NCERT Solutions For Class 9 Science Chapter 6 - How Forces Affect Motion

Class 9 Science(Exploration) Chapter 6 - How Forces Affect Motion Question Answers: Download Solution PDF

NCERT Science Class 9 Chapter 6 How Forces Affect Motion Question Answers PDF is made for students so they can get the idea of the chapter in a clear, easy manner. In this NCERT solution you get answers that are well explained for each question, also covering key parts like force, the laws of motion, inertia, and momentum. This PDF helps a lot for daily practice and it works for quick revision too before exams, even when there is less time. It is a good resource to make understanding stronger and score better marks in tests.

Class 9 Science Chapter 6 - How Forces Affect Motion Revise, Reflect, Refine Question Answer:

In this section, you will find easy and clear answers to the Revise, Reflect, Refine questions. These help you understand the chapter better, check your learning, and quickly revise important concepts before exams.

1. Using a horizontal force F, a table is moved across the floor at a constant velocity. How much is the frictional force exerted by the floor on the table?

Answer:

When the table moves with constant velocity, its acceleration is zero. Therefore, the net force acting on it is zero. So, the frictional force must be equal in magnitude and opposite in direction to the applied force F.

2. For a ball moving on a smooth frictionless surface, choose the appropriate option that will make the following statements physically correct.
(i) If no net force is applied on the ball, the velocity of the ball will remain the same/increase/decrease.
(ii) If a net force is applied on the ball in the direction of its motion, the magnitude of the velocity of the ball will remain the same/ increase/decrease.
(iii) If a net force is applied on the ball in a direction opposite to the direction of its motion, the magnitude of the velocity of the ball will remain the same/increase/decrease

Answer:

(i) If no net force is applied on the ball, the velocity of the ball will remain the same.

(ii) If a net force is applied in the direction of motion, the velocity of the ball will increase.

(iii) If a net force is applied opposite to the direction of motion, the velocity of the ball will decrease.

3. Two blocks P and Q on a smooth horizontal surface are shown in Fig. 6.36a and Fig. 6.36b. Two forces of magnitudes 4 N and 5 N are acting in opposite directions on block P, while block Q is moving with a constant velocity.

Which of the following statement is correct?
(i) P experiences a net force and Q does not experience a net force.
(ii) P does not experience a net force and Q experiences a net force.
(iii) Both P and Q experience a net force.
(iv) Neither P nor Q experiences a net force.

Answer:
On block P, two forces of 4 N and 5 N act in opposite directions. Net force on $\mathrm{P}=5-4=1\mathrm{N}$, so P experiences a net force.

Block $Q$ is moving with constant velocity, so its acceleration is zero. Hence, net force on $\mathrm{Q}=0$.

Correct option: (i) P experiences a net force and Q does not experience a net force.

4. While practising for the snake boat race (Vallum kalli in Kerala), 100 oarsmen are rowing a boat together. Out of these, 95 row backwards to propel the boat forward. But by mistake, 5 oarsmen row in the opposite direction. If each oarsman applies a horizontal force of 200 N, what is the net force on the snake boat (Ignore drag forces, air friction, etc.)

Answer:
Force by each oarsman $=200\mathrm{N}$
Force by 95 oarsmen (forward direction):

$95\times 200=19000\mathrm{N}$

Force by 5 oarsmen (opposite direction):

$5\times 200=1000\mathrm{N}$

Net force on the boat:

$19000-1000=18000\mathrm{N}$
5. When a net force acts on an object, we observe that the object accelerates: 
(i) opposite to the direction of force, with acceleration proportional to the force acting on the object.
(ii) opposite to the direction of force, with acceleration proportional to the mass of the object.
(iii) in the direction of force, with acceleration inversely proportional to the force acting on the object.
(iv) in the direction of force, with acceleration proportional to the force acting on the object.

Answer:
(iv) in the direction of force, with acceleration proportional to the force acting on the object.

