NCERT Solutions Class 9 Science Chapter 6 — How Forces Affect Motion

NCERT Class 9 Science — How Forces Affect Motion, Chapter 6, Revise, Reflect, Refine. All 16 questions solved.

Also here. Solutions to the in-text questions for this chapter and the activities for this chapter are on their own pages.

These are the end-of-chapter questions from Chapter 6 of the new Exploration textbook for Class 9 (2026). They test one idea from three sides. Questions 1 to 4 ask what the net force on an object is, and what happens when it is zero. Questions 5, 6 and 10 to 15 use Newton’s second law to connect force, mass and acceleration, often together with the kinematic equations of Chapter 4. Questions 7, 9 and 16 are about Newton’s third law, and Question 8 applies the second law to a safe landing. Everything rests on three results from the chapter:

• Net force and the first law — forces in the same direction add and forces in opposite directions subtract. If the net force on an object is zero, it stays at rest, or keeps moving with a constant velocity.
• Newton’s second law — $F = ma$, or $a = \dfrac{F}{m}$. The acceleration is in the direction of the net force. For objects moving together as one system, $a = \dfrac{F}{m_1 + m_2}$ (Eq. 6.4).
• Newton’s third law — forces come in pairs. When one object pushes or pulls a second, the second pushes or pulls the first with an equal and opposite force, and the two forces act on different objects.

The kinematic equations used here are $v = u + at$ and $v^2 = u^2 + 2as$. Every mass is converted to kilograms, every distance to metres and every speed to $\text{m s}^{-1}$ before it goes into a formula.

Key insight. Always ask two questions: which object am I looking at, and what is the net force on it? If the net force is zero, its velocity does not change — which is why a table pushed at constant velocity, a ball rolling on a frictionless floor and a straight position–time graph all have no net force (Questions 1, 2, 3 and 6). If the net force is not zero, the object accelerates at $a = F/m$ in the direction of that force (Questions 10 to 15). And a third-law pair never cancels, because its two forces act on two different objects: the forces are equal, but the object with less mass gets the larger acceleration (Questions 7, 9 and 16).

Question 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?

Solution. “Constant velocity” is the clue. If the velocity does not change, the acceleration is zero, and by Newton’s first law the net force on the table must be zero.

Two horizontal forces act on the table: the applied force $F$ in the direction of motion, and the force of friction from the floor, which always acts opposite to the direction of motion. (The table’s weight and the normal force from the floor are vertical and balance each other.) For the two horizontal forces to add up to zero, they must be equal in size and opposite in direction:

$$F – f = 0 \qquad\Rightarrow\qquad f = F$$

So the person is not overcoming friction with a larger force; they are only matching it. A larger force would be needed to start the table moving or to speed it up.

Question 2

For a ball moving on a smooth frictionless surface, choose the appropriate option that will make the following statements physically correct.

Solution. The surface is frictionless, so no friction acts on the ball. The only horizontal force is whatever force is applied, and that is the net force. What happens to the velocity then depends only on whether there is a net force, and which way it points.

Question 2 (i)

If no net force is applied on the ball, the velocity of the ball will remain the same/increase/decrease.

Solution. With no net force, Newton’s first law says an object in motion continues to move with a constant velocity. Nothing is pushing the ball forward and nothing, not even friction, is holding it back, so it keeps the same velocity.

Question 2 (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.

Solution. By Newton’s second law, the ball accelerates in the direction of the net force. Here that is the direction in which it is already moving, so each second its velocity grows in the same direction — the ball speeds up.

Question 2 (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.

Solution. Now the acceleration points opposite to the motion. An acceleration opposite to the velocity makes the ball slow down, just as friction slows a rolling ball in everyday life.

Question 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 $4\ \text{N}$ and $5\ \text{N}$ are acting in opposite directions on block P, while block Q is moving with a constant velocity.

Which of the following statement is correct?

Solution. Block P. The two forces act in opposite directions, so the net force is their difference. The $5\ \text{N}$ force pushes to the right and the $4\ \text{N}$ force pushes to the left:

$$\text{net force on P} = 5\ \text{N} – 4\ \text{N} = 1\ \text{N} \ \text{to the right}$$

The forces are opposite but not equal, so they are unbalanced and P does experience a net force. It will accelerate to the right.

Block Q. Q is moving, but it is moving with a constant velocity. By Newton’s first law, an object keeps a constant velocity only when the net force on it is zero. Motion itself does not need a force; only a change in motion does.

Question 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 $200\ \text{N}$, what is the net force on the snake boat? (Ignore drag forces, air friction, etc.)

