NCERT Solutions for Class 9th Science Chapter 6 End-of-chapter exercise — Revise, Reflect, Refine

Book page 112 – 114 Updated on2026-09-08

Q1.

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

The frictional force is F — equal in magnitude to the applied force, and opposite in direction.

Velocity is constant → a = 0 m s^–2
Net horizontal force = ma = m × 0 = 0 N

Horizontal forces: applied force F (forwards), friction F_f (backwards)
F – F_f = 0
F_f = F, directed opposite to the motion

Why it happens: constant velocity is the signature of zero net force (Newton's first law). Since the only two horizontal forces on the table are your push and friction, they must cancel exactly. Note that friction here is not "less than" your push — if it were, the table would be speeding up.

Tip: if F were larger than friction the table would accelerate; if it were smaller, the table would slow down. Constant velocity pins them to being equal.

Q2.

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

Part	Correct option	Reason
(i) No net force	remain the same	a = F/m = 0 ÷ m = 0 m s^–2; velocity cannot change (Newton's first law)
(ii) Net force along the motion	increase	Acceleration is along the velocity, so the ball speeds up
(iii) Net force opposite to the motion	decrease	Acceleration is opposite to the velocity — a retardation, so the ball slows down

Why the direction of the force decides everything: Newton's second law says the acceleration is always along the net force. If that direction matches the velocity, each second adds to the speed; if it opposes the velocity, each second subtracts from it. "Frictionless" is stated so that you know no hidden retarding force is acting — whatever happens is entirely due to the net force named in the question.

Q3.

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

The correct statement is (i) — P experiences a net force and Q does not experience a net force.

Block P: 5 N to the right, 4 N to the left (Fig. 6.36a)
The forces are opposite, so subtract them
Net force = 5 N – 4 N = 1 N to the right → non-zero

Block Q: moving with constant velocity (Fig. 6.36b)
a = 0 m s^–2 → Net force = ma = 0 N

P has unbalanced forces (net 1 N to the right); Q moves at constant velocity, so its net force is zero.

Why Q has no net force even though it is moving: motion does not require a force — only a change of motion does. Q's velocity is constant, so its acceleration is zero, so by F = ma the net force on it must be zero.

Q4.

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

The net force is 18 000 N in the forward direction.

Force by each oarsman = 200 N

Forward group: 95 oarsmen
F_forward = 95 × 200 N = 19 000 N

Opposing group: 5 oarsmen
F_backward = 5 × 200 N = 1 000 N

The two groups push in opposite directions, so subtract
Net force = 19 000 N – 1 000 N = 18 000 N, forwards

Why it happens: forces in the same direction add and forces in opposite directions subtract, with the net force along the larger total. The 5 mistaken oarsmen do not merely fail to help — they actively cancel 5 of the correct oarsmen, so the crew loses 2 × 1 000 N = 2 000 N compared with all 100 rowing together (which would give 20 000 N).

Check it yourself: the effective number of oarsmen is 95 – 5 = 90, and 90 × 200 N = 18 000 N — the same answer, reached faster.

Q5.

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

The correct option is (iv) — in the direction of force, with acceleration proportional to the force acting on the object.

Newton's second law: a = F ÷ m, i.e. F = ma
Direction of a = direction of the net force F
For fixed m: a ∝ F (double the force, double the acceleration)
For fixed F: a ∝ 1 ÷ m (double the mass, halve the acceleration)

Option	Verdict	What is wrong
(i)	Wrong	Direction is wrong; acceleration is along the net force
(ii)	Wrong	Direction is wrong, and a is inversely — not directly — related to mass
(iii)	Wrong	a is directly proportional to force, not inversely
(iv)	Correct	Matches a = F/m exactly

The evidence behind it: Activity 6.3 showed that with the mass fixed, doubling the force roughly doubled the acceleration. Activity 6.4 showed that with the force fixed, doubling the mass roughly halved it. Together they give a = F/m.

Q6.

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

The correct option is (iii) — Object C.

Object	Shape of the position–time graph	What the slope tells you	Net force
A	Straight line, sloping upwards	Constant positive velocity	Zero
B	Horizontal straight line	Position not changing — the object is at rest	Zero
C	Curve that gets steeper and steeper	Velocity increasing — the object is accelerating	Non-zero
D	Straight line, sloping downwards	Constant negative velocity (moving back at a steady rate)	Zero

Slope of a position–time graph = velocity
Straight line (A, B, D) → constant slope → constant velocity → a = 0 → net F = ma = 0 N
Curved line (C) → changing slope → changing velocity → a ≠ 0 → net F ≠ 0

Only C is curved. A changing slope means a changing velocity, and only a changing velocity needs a net force.

The trap in this question: D slopes downwards, which looks dramatic, but a straight line of any slope means a constant velocity — the object is simply moving in the negative direction at a steady rate. Steepness is speed; curvature is acceleration.

Q7.

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 moves backwards, away from the shore.