6. The position-time graph for four objects A, B, C and D moving along a straight line are given in Fig. 6.37. A net force acts on:
(i) Object A
(ii) Object B
(iii) Object C
(iv) Object D

Answer:

A net force acts only when there is acceleration, i.e., when velocity changes.

Object A shows a straight-line graph with constant slope, so velocity is constant and no net force acts.
Object B shows a horizontal line, so the object is at rest and no net force acts.
Object C shows a curved graph, which means velocity is changing and acceleration is present, so a net force acts.
Object D shows a straight-line graph with constant slope (negative), so velocity is constant and no net force acts.

Correct answer: (iii) Object C

7. A sailor jumps out from a small boat to the shore (Fig. 6.38). As the sailor jumps forward, will the boat move? If yes, in which direction and why.

Answer:
Yes, the boat will move.
When the sailor jumps forward towards the shore, he pushes the boat backward. According to Newton’s third law, every action has an equal and opposite reaction.
As a result, the boat moves in the opposite direction to the sailor’s motion.

8. During a high jump event, a landing mat or sand bed is placed for the athlete to fall upon (Fig. 6.39). Explain the reason behind it.

Answer:
A landing mat or sand bed is used to increase the time of impact when the athlete lands.
When the athlete falls, their momentum changes to zero. According to the concept of impulse,

$F=\frac{\mathrm{\Delta }p}{\mathrm{\Delta }t}$
By increasing the time of impact, the force acting on the athlete decreases.

9. A hand cart loaded with vegetables collides with an identical but empty hand cart. During the collision:
(i) the loaded cart exerts a force of larger magnitude on the empty cart.
(ii) the empty cart exerts a force of larger magnitude on the loaded cart.
(iii) neither cart exerts a force on the other.
(iv) the loaded cart and the empty cart, both exert an equal magnitude of force on each other.

Answer:

(iv) the loaded cart and the empty cart, both exert an equal magnitude of force on each other.

Explanation:
According to Newton’s third law, action and reaction forces are equal in magnitude and opposite in direction, regardless of the masses of the objects.

10. The acceleration-mass graph for the acceleration produced by a force on objects of different masses is plotted in Fig. 6.40. Plot the force-mass graph for this case.

Answer:

According to Newton's second law of motion: $F=m\cdot a$
By taking specific data points from the curve in Fig. 6.40:

• At $m=1\mathrm{kg},a=10.0\mathrm{m}{\mathrm{s}}^{-2}$
  $F=1\times 10.0=10\mathrm{N}$
• At $m=2\mathrm{kg},a=5.0\mathrm{m}{\mathrm{s}}^{-2}$
  $F=2\times 5.0=10\mathrm{N}$
• At $m=4\mathrm{kg},a=2.5\mathrm{m}{\mathrm{s}}^{-2}$
  $F=4\times 2.5=10\mathrm{N}$

Since the product of mass and acceleration is constant ( 10 N ) at every point, a constant force of 10 N is applied in all cases.

11. The velocity-time graph of an object of mass 10 kg moving along a straight line is shown in Fig. 6.41. Calculate the force acting on the object by using the graph.

Answer:
From the graph:
Initial velocity, $u=10\mathrm{m}/\mathrm{s}$
Final velocity, $v=30\mathrm{m}/\mathrm{s}$
Time, $t=8\mathrm{s}$
Acceleration:

$a=\frac{v-u}{t}=\frac{30-10}{8}=\frac{20}{8}=2.5\mathrm{m}/{\mathrm{s}}^2$

Force:

$F=ma=10\times 2.5=25\mathrm{N}$

12. A bullet of mass 50 g moving with a speed of $100\mathrm{m}{\mathrm{s}}^{-1}$ enters a heavy stationary wooden block and stops after penetrating a distance of 50 cm . Estimate the stopping force acting on the bullet (assume that the bullet undergoes constant acceleration within the block).