Solution. Rowing works by the third law, like the canoe in the chapter: an oarsman pushes the water backwards with the oar, and the water pushes the oar, and so the boat, forward. So the $95$ oarsmen who row the correct way give forward forces, and the $5$ who row the opposite way push the water forward and so give backward forces on the boat.

Forces in the same direction add:

$$\text{forward force} = 95\times200 = 19\,000\ \text{N}, \qquad \text{backward force} = 5\times200 = 1000\ \text{N}$$

Forces in opposite directions subtract, and the net force points the way of the larger one:

$$\text{net force} = 19\,000 – 1000 = 18\,000\ \text{N} \ \text{forward}$$

The same answer comes in one step: $5$ of the forward forces are cancelled by the $5$ backward ones, leaving $90$ unopposed oarsmen, and $90\times200 = 18\,000\ \text{N}$.

Question 5

When a net force acts on an object, we observe that the object accelerates:

Solution. Newton’s second law says two things, and each option can be checked against both. First, the object accelerates in the direction of the net force; that rules out (i) and (ii). Second, the acceleration is proportional to the net force and inversely proportional to the mass, $a = F/m$: double the force and the acceleration doubles. Option (iii) says the acceleration is inversely proportional to the force, which has it the wrong way round. Only (iv) gets both parts right.

Question 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:

Solution. A net force shows up as a change in velocity, so look for the graph in which the velocity changes. From Chapter 4, the slope of a position–time graph is the velocity.

• Object A: a straight line rising from the origin. The slope is the same everywhere, so the velocity is constant. No net force.
• Object B: a horizontal line. The position does not change, so the object is at rest. No net force.
• Object C: a curve that starts flat and gets steeper and steeper. The slope, and so the velocity, keeps increasing: C is accelerating, so a net force acts on it.
• Object D: a straight line sloping downwards. The slope is constant (negative), so D moves back towards the starting point with a constant velocity. No net force.

Question 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?

Solution. To jump forward, the sailor must push on something, and the only thing under their feet is the boat. So the sailor’s feet push the boat backwards. By Newton’s third law, the boat pushes the sailor forward with an equal force at the same instant, and that forward force is what carries the sailor towards the shore.

The two forces are equal and opposite, but they act on different objects — one on the sailor, one on the boat — so they do not cancel. The backward force acts on the boat, and the boat floats on water, which does not hold it in place the way the ground holds us. A small boat also has a fairly small mass, so the force gives it a noticeable acceleration, $a = F/m$. The boat therefore moves backwards, away from the shore. This is why a person stepping off a small boat carelessly can find the boat sliding away and land in the water.

Question 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.

Solution. An athlete coming down from above the bar reaches the ground with a large speed, and landing means that speed has to be brought down to zero. That change in velocity is the same whatever they land on. What the landing surface changes is the time the stopping takes.

On hard ground the body stops almost instantly. A given change in velocity in a very short time is a very large acceleration, $a = \dfrac{v – u}{t}$ with $t$ small, and by $F = ma$ a large acceleration needs a large force. That large force from the ground on the body is what injures bones and joints.

A thick mat or a sand bed gives way under the athlete. The body sinks into it and slows down gradually, over a much longer time. The same change in velocity over a longer time is a smaller acceleration, so the force on the body is much smaller. This is the same reason a fielder pulls the hands back while catching a fast cricket ball, and the reason cars have airbags.

To see how large the effect is, take some illustrative values: a $60\ \text{kg}$ athlete who reaches the ground at $6\ \text{m s}^{-1}$. If hard ground stops them in $0.05\ \text{s}$, the net force needed is $60\times\dfrac{6}{0.05} = 7200\ \text{N}$. If the mat stops them in $0.5\ \text{s}$, it is $60\times\dfrac{6}{0.5} = 720\ \text{N}$ — ten times smaller, because the time is ten times longer.

Question 9

A hand cart loaded with vegetables collides with an identical but empty hand cart. During the collision:

Solution. The two carts push on each other during the collision, so (iii) is wrong. These two forces are an action–reaction pair, and Newton’s third law says the forces of such a pair are always equal in magnitude and opposite in direction. It does not matter that one cart is heavier, or which one was moving; so (i) and (ii) are wrong too.

What the difference in mass does change is the effect of the force. The empty cart has the smaller mass, so the same force gives it the larger acceleration, $a = F/m$. That is why the empty cart is jolted more — not because it is pushed harder.

Question 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.