Sailor pushes the boat backwards with force F (in order to jump forward)
By Newton's third law, the boat pushes the sailor forwards with force F
Water offers very little friction, so the backward force on the boat is nearly unopposed
a_boat = F ÷ m_boat, directed away from the shore

Why the boat moves so noticeably: the boat is small, so its mass is not very much larger than the sailor's, and it floats on water where friction is almost negligible. The same equal-and-opposite force therefore produces an easily visible acceleration of the boat. If the sailor jumped from a heavy ship tied to a pier, the same force would give a far smaller acceleration and you would not notice it.

Tip: this is why a sailor is told to moor the boat first. If the boat slides back as you push off, part of your push is wasted and you may fall into the water.

Q8.

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

The mat increases the time over which the athlete's velocity falls to zero, which reduces the acceleration and therefore the force on the athlete.

Athlete lands with downward velocity u and is brought to v = 0
a = (v – u) ÷ t = –u ÷ t
F = ma = –mu ÷ t

m and u are the same whether the surface is hard or soft
Soft mat → much larger t → much smaller |a| → much smaller |F|

Feel the size of the effect. Take m = 60 kg and u = 6 m s^–1.

Landing surface	Stopping time t	a = u ÷ t	F = ma
Hard ground	0.02 s	300 m s^–2	18 000 N
Foam mat / sand bed	0.50 s	12 m s^–2	720 N

Why it happens: the mat cannot change how fast the athlete arrives, only how gently they are brought to rest. Because the foam or sand compresses, the same change of velocity is spread over a far longer time, so the acceleration — and by F = ma the force on bones and joints — drops by a factor of about 25 in this example. The same reasoning explains a fielder pulling his hands back while catching, an airbag in a car and bubble wrap round glassware.

Q9.

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

The correct option is (iv) — both carts exert forces of equal magnitude on each other.

Newton's third law: whenever one object exerts a force on a second,
the second simultaneously exerts an equal and opposite force on the first

F (loaded on empty) = F (empty on loaded) in magnitude
The two forces are opposite in direction and act on different carts

But the accelerations are not equal:
a_empty = F ÷ m_empty, a_loaded = F ÷ m_loaded
m_loaded > m_empty → a_empty > a_loaded

Why the answer feels wrong at first: we watch the empty cart get flung away and conclude it was hit harder. What we are actually seeing is its larger acceleration, not a larger force. The forces are equal; the masses are not, and a = F/m does the rest. Newton's third law makes no reference at all to the masses of the interacting bodies.

Did you know? The same reasoning explains Example 6.7 — the Earth and a falling fruit pull each other with equal forces, but the Earth's enormous mass makes its acceleration far too small to notice.

Q10.

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

The force–mass graph is a horizontal straight line at F = 10 N, because the same force was applied to every object.

Read pairs of values off Fig. 6.40, then use F = ma:

m = 1 kg, a = 10.0 m s^–2 → F = 1 kg × 10.0 m s^–2 = 10 N
m = 2 kg, a = 5.0 m s^–2 → F = 2 kg × 5.0 m s^–2 = 10 N
m = 4 kg, a = 2.5 m s^–2 → F = 4 kg × 2.5 m s^–2 = 10 N
m = 5 kg, a = 2.0 m s^–2 → F = 5 kg × 2.0 m s^–2 = 10 N

F = 10 N for every mass

Force–mass graph for Fig. 6.40: a straight line parallel to the mass axis at F = 10 N.

Why the two graphs look so different: the acceleration–mass graph in Fig. 6.40 is a falling curve because a = F/m — with F fixed, a is inversely proportional to m, so the curve drops steeply and then flattens. Multiplying each point by its own mass undoes that inverse relation exactly, and the product F comes out the same everywhere. A horizontal force–mass line is therefore the graphical statement of "the same force was used on every object".

Q11.

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

The force acting on the object is 25 N, in the direction of motion.

From Fig. 6.41, the graph is a straight line through the points
(t = 0 s, v = 10 m s^–1), (t = 4 s, v = 20 m s^–1), (t = 8 s, v = 30 m s^–1)

Acceleration = slope of the velocity–time graph = (v – u) ÷ t
a = (30 m s^–1 – 10 m s^–1) ÷ (8 s – 0 s)
a = 20 m s^–1 ÷ 8 s = 2.5 m s^–2

Newton's second law: F = ma
F = 10 kg × 2.5 m s^–2
F = 25 kg m s^–2 = 25 N

Check it yourself: use the other pair of points — a = (20 – 10) m s^–1 ÷ (4 – 0) s = 2.5 m s^–2. Same answer, which confirms the line really is straight and the acceleration constant.

Why the slope gives the acceleration: acceleration is defined as change of velocity divided by the time taken, and that is exactly what "rise over run" measures on a velocity–time graph. Because the line is straight, the slope — and therefore the force — is the same throughout the motion. A curved velocity–time graph would mean a changing force.