Answer:
Given:
Mass, $m=50\mathrm{g}=0.05\mathrm{kg}$
Initial velocity, $u=100\mathrm{m}/\mathrm{s}$
Final velocity, $v=0$
Distance, $s=50\mathrm{cm}=0.5\mathrm{m}$

$\begin{array}{c}{v}^2=u^2+2as\\ 0=(100{)}^2+2a(0.5)\\ 0=10000+a\Rightarrow a=-10000\mathrm{m}/{\mathrm{s}}^2\end{array}$

Force: $F=ma=0.05\times (-10000)=-500\mathrm{N}$

13. An ace footballer converted a penalty shot by kicking the football with a speed of $108\mathrm{km}{\mathrm{h}}^{-1}$. The estimated force they imparted was 800 N . The mass of the football was 0.4 kg . Calculate the time of contact between their foot and the ball.

Answer:
Given:
Mass, $m=0.4\mathrm{kg}$
Final velocity, $v=108\mathrm{km}/\mathrm{h}=30\mathrm{m}/\mathrm{s}$
Initial velocity, $u=0$
Force, $F=800\mathrm{N}$

$\begin{array}{c}F=\frac{m(v-u)}{t}\\ 800=\frac{0.4\times 30}{t}\\ 800=\frac{12}{t}\Rightarrow t=\frac{12}{800}=0.015\mathrm{s}\end{array}$

14. An object of mass 2 kg moving with a constant velocity of $10\mathrm{m}{\mathrm{s}}^{-1}$ encounters a rough patch where the force of friction on the object is 7 N . At the same time, an additional constant force of 3 N opposing the motion is applied on the object. After entering the rough patch, how much distance does the object travel before coming to rest?

Answer:
Given:
Mass, $m=2\mathrm{kg}$
Initial velocity, $u=10\mathrm{m}/\mathrm{s}$
Final velocity, $v=0$
Friction force $=7\mathrm{N}$
Additional opposing force $=3\mathrm{N}$

Total opposing force:

$F=7+3=10\mathrm{N}$

Acceleration:

$a=\frac{F}{m}=\frac{10}{2}=5\mathrm{m}/{\mathrm{s}}^2$$

(Since force is opposite, $a=-5\mathrm{m}/{\mathrm{s}}^2$ )

$\begin{array}{c}{v}^2=u^2+2as\\ 0=(10{)}^2+2(-5)s\\ 0=100-10s\Rightarrow s=10\mathrm{m}\end{array}$
15. A tractor pulls a harrow (a ploughing tool) of mass $m_1$ with a net force $F$ resulting in an acceleration of $a_1$. The same tractor pulls a trolley of mass $m_2$ with a force $F$ producing an acceleration of $a_2$. If the tractor now pulls the trolley with the harrow placed on it (with the same force $F$ ), then obtain an expression for the resulting acceleration in terms of $a_1$ and $a_2$. Ignore friction.

Answer:
For harrow:

$F=m_1{a}_1\Rightarrow m_1=\frac{F}{a_1}$
For trolley:

$F=m_2{a}_2\Rightarrow m_2=\frac{F}{a_2}$
When both are together:

$\text{Total mass}=m_1+m_2=\frac{F}{a_1}+\frac{F}{a_2}$
Resulting acceleration:

$\begin{array}{c}a=\frac{F}{m_1+m_2}=\frac{F}{\frac{F}{a_1}+\frac{F}{a_2}}\\ a=\frac{1}{\frac{1}{a_1}+\frac{1}{a_2}}\\ a=\frac{a_1{a}_2}{a_1+a_2}\end{array}$

16. When the pole of a bar magnet is brought close to a magnetic compass, the bar magnet and the compass needle (which is also a magnet) exert a magnetic force on each other. As per Newton's third law of motion, both the forces are equal in magnitude and opposite in direction. However, the compass needle moves, whereas the bar magnet does not move (Fig. 6.42). Explain why.

Answer:

According to Newton’s third law, the forces on the bar magnet and the compass needle are equal and opposite. However, their masses are very different.

The compass needle has very small mass, so even a small force produces a noticeable acceleration and it moves easily. The bar magnet has a much larger mass, so the same force produces very little acceleration, and its motion is not noticeable.

