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:

A same amount of force will act in the opposite direction, according to Newton's third law of motion.

Friction is the name of this force. As a result, the cabinet is subjected to a \[200\] N frictional force.


11. Two objects, each of mass \[1.5\] kg are moving in the same straight line but in opposite directions. The velocity of each object is \[2.5\] m s−1 before the collision during which they stick together. What will be the velocity of the combined object after collision?

Given:

Mass of first object , m1 \[ = 1.5\] kg

Mass of second object , m2 \[ = 1.5\] kg

Velocity of m1 before collision, v1 \[ = 2.5\] m/s

Velocity of m2, (moving in opposite direction ) before collision, v2 \[ =  - 2.5\] m/s

Ans: 

After collision, the two objects stick together.

Total mass of the combined object \[ = {m_1} + {m_2}\]

\[ = 1.5\] kg \[ + 1.5\] kg

\[ = 3\]kg

Velocity of the combined object \[ = v\]

According to the law of conservation of momentum:

Total momentum before collision = Total momentum after collision

\[{m_1}{v_1} + {m_2}{v_2} = ({m_1} + {m_2})v\]

\[ \Rightarrow 1.5 \times 2.5 + 1.5\left( { - 2.5} \right) = \left( {1.5 + 1.5} \right)v\]

\[ \Rightarrow 3.75 - 3.75 = 3v\]

\[ \Rightarrow v = 0\]

Hence, the velocity of the combined object after collision \[v = 0\] m/s.


12. 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: 

The static friction force is quite strong due to the truck's massive bulk. Because the student's effort is insufficient to overcome the static friction, the truck cannot be moved. In this circumstance, the net imbalanced force in either direction is zero, which explains why there is no movement. The force exerted by the learner and the force exerted owing to static friction cancel each other out.

As a result, the student is correct in claiming that the two equal and opposing forces cancel each other out.


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

Ans: Mass of the hockey ball, m \[ = 200\] g \[ = 0.2\] kg

velocity of the ball , \[{v_1} = 10\] m/s

Initial momentum \[ = m{v_1}\]

velocity of the ball after struck by the stick, \[{v_2} =  - 5\] m/s

Final momentum \[ = m{v_2}\]

Change in momentum 

\[ = m{v_1} - m{v_2}\]

\[ = m\left( {{v_1} - {v_2}} \right)\]

\[ = 0.2\left( {10 - \left( { - 5} \right)} \right)\]

\[ = 0.2 \times 15\]

\[ = 3\] kg \[m{s^{ - 1}}\]

Hence, the change in momentum of the hockey ball \[ = 3\] kg \[m{s^{ - 1}}\]


14. A bullet of mass \[10\] g travelling horizontally with a velocity of \[150\] \[m{s^{ - 1}}\] strikes a stationary wooden block and comes to rest in \[0.03\] s. 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.

Given:

Initial velocity of the bullet , u \[ = 150\] m/s

Final velocity, \[v = 0\]Time, \[t = 0.03\] s

Ans:

According to the first equation of motion, \[v = u + at\]

Acceleration of the bullet, a

\[0 = 150 + \left( {a \times 0.03s} \right)\]

\[a =  - \dfrac{{150}}{{0.03}}\]

\[a =  - 5000\] m/s2

(Negative sign indicates that the velocity of the bullet is decreasing.)

According to the third equation of motion:

\[{v^2} = {u^2} + 2as\]

\[{0^2} = {\left( {150} \right)^2} + 2\left( { - 5000} \right)s\]

\[0 = 22,500 - 10000s\]

\[10000s = 22,500\]

\[s = \dfrac{{22,500}}{{10000}}\]

\[s = 2.25\] m

Hence, the distance of penetration of the bullet into the block \[s = 2.25\] m

From Newton’s second law of motion:

Force, F = Mass × Acceleration

Mass of the bullet, m \[ = 10\] g \[ = 0.01\] kg

Acceleration of the bullet, a \[ =  - 5000\] \[\dfrac{m}{{{s^2}}}\]

F = ma 

\[ = 0.01 \times  - 5000\]

\[ =  - 50\]N

Hence, the magnitude of force exerted by the wooden block on the bullet \[ =  - 50\]N

but it acts in opposite direction.