Solution. First read the marked points off the graph, then use Newton’s second law, $F = ma$, to find the force at each one:

$$\begin{array}{l|cccc} \text{Mass } m\ (\text{kg}) & 1 & 2 & 4 & 5 \\ \hline \text{Acceleration } a\ (\text{m s}^{-2}) & 10.0 & 5.0 & 2.5 & 2.0 \\ \hline \text{Force } F = ma\ (\text{N}) & 10 & 10 & 10 & 10 \end{array}$$

Every point gives the same force, $10\ \text{N}$. So the graph shows one and the same force acting on objects of different masses: doubling the mass from $1\ \text{kg}$ to $2\ \text{kg}$, or from $2\ \text{kg}$ to $4\ \text{kg}$, halves the acceleration, exactly as $a = F/m$ says for a fixed force. (Any other point on the curve agrees; at $3\ \text{kg}$ it reads about $3.3\ \text{m s}^{-2}$, and $3\times3.33 = 10\ \text{N}$.)

A force that stays at $10\ \text{N}$ whatever the mass plots as a straight line parallel to the mass axis:

Question 11

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

Solution. The graph is a straight line inclined to the time axis, so the object has a constant acceleration, as in the chapter’s Example 6.6. Read two points off the line: at $t = 0\ \text{s}$ the velocity is $u = 10\ \text{m s}^{-1}$, and at $t = 8\ \text{s}$ it is $v = 30\ \text{m s}^{-1}$. Using $v = u + at$:

$$30 = 10 + a\times8 \qquad\Rightarrow\qquad a = \frac{30 – 10}{8} = \frac{20}{8} = 2.5\ \text{m s}^{-2}$$

The other marked point gives the same slope: from $0$ to $4\ \text{s}$ the velocity rises from $10$ to $20\ \text{m s}^{-1}$, and $10/4 = 2.5\ \text{m s}^{-2}$. Now Newton’s second law gives the force:

$$F = ma = 10\ \text{kg}\times2.5\ \text{m s}^{-2} = 25\ \text{N}$$

The velocity is increasing, so the force acts in the direction of motion.

Question 12

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

Solution. To use $F = ma$ we need the bullet’s acceleration inside the block. We know its starting speed, its final speed and the distance it travels, but not the time, so the kinematic equation to use is the one without $t$: $v^2 = u^2 + 2as$. The block is heavy and does not move, so the $50\ \text{cm}$ is the distance over which the bullet stops.

Convert to SI units first: $m = 50\ \text{g} = 0.05\ \text{kg}$ and $s = 50\ \text{cm} = 0.5\ \text{m}$. With $u = 100\ \text{m s}^{-1}$ and $v = 0$:

$$0^2 = 100^2 + 2\times a\times0.5 \qquad\Rightarrow\qquad a = -\frac{10\,000}{1} = -10\,000\ \text{m s}^{-2}$$

The negative sign means the acceleration is opposite to the bullet’s motion; it is slowing down. Then

$$F = ma = 0.05\times(-10\,000) = -500\ \text{N}$$

Question 13

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

Solution. For a penalty the ball starts at rest on the spot, so $u = 0$, and it leaves the foot at $v = 108\ \text{km h}^{-1}$. Treat the $800\ \text{N}$ as a steady force during the contact. The plan is to find the acceleration from $F = ma$, then the time from $v = u + at$.

The speed must be in $\text{m s}^{-1}$:

$$v = 108\ \text{km h}^{-1} = \frac{108\times1000\ \text{m}}{60\times60\ \text{s}} = 30\ \text{m s}^{-1}$$

The acceleration of the ball while the foot pushes it is

$$a = \frac{F}{m} = \frac{800}{0.4} = 2000\ \text{m s}^{-2}$$

Now $v = u + at$ gives

$$30 = 0 + 2000\times t \qquad\Rightarrow\qquad t = \frac{30}{2000} = 0.015\ \text{s}$$

The contact lasts only fifteen thousandths of a second, which is why so large a force is needed to give the ball its speed.

Question 14

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

Solution. Before the patch the velocity is constant, so no net force acts. On the patch two forces act, and both oppose the motion: friction ($7\ \text{N}$) and the extra applied force ($3\ \text{N}$). Forces in the same direction add, so taking the direction of motion as positive,

$$\text{net force} = -(7 + 3) = -10\ \text{N}$$

By Newton’s second law, the acceleration is

$$a = \frac{F}{m} = \frac{-10}{2} = -5\ \text{m s}^{-2}$$

The object slows from $u = 10\ \text{m s}^{-1}$ to $v = 0$. The distance is asked and the time is not given, so use $v^2 = u^2 + 2as$:

$$0^2 = 10^2 + 2\times(-5)\times s \qquad\Rightarrow\qquad 10s = 100 \qquad\Rightarrow\qquad s = 10\ \text{m}$$

(It takes $t = \dfrac{0 – 10}{-5} = 2\ \text{s}$ to stop.)