Class 9 Science Chapter 6 - How Forces Affect Motion Think It Over Question Answer(Page 94):

This section provides simple and clear answers to the Think It Over questions, helping students think deeply about the concepts. It is useful for improving understanding and quick revision before exams.

1. Why does a canoe move forward when the canoeist pushes water backwards with their paddle and why does it move faster when they push harder?

Answer:

When the canoeist pushes water backward with the paddle, the water exerts an equal and opposite force on the canoe (Newton’s third law). This reaction force pushes the canoe forward.

When the canoeist pushes harder, a greater force is applied on the water. As a result, the reaction force on the canoe also increases, producing greater acceleration.

2. Suppose the same canoeist uses the same paddle force in two different canoes, one empty and one carrying another passenger. In which case will the canoe move faster?

Answer:
The canoe will move faster when it is empty.
When the same force is applied, acceleration depends on mass:

$a=\frac{F}{m}$

The empty canoe has less mass, so it gets more acceleration and moves faster. The canoe with an extra passenger has more mass, so acceleration is less.

NCERT Class 9 Science Chapter 6 - How Forces Affect Motion Pause and Ponder Questions and Answers:

This section includes simple and thoughtful answers to the Pause and Ponder questions, helping students think deeply about the concepts. It supports better understanding and strengthens conceptual clarity.

Class 9 Science NCERT Chapter 6 Pause and Ponder (Page - 97)

1. A weightlifter lifts a barbell (Fig. 6.8). List two forces that are acting on the barbell. Are these forces balanced if the weightlifter keeps the barbell steady?

Answer:

Two forces acting on the barbell are:

• Gravitational force (weight) acting downward
• Upward force applied by the weightlifter

If the weightlifter holds the barbell steady, these forces are balanced.

2. Two players R and S are participating in an arm-wrestling match (Fig. 6.9). At the instant, when the arms tilt to the front direction (out of the page towards you), are the forces exerted by the players balanced? If not, which player exerted the larger force?

Answer: No, the forces are not balanced. Player S exerted the larger force.

NCERT Class 9 Science Chapter 6 Pause and Ponder(Page 101)

3. An object is moving with a constant velocity. Is there a net force acting upon it?

Answer:

No, there is no net force acting on the object.

When an object moves with constant velocity, it means both its speed and direction are not changing. Therefore, there is no acceleration.

According to Newton’s first law, if there is no acceleration, the net force must be zero.

4. Suppose, no net force is acting on an object. Which of the following situations are possible?
(i) Object remains at rest if at rest.
(ii) Object keeps moving with a constant velocity if already moving.
(iii) Object is moving with a constant acceleration.

Answer:

(i) Object remains at rest if at rest — True
(ii) Object keeps moving with a constant velocity if already moving — True
(iii) Object is moving with a constant acceleration — False

Answer:

In real life, many forces act on an object at the same time. However, if these forces are equal in magnitude and opposite in direction, their effect cancels out and the net force becomes zero.

For example, when a book is kept on a table, the gravitational force pulls it downward, while the normal force from the table pushes it upward with equal magnitude. As a result, the net force is zero and the book remains at rest.

Class 9 Science NCERT Chapter 6 Pause and Ponder(Page 106)

6. A toy car of mass 100 g is moving with a constant velocity of $0.5\mathrm{m}{\mathrm{s}}^{-1}$. What is the net force acting on the toy car?

Answer:
Since the toy car is moving with constant velocity, its acceleration is zero.
According to Newton's second law,

$\begin{array}{c}F=ma\\ F=0.1\times 0=0\mathrm{N}\end{array}$

7. Two children of different masses are sitting on identical swings. To impart identical initial acceleration, for which child would you require to apply a larger force? Explain why.

Answer:
A larger force is required for the child with greater mass.
According to Newton's second law,

$F=ma$

For the same acceleration, force is directly proportional to mass. Therefore, the child with greater mass needs more force to achieve the same acceleration.