15. An object of mass \[1\] kg travelling in a straight line with a velocity of \[10\] \[m{s^{ - 1}}\]collides with, and sticks to, a stationary wooden block of mass \[5\] kg. 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.

Given:

Mass of the object, \[{m_1} = 1\] kg

Velocity of the object before collision, \[{v_1} = 10\]m/s

Mass of the wooden block, \[{m_2} = 5\] kg

Velocity of the wooden block before collision, \[{v_2} = 0\] m/s

Ans: 

∴ Total momentum before collision

$ = {m_1}{v_1} + {m_2}{v_2}$

\[ = 1\left( {10} \right) + 5(0)\]

\[ = 10\] kg \[m{s^{ - 1}}\]

It is given that after collision, the object and the wooden block stick together.

Total mass of the combined system, 

\[m = {m_1} + {m_2}\]

\[ = 1\] kg \[ + 5\] kg

\[ = 6\] kg

Velocity of the combined object \[ = v\]

According to the law of conservation of momentum:

Total momentum before collision \[ = \] Total momentum after collision

$ \Rightarrow {m_1}{v_1} + {m_2}{v_2}$ $ = \left( {{m_1} + {m_2}} \right)v$

$ \Rightarrow 1\left( {10} \right) + 5\left( 0 \right) = \left( {1 + 5} \right)v$

$ \Rightarrow 10 = 6v$

$ \Rightarrow v = \dfrac{{10}}{6}$

$ \Rightarrow v = \dfrac{5}{3}$ m/s

$v = 1.66$ m/s

Total momentum after collision

\[{m_1}v + {m_2}v\]

\[ = v\left( {{m_1} + {m_2}} \right)\]

\[ = 10\left( {6 \times 6} \right)\]

\[ = 10\]kg m/s

The total momentum after collision is also \[10\] kg m/s.

Total momentum just before the impact \[ = 10\]kg m/s .

Total momentum just after the impact \[ = 10\]kg m/s .

Hence, velocity of the combined object after collision \[ = \dfrac{5}{3}\] m/s .


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

Given:

Initial velocity of the object, u \[ = 5\] m/s

Final velocity of the object, v \[ = 8\] m/s

Mass of the object, m \[ = 100\] kg

Time taken by the object to accelerate, t \[ = 6\] s

Ans:

Initial momentum \[ = \] mu

\[ = 100 \times 5\]

\[ = 500\] kg \[m{s^{ - 1}}\]

Final momentum \[ = \] mv 

\[ = 100 \times 8\]

\[ = 800\] kg \[m{s^{ - 1}}\]

Force exerted on the object, 

F \[ = \] mv-mu/t

F \[ = \left( {\dfrac{{800 - 500}}{6}} \right)\]

F \[ = \dfrac{{300}}{6}\]

F \[ = 50\] N

Initial momentum of the object is \[500\] kg \[m{s^{ - 1}}\].

Final momentum of the object is \[800\] kg \[m{s^{ - 1}}\].

Force exerted on the object is \[50\] N.


17. Akhtar, Kiran and Rahul were riding in a motor car 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:

As a result, the vehicle and insect systems have no change in momentum.

In this case, the insect experiences a bigger change in velocity, which results in a greater shift in momentum. Kiran's assessment is correct from this perspective.

The motorcar travels at a faster speed and has a bigger mass than the insect.

Furthermore, the motorcar continues to travel in the same direction after the collision, indicating that the motorcar has the least amount of momentum change, whilst the insect has the most. As a result, Akhtar's statement is likewise correct.

Because the momentum acquired by the bug is equal to the momentum lost by the motorcar, Rahul's observation is likewise true. This is also in agreement with the conservation of momentum law. However, he committed an error since the system suffers from a flaw. Because the momentum before the collision is identical to the momentum after the impact, there is no change in momentum following the accident.


18. How much momentum will a dumbbell of mass \[10\] kg transfer to the floor if it falls from a height of \[80\] cm? Take its downward acceleration to be \[10\] \[m{s^{ - 2}}\].

Given:

Mass of the dumbbell, m \[ = 10\] kg

Distance covered by the dumbbell, s \[ = 80\] cm \[ = 0.8\] m

Acceleration in the downward direction, a \[ = 10\] \[\dfrac{m}{{{s^2}}}\]

Initial velocity of the dumbbell, u \[ = 0\]

Ans: 

Final velocity of the dumbbell v = ?