Question 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.

Solution. The masses are not given, but each can be written in terms of $F$ and its acceleration using $m = F/a$:

$$m_1 = \frac{F}{a_1}, \qquad m_2 = \frac{F}{a_2}$$

With the harrow on the trolley, the two move together as one system of mass $m_1 + m_2$, and the same force $F$ acts on the system. By Eq. (6.4),

$$a = \frac{F}{m_1 + m_2} = \frac{F}{\dfrac{F}{a_1} + \dfrac{F}{a_2}}$$

$F$ appears in every term, so it cancels. Dividing the top and bottom by $F$ and adding the fractions:

$$a = \frac{1}{\dfrac{1}{a_1} + \dfrac{1}{a_2}} = \frac{1}{\dfrac{a_2 + a_1}{a_1a_2}} = \frac{a_1a_2}{a_1 + a_2}$$

A quick check: if the harrow and the trolley had the same mass, then $a_1 = a_2$ and the formula gives $a = a_1/2$. That is right, because the same force on twice the mass gives half the acceleration. The combined acceleration also comes out smaller than both $a_1$ and $a_2$, as it must, since the load is heavier than either part.

Question 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.

Solution. Newton’s third law says the two forces are equal. It says nothing about the two objects responding equally. What each force does depends on the object it acts on, through $a = F/m$ — the same reasoning as the chapter’s Example 6.7, where the Earth and a falling fruit pull each other equally but only the fruit visibly moves.

The compass needle is a thin, very light piece of magnetised metal, balanced on a pivot so that it can turn almost without friction. A small force on a small mass gives it a large acceleration, so the needle swings round.

The bar magnet has a much larger mass, so the same force would give it a much smaller acceleration. And the magnet rests on the table (or is held in the hand), where friction easily balances a force this small. With the magnetic force balanced by friction, the net force on the magnet is zero, and by the first law it stays at rest.

Common mistakes

• Question 1: thinking the push must be larger than friction to keep the table moving. This is the old idea the chapter describes Galileo correcting. A larger push is needed to start the table or speed it up; to keep it moving at a constant velocity the push only has to equal friction, because the net force must be zero.
• Question 3: deciding that Q must have a net force because it is moving. Motion does not need a net force; only a change in motion does. Students also sometimes call P’s forces “balanced” because they are opposite, but opposite forces balance only when they are also equal, and $5\ \text{N}$ and $4\ \text{N}$ are not.
• Question 6: choosing Object D because its graph goes down. A downward-sloping straight line is not slowing down; it is moving back towards the start at a constant velocity. What matters is whether the slope changes, and only C’s slope does.
• Question 9: believing the loaded cart pushes harder because it is heavier. Third-law forces are always equal. The heavier cart is affected less because the same force gives a larger mass a smaller acceleration, not because it feels a smaller force.
• Question 12: leaving $50\ \text{g}$ and $50\ \text{cm}$ unconverted. Using $50$ for both gives $a = 10\,000/100 = 100\ \text{m s}^{-2}$ and $F = 50\times100 = 5000\ \text{N}$, ten times the correct answer. With $0.05\ \text{kg}$ and $0.5\ \text{m}$ the force is $500\ \text{N}$.
• Question 13: using $108$ directly as the speed. That gives $t = 0.4\times108/800 = 0.054\ \text{s}$. The speed must first be converted: $108\ \text{km h}^{-1} = 30\ \text{m s}^{-1}$.
• Question 14: using only the $7\ \text{N}$ of friction, or subtracting the two forces. Both forces oppose the motion, so they add to $10\ \text{N}$. Using $7\ \text{N}$ alone gives $a = 3.5\ \text{m s}^{-2}$ and $s = 14.3\ \text{m}$; subtracting gives $4\ \text{N}$ and $s = 25\ \text{m}$.
• Question 15: adding or averaging the accelerations. When the harrow is put on the trolley it is the masses that add, not the accelerations. Write each mass as $F/a$, add them, and only then use $a = F/m$; the answer, $a_1a_2/(a_1 + a_2)$, is smaller than either $a_1$ or $a_2$.

Practise next

• The in-text Pause and Ponder questions of this chapter, ten across four boxes, which follow the argument one law at a time — balanced forces, then a zero net force, then the second law, then the third — instead of mixing them as the questions here do.
• Chapter 7, Work, Energy, and Simple Machines — the next physics chapter: what a force achieves when it moves an object through a distance, and how that is measured as work and energy.
• Chapter 4, Describing Motion Around Us — for revising velocity, acceleration, position–time and velocity–time graphs, and the kinematic equations used in Questions 6 and 11 to 14.