8. How are glass items packed for transportation using a bubble wrap or hay protected from damage?

Answer:
Glass items are packed with bubble wrap or hay to increase the time of impact when they experience a shock or fall.

According to the concept of impulse,

$F=\frac{\mathrm{\Delta }p}{\mathrm{\Delta }t}$

If the time of impact increases, the force acting on the object decreases. Bubble wrap or hay provides a soft cushioning, which increases the time during which the glass comes to rest.

NCERT Chapter 6 Science Class 9 Pause and Ponder (Page 110)

Answer:

A fireperson sometimes struggles while holding the pipe because water comes out with high speed and large momentum. When water is expelled from the pipe, an equal and opposite force acts on the pipe (Newton’s third law). This backward force is quite large due to the high velocity of water.

10. Suppose a spacecraft is moving in a region of space where the gravitational force acting upon it is negligible. Suggest how can it change its velocity.

Answer:
Even if gravitational force is negligible, the spacecraft can change its velocity by expelling gas in the opposite direction. When the spacecraft ejects gas backward, an equal and opposite force acts on it (Newton’s third law), which changes its velocity.

Class 9 Science NCERT Chapter 6 - How Forces Affect Motion, Activity Question Answer:

Class 9 Chapter 6 Science Activity 6.1: Let us investigate

1. Collect four coins of ` 10, one large strong rubber band and an adhesive tape. Locate horizontal surfaces of different materials, such as wooden table top, cemented floor, laminated table top, and polished marble or tiled floor (you may also choose other surfaces). Check that the surfaces are level.
2. Stack the four coins on top of each other and secure them together with an adhesive tape around the sides.
3. Hold the rubber band slightly stretched between your forefinger and thumb on the wooden table top (Fig. 6.12a). Mark points A and B at its ends as shown in Fig. 6.12b. Make another mark C up to which you will
stretch the rubber band.

4. Holding the ends of the rubber band at A and B, place the stack of coins near the middle of A and B . Now, using a finger of your other hand, push back the stack of coins till the rubber band is pulled back to the mark C (Fig. 6.12c). Then, release the stack of coins and observe its motion. Do you find that after losing contact with the rubber band, the velocity of the stack of coins decreases gradually and it comes to rest after travelling some distance? Measure the distance travelled from C and record it. Repeat this step twice.
5. Repeat steps 3 and 4 for laminated table top while ensuring that the points $\mathrm{A},\mathrm{B}$ and C are marked at the same distances as earlier. Does the stack of coins travel a larger distance than it did on the wooden table top before coming to rest? Does its velocity decrease more slowly now?
6. Next, repeat step 5 on a horizontal polished marble or tile floor. Does the stack of coins travel an even larger distance and its velocity decrease even more slowly? What conclusion do you draw from your observations?

Observations

• Wooden table: Coins travel a short distance (about 20–30 cm) and stop quickly. This shows more friction.
• Laminated surface: Coins travel a larger distance (about 35–50 cm). This shows less friction.
• Marble/tile floor: Coins travel the maximum distance (about 55–70 cm) and slow down very slowly. This shows the least friction.

Explanation

The stretched rubber band applies a forward force and sets the coins in motion. After losing contact with the rubber band, only friction acts opposite to the motion. This friction gradually reduces the velocity and finally brings the coins to rest.

Conclusion

Friction depends on the nature of the surface. Rough surfaces have more friction and stop the motion quickly, while smooth surfaces have less friction and allow the object to travel a longer distance before coming to rest.

NCERT Class 9 Chapter 6 Science Activity 6.2: Let us measure

1. Take a spring balance and a wooden block.
2. Place the spring balance in a horizontal position on one of the surfaces used in Activity 6.1 and check that its scale reading is zero. Attach the wooden block to the hook of the spring balance as shown in Fig. 6.14

3. Pull the spring balance with gradually increasing force and note down the reading on it when the block just starts moving. What does this reading indicate? The forces acting on the block are the force applied by the spring on it and the force of friction. If the velocity of the block is neither increasing nor decreasing, what can you say about the net force acting on the block? Does the reading of the spring balance indicate the magnitude of the force of friction acting on the wooden block?
4. Now repeat step 3 on the remaining three surfaces from Activity 6.1.
5. Compare the readings of the spring balance for all surfaces. Are the readings different? Is the reading smallest for the surface on which the stack of coins travelled the largest distance? Is the reading largest for which the distance travelled was the smallest?