According to the third equation of motion:

\[{v^2} = {u_2} + 2as\]

\[{v^2} = 0 + 2\left( {10} \right)0.8\]

\[{v^2} = 20 \times 0.8\]

\[{v^2} = 16\]

\[v = \sqrt {16} \]

\[v = 4\] m/s

Hence, the momentum with which the dumbbell hits the floor is

\[ = \]mv

\[ = 10 \times 4\]

\[ = 40\] kg \[m{s^{ - 1}}\]


ADDITIONAL EXERCISE:

1. The following is the distance-time table of an object in motion:

Time in seconds

Distance in metres

\[0\]

\[0\]

\[1\]

\[1\]

\[2\]

\[8\]

\[3\]

\[27\]

\[4\]

\[64\]

\[5\]

\[125\]

\[6\]

\[216\]

\[7\]

\[343\]


(a) What conclusion can you draw about the acceleration? Is it constant, increasing,

decreasing, or zero?

Ans:

From the table, we can see that the distance changes unequally in equal intervals of time. Thus the object is said to be in non- uniform motion. Since, velocity of the object is increasing with time, the acceleration is also increasing.


(b)What do you infer about the forces acting on the object?

Ans: 

According to Newton’s second law of motion, \[F = mat\]. In the given case, acceleration is increasing , which indicates that the force is also increasing.


2. Two persons manage to push a motorcar of mass \[1200\]kg at a uniform velocity along a level road. The same motorcar can be pushed by three persons to produce an acceleration of \[0.2\] \[m{s^{ - 2}}\]. With what force does each person push the motorcar?

(Assume that all persons push the motorcar with the same muscular effort)

Given:

Mass of the motor car \[ = 1200\] kg

Acceleration produced by the car, when it is pushed by the third person, a \[ = 0.2\] \[\dfrac{m}{{{s^2}}}\]

Ans: 

Let the force applied by the third person be F.

From Newton’s second law of motion:

Force = Mass × Acceleration

F \[ = 1200 \times 0.2\]

F \[ = 240\] N

Thus, the third person applies a force of magnitude \[240\]N.

Hence, each person applies a force of \[240\] N to push the motor car.


3. A hammer of mass \[500\]g, moving at \[50\] \[m{s^{ - 1}}\], strikes a nail. The nail stops the hammer in a very short time of \[0.01\] s. What is the force of the nail on the hammer?

Given:

Mass of the hammer, m \[ = 500\] g \[ = 0.5\] kg

Initial velocity of the hammer, u \[ = 50\] m/s

Time taken by the nail to the stop the hammer, t \[ = 0.01\] s

Velocity of the hammer, v \[ = 0\]

Ans:

From Newton’s second law of motion:

Force, F =m(v-u)/t

F \[ = \dfrac{{0.5\left( {0 - 50} \right)}}{{0.01}}\] 

F \[ =  - 2500\]N

The hammer strikes the nail with a force F \[ =  - 2500\]N.

Hence, from Newton’s third law of motion, the force of the nail on the hammer is equal and opposite, i.e.,\[ + 2500\] N.


4. A motorcar of mass \[1200\] kg is moving along a straight line with a uniform velocity of \[90\] km/h. Its velocity is slowed down to \[18\] km/h in \[4\] s by an unbalanced external force. Calculate the acceleration and change in momentum. Also calculate the magnitude of the force required.

Given:

Mass of the motor car, m \[ = 1200\] kg

Initial velocity of the motor car, u \[ = 90\]km/h \[ = 25\] m/s

Final velocity of the motor car, v \[ = 18\] km/h \[ = 5\]m/s

Time taken, t \[ = 4\] s

Ans: 

According to the first equation of motion:

\[v = u + at\]

\[5 = 25 + a\left( 4 \right)\]

\[5 - 25 = a\left( 4 \right)\]

\[20 = a\left( 4 \right)\]

\[a = \dfrac{{20}}{4}\]

\[a =  - 5\] m/s2

Change in momentum 

= mv − mu 

= m (v−u)

\[ = 1200\left( {5 - 25} \right)\]

\[ = 1200\left( { - 20} \right)\]

\[ =  - 24000\] kg \[m{s^{ - 1}}\] 

Force = Mass × Acceleration

Force \[ = 1200 \times  - 5\]

Force \[ =  - 6000\] N

Acceleration of the motor car \[ =  - 5\] m/s2

Change in momentum of the motor car \[ =  - 24000\] kg \[m{s^{ - 1}}\] 

Hence, the force required to decrease the velocity \[ =  - 6000\] N.