Class 9 NCERT Chapter 6 Science Activity 6.2 – Let us measure

Observations

• When the block just starts moving, the spring balance shows a certain reading.
• This reading is different for different surfaces.
• The reading is maximum for rough surfaces like a wooden table.
• The reading is minimum for smooth surfaces like marble or tile.

Explanation

As the spring balance is pulled, it applies a force on the wooden block. Initially, the block does not move because friction balances this applied force. As the applied force increases, a point is reached when the block just begins to move.

At this moment, the applied force is equal to the maximum static friction. Hence, the reading of the spring balance indicates the force of friction acting on the block.

After the block starts moving, friction continues to oppose the motion.

Conclusion

The reading of the spring balance gives the magnitude of the force of friction. Greater the roughness of the surface, greater is the frictional force and hence higher is the reading. Smooth surfaces have less friction and show smaller readings.

Class 9 Chapter 6 Science Activity 6.3: Let us experiment (Demonstration activity)

This activity is recommended to be performed as a classroom group activity facilitated by the teacher.
1. Take four ball bearing wheels, two pencils, an empty cardboard box (to make a cart), a paper cup, a piece of pipe (to use as a pulley), a length of thread, some coins or other objects (to place in cup) and a weighing scale to measure mass.
2. Insert two pencils through the sides of the box near the bottom, to function as axles and attach a wheel to each of their free ends as shown in Fig. 6.17a (if the wheels are loose, wrap some adhesive tape at the pencil ends to fit the wheels tightly). Attach a thread to the front end of the box with which you can pull the cart.

3. Draw a line at one end of the table, which will mark the starting point for the cart. Put the thread over a small pipe attached at the other end of the table (Fig. 6.17b). To this thread attach a cup in which you can put some objects. As you let the system go, the cup will move down due to the gravitational force by the Earth on it, and the thread will pull the cart with a constant force.
4. Measure the mass of the cup along with any other objects put inside it with the weighing scale.
5. Start recording a video of the cart in slow motion. Release the cart from the start line, and record the video until it reaches the pipe at the other end of the table.
6. Read the time when you released the cart and when it reached the end of the table by seeing the slow-motion video, and record the difference as time $T_1$.
7. Now double the mass of the cup with the objects inside it, and repeat steps 5 and 6 to record the time difference $T_2$.

Observations

• When a certain mass is placed in the cup, the cart moves and takes time $T_1$ to reach the end.
• When the mass in the cup is increased, the cart moves faster and takes less time $T_2$.
• Thus, $T_2<T_1$.

Explanation
The hanging cup pulls the cart due to gravitational force. This force acts as a pulling force on the cart through the thread.

When the mass in the cup is increased, the gravitational force increases. As a result, the pulling force on the cart also increases. A greater force produces greater acceleration, so the cart moves faster and reaches the end in less time.

Conclusion
The experiment shows that greater the applied force, greater is the acceleration produced. Hence, when the mass (and therefore force) increases, the cart moves faster and takes less time to cover the same distance.

Activity 6.4: Let us experiment (Demonstration activity)

This activity is recommended to be performed as a classroom group activity facilitated by the teacher.
1. Repeat Activity 6.3 with a variation. Keep the mass of the cup and objects inside it constant. Double the mass of the cart by adding more objects in it.
2. Measure the mass of the cart along with the objects inside it with a weighing scale.
3. Carry out steps 5 and 6 of Activity 6.3.

Analysis
From the measured times, the ratio of accelerations is given by:

$a_1/a_2=T_2^2/T_1^2$

When the mass of the cart is increased while keeping the applied force the same, the time taken increases $(T_2>T_1)$, which means the acceleration decreases ( $a_2<a_1$ ).

Observation
For a given force, the acceleration produced is inversely proportional to the mass of the object. When the mass is doubled, the acceleration approximately becomes half.

Conclusion
This experiment verifies Newton's Second Law of Motion. It states that the acceleration of an object is directly proportional to the applied force and inversely proportional to its mass.

$F=ma$

The acceleration is in the direction of the net force acting on the object.

Class 9 Chapter 6 Science Activity 6.5: Let us explore

1. Locate a chair with wheels and a large heavy table.
2. Sit on the chair with your legs raised above the floor. Now, using both your hands, push the table away from you, i.e., apply a force on the table in the forward direction as shown in Fig. 6.23a. What happens to you? Does the chair you are sitting upon move in the opposite direction?
3. Now, try to pull the table towards you, i.e., apply a force on the table in the direction opposite to that in step 2 (Fig. 6.23b). In which direction does your chair move now?

Observations
Pushing the table:
When you push the table forward, it exerts an equal and opposite force on you. As a result, the chair moves backward, while the table moves (or tends to move) forward.

Pulling the table:
When you pull the table towards yourself, it exerts an equal and opposite force on you in the opposite direction. The chair moves forward, while the table moves towards you.

Conclusion
This activity demonstrates Newton's Third Law of Motion. For every action, there is an equal and opposite reaction. The forces always occur in pairs and act on two different objects.

NCERT Class 9 Chapter 6 Science Activity 6.6: Let us verify

1. Take two identical spring balances.
2. Place them in horizontal position on a table and connect them by their hooks as shown in Fig. 6.26. Fix the free end of one of the spring balances to an immovable object or hold it fixed by your hand.
3. Imagine that you are pulling the free end of the other spring balance with your other hand. Predict what will be the readings of their scales if the spring balances are stationary.
4. Now, carry out step 3. Repeat it multiple times by varying the magnitude of the force applied by you. Is your observation same as your prediction?

Prediction
If two spring balances are connected and kept stationary, both should show the same reading because the forces acting are equal and opposite.

Observations
The readings of both spring balances are always equal. This shows that the forces they exert on each other are equal in magnitude and opposite in direction.

Explanation
One spring balance measures the force applied by you, while the other measures the force transmitted through the connection. Since the system is at rest, these forces are equal, so both balances show the same reading.

Conclusion
This experiment verifies Newton's Third Law of Motion. Whenever one object exerts a force on another, the second object exerts an equal and opposite force on the first.

Class 9 Chapter 6 Science NCERT Activity 6.7: Let us understand

1. Collect a large balloon, a piece of drinking straw, adhesive tape, a long thread, and two nails or hooks on two walls.
2. Inflate the balloon and tie its neck with a small piece of thread.
3. Stick the piece of straw with an adhesive tape on the surface of the balloon such that, one end of the straw points towards the neck of the balloon, as shown in Fig. 6.29.
4. Pass the thread through the straw and tie its two ends to the nails, keeping the thread taut (Fig. 6.29).
5. Remove the thread tied to the neck of the balloon and observe in which direction the straw and the balloon move.

Observations
When the thread at the neck of the balloon is removed, air rushes out of the balloon in one direction. The balloon, along with the straw, moves in the opposite direction along the thread.

Explanation
The air escaping from the balloon exerts a force in one direction. In response, the balloon experiences an equal and opposite force. This force pushes the balloon forward along the thread.

Conclusion
This activity demonstrates Newton's Third Law of Motion. When air is expelled in one direction, the balloon moves in the opposite direction due to an equal and opposite reaction force.

NCERT Solution for Class 9 Science Chapter 4: Topics

6.1 The Concept of Force

• 6.1.1 Measuring the magnitude of a force

6.2 Balanced and Unbalanced Forces
6.3 The Force of Friction: Often Overlooked but Always Present
6.4 Newton’s First Law of Motion
6.5 Newton’s Second Law of Motion
6.6 Newton’s Third Law of Motion
6.7 Forces Acting on a System of Objects