5. A large truck and a car, both moving with a velocity of magnitude v, have a head-on collision and both of them come to a halt after that. If the collision lasts for \[1\] s:

Let the mass of the truck be M and that of the car be m.

Thus, M > m

Initial velocity of both vehicles, v

Final velocity of both vehicles, v’ = 0 (since the vehicles come to rest after collision)

Time of impact, t \[ = 1\]s


(a) Which vehicle experiences the greater force of impact?

Ans: 

From Newton’s second law of motion, the net force experienced by each vehicle is given by the relation:

\[{F_{car}} = m\left( {v' - v} \right)/t =  - mv\]

\[{F_{truck}} = m\left( {v' - v} \right)/t =  - Mv\]

Since the mass of the truck is greater than that of the car, it will experience a greater force of impact.


(b) Which vehicle experiences the greater change in momentum?

Ans: 

Initial momentum of the car = mv

Final momentum of the car = 0

Change in momentum = mv - 0

= mv

Initial momentum of the truck = Mv

Final momentum of the truck = 0

Change in momentum = Mv -0

= Mv

Since the mass of the truck is greater than that of the car, it will experience a greater change in momentum.


(c) Which vehicle experiences the greater acceleration?

Ans: 

By Newton's third law of motion, for every action there is an equal and opposite reaction that acts on different bodies. Since the truck experiences a greater force of impact (action), this larger impact force is also experienced by the car (reaction). Thus, the car is likely to suffer more damage than the truck.


(d) Why is the car likely to suffer more damage than the truck?

Ans: 

Truck experiences a greater force of impact ( action), this larger impact force is also

experienced by the car ( reaction).Thus, the car is likely to suffer more damage than the truck.


NCERT Solutions for Class 9 Science Chapter 9 - Force and Laws of Motion

You can opt for Chapter 9 - Force and Laws of Motion NCERT Solutions for Class 9 Science PDF for Upcoming Exams and also You can Find the Solutions of All the Maths Chapters below.

NCERT Solutions for Class 9 Science


NCERT Solutions For Class 9 Science- Free PDF Download

Technology doesn't act as a barrier with NCERT Solutions if you are worried about internet connectivity. Our NCERT Solutions Class 9 is available in pdf format and is easy to download.  Get them from our website or app and they can be accessed anytime and anywhere upon download.  NCERT Solutions Class 9 is entirely free of cost. So if you are going to have a test or exam near, NCERT  Solutions Class 9 is there for you. Solutions are made in such a way that all students, whether bright or average can rely on them.


An Overview of Class 9 Science Chapters 9- Force And Laws Of Motion

In the curriculum of Class 9 Science, Chapter 9 is the Force and Laws of Motion This chapter belongs to Unit II- Motion, Force, and Work. If you are a student of Class 9 then you might be well aware of the chapter. This chapter is all about Force, Types of Forces, and Laws of motion given by Sir Isaac Newton. The topics of this chapter are Force and its types, First law of motion, Inertia and mass, Second law of motion, Mathematical formulation of the second law of motion, Third law of motion, Conservation of momentum, etc. All these concepts are explained in a simple language combined with diagrams, activities/experiments involved, and an explanation of the numerical problems if any.

Our subject matter experts have prepared these NCERT Solutions Class 9 Chapter 9 in an efficient manner which not only makes the study interesting but also builds a strong foundation for students.


Class 9 Science Chapter 9 Force and Laws of Motion Weightage

Chapter 9 belongs to Unit II of the Class 9 curriculum and this unit has a weightage of 27 marks. Many questions of the Physics section are formed from this Chapter.  Preparing with these NCERT Solutions will help the student to score better in their exams.


Here is More Detail About The Contents of Chapter 9

9.1 Balanced and Unbalanced forces

9.2 First Law of Motion

9.3 Inertia and mass

9.4 Second law of motion

9.4.1 Mathematical formulation of the second law of motion

9.5 Third law of motion

9.6 Conservation of motion


Benefits of NCERT Solutions Class 9 Chapter 9

  • Preparing from our NCERT Solutions Class 9 is a great way for students through which they have a strong grip on the topics of the chapter

  • These solutions not only build concepts but also help in strategy formation for students to excel in exams.

  • Detailed analysis of topics with weightage is given which helps the students in better preparation.

  • Highly simplified language is used by our experts to prepare these NCERT Solutions which makes it understandable for the students.

  • Students without any hesitation can rely upon these NCERT solutions for last-minute preparation or revision starting from the zero levels. 

FAQs (Frequently Asked Questions)

Q1. What are force and its effects?

Ans: A push or a pull on anybody is called Force. The direction in which a body is pushed or pulled is called the direction of the force. For example, if a horse cart is pulled by a horse in the east direction then that ‘pull’ is the force and east is the direction of the force.


Effects of Force

We cannot see the force but through its effect, we can identify the force. There are various effects of force as explained below-

  • Making a stationary body move. For example, Kicking a ball at rest. 

  • A force can stop a moving body. For example, Brakes applied on a moving cycle.  

Q2. What are the 3 Laws of Motion?

Ans: Newton gave the 3 laws of motion that describe the motion of moving bodies. 

  • First Law of motion:- A body at rest will remain at rest, and a body in motion will continue in motion with uniform speed unless an external force is applied on the body to change its state of rest or uniform motion.

  • Second Law of motion:-  The rate of change of momentum is directly proportional to the applied force, and takes place in the direction in which the force acts. 

  • Third Law of motion: To every action, there is an equal and opposite reaction. Example: Firing of Gun.

Q3. Which concepts in the NCERT Solutions for Class 9 Science Chapter 9 are important from the exam perspective?

Ans: Class 9 Science Chapter 9 Force and Laws of Motion is a practical chapter that carries high weightage in the exam. This chapter carries 27 marks, hence you need to know the important topics that you should prepare well. The following are some of the important topics from this chapter that you should prepare thoroughly:

  • Balanced and Unbalanced forces

  • First Law of Motion 

  • Inertia and mass

  • Second Law of Motion

  • Mathematical Formulation of the second law of motion

  • Third Law of Motion

  • Conservation of motion.

Q4. What is Force Class 9th NCERT?

Ans: Force is referred to as the frequency of action to change the motion of any object or person. You apply force to change the motion of an object from the resting stage to motion or vice versa. Several characteristics, such as the weight of the object, the height at which the object is placed, and the slope of the path, determine the force needed to be applied on an object. Force is applied to accelerate or develop the motion in an object or to decline the already induced motion of the object. 

Q5. How many laws of motion are there and what do they imply?

Ans: There are three laws of motion described by Newton. These are:

  • First Law of Motion - If an object is at rest, it will stay at rest unless a net force is applied to it. If an object is in motion, it will stay in motion unless a net force is applied to it.

  • Second Law of Motion - More force applied, more acceleration.

  • Third Law of Motion - For every action, there is an equal and opposite reaction.

Q6. Where can I find the downloadable solutions for Class 9 NCERT Chapter 9?

Ans: To find the downloadable solutions for NCERT Class 9 chapter 9, follow these steps -

  • Click on the link NCERT Solutions for Class 9 Science (Physics) Chapter 9 

  • You will land on the Vedantu Solutions page for NCERT Class 9 Chapter 9 “Force and Laws of Motions”.

  • At the top of the page, you will see an option to download the PDF of the Solutions for NCERT Chapter 9.

  • You can also get important questions here to practice more questions for the exam.

Q7. What are the key points to choose NCERT Solutions for Class 9 Science Chapter 9? 

Ans: In NCERT Class 9 Science Chapter 9, Force and Laws of Motion, you will find many identities and formulae that you need to keep in mind while solving the numericals. You should have a guide with yourself to understand the tricks to solve these questions faster. NCERT Solutions for Class 9 Chapter 9 are prepared by subject specialists, and they are highly accurate. You will get many tricks to solve your question even faster than before. You can have a deep study about these on the Vedantu Mobile app and for free of cost.

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