Q12.

A bullet of mass 50 g moving with a speed of 100 m 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

The stopping force is 500 N, acting opposite to the bullet's motion.

Convert to SI units first
m = 50 g = 0.05 kg, u = 100 m s^–1, v = 0 m s^–1, s = 50 cm = 0.5 m

Step 1 — find the acceleration using v² = u² + 2as
(0 m s^–1)² = (100 m s^–1)² + 2 × a × 0.5 m
0 = 10 000 m² s^–2 + (1.0 m) a
a = –10 000 m² s^–2 ÷ 1.0 m = –10 000 m s^–2

Step 2 — find the force using F = ma
F = 0.05 kg × (–10 000 m s^–2)
F = –500 N
The minus sign means the force opposes the motion; its magnitude is 500 N

Why the force is so large: the bullet loses all 100 m s^–1 of its velocity within just half a metre of wood. Such an enormous change of velocity over such a short distance means a huge retardation, and F = ma turns that into a large force even though the bullet weighs only 50 g — about 0.49 N. The stopping force is more than a thousand times the bullet's own weight.

Tip: always convert grams to kilograms and centimetres to metres before substituting, otherwise the newton will not come out. 50 g = 0.05 kg and 50 cm = 0.5 m.

Q13.

An ace footballer converted a penalty shot by kicking the football with a speed of 108 km 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

The time of contact is 0.015 s, that is 15 milliseconds.

Step 1 — convert the speed to SI units
v = 108 km h^–1 = 108 × 1000 m ÷ 3600 s = 30 m s^–1
The ball starts from rest, so u = 0 m s^–1

Step 2 — find the acceleration from F = ma
a = F ÷ m = 800 N ÷ 0.4 kg
a = 2000 kg m s^–2 ÷ kg = 2000 m s^–2

Step 3 — find the contact time from v = u + at
30 m s^–1 = 0 m s^–1 + (2000 m s^–2) t
t = 30 m s^–1 ÷ 2000 m s^–2
t = 0.015 s = 15 ms

Why the time is so short: 800 N acting on a mass of only 0.4 kg gives a colossal acceleration of 2000 m s^–2 — about 200 times g. At that rate the ball reaches 30 m s^–1 in a fraction of a heartbeat. This is why a football leaves the boot almost instantly and why high-speed cameras are needed to see the ball deform against the foot.

Tip: to convert km h^–1 to m s^–1, multiply by 5/18. Here 108 × 5/18 = 30 m s^–1.

Q14.

An object of mass 2 kg moving with a constant velocity of 10 m 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

The object travels 10 m in the rough patch before stopping.

Step 1 — find the net force. Both forces oppose the motion, so they add.
F_net = 7 N + 3 N = 10 N, directed opposite to the motion

Step 2 — find the acceleration using F = ma
a = F ÷ m = –10 N ÷ 2 kg
a = –5 m s^–2 (negative because it retards the motion)

Step 3 — find the distance using v² = u² + 2as, with u = 10 m s^–1, v = 0
(0 m s^–1)² = (10 m s^–1)² + 2 × (–5 m s^–2) × s
0 = 100 m² s^–2 – (10 m s^–2) s
s = 100 m² s^–2 ÷ 10 m s^–2 = 10 m

Why the two forces add: friction acts opposite to the motion, and the extra 3 N is stated to oppose the motion too. Forces pointing the same way always add in magnitude, so the object feels a single 10 N retarding force. Before the rough patch the object had constant velocity, which tells you its net force was zero there — the rough patch is what makes the net force non-zero.

Check it yourself: the stopping time is t = (v – u)/a = (0 – 10 m s^–1) ÷ (–5 m s^–2) = 2 s, and the average velocity is (10 + 0)/2 = 5 m s^–1. Distance = 5 m s^–1 × 2 s = 10 m ✓

Q15.

A tractor pulls a harrow (a ploughing tool) of mass m1 with a net force F resulting in an acceleration of a1. The same tractor pulls a trolley of mass m2 with a force F producing an acceleration of a2. 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 a1 and a2. Ignore friction.

Answer

The resulting acceleration is a = a₁a₂ ÷ (a₁ + a₂).

Step 1 — write each mass in terms of F and its acceleration using F = ma
Harrow alone: F = m₁a₁ → m₁ = F ÷ a₁
Trolley alone: F = m₂a₂ → m₂ = F ÷ a₂

Step 2 — treat harrow + trolley as one system (Eq. 6.4)
a = F ÷ (m₁ + m₂)

Step 3 — substitute the masses
a = F ÷ (F/a₁ + F/a₂)
a = F ÷ [ F (1/a₁ + 1/a₂) ]
a = 1 ÷ (1/a₁ + 1/a₂)
a = a₁a₂ ÷ (a₁ + a₂)

A quick sanity check with numbers. Let a₁ = 6 m s^–2 and a₂ = 3 m s^–2.

a = (6 × 3) ÷ (6 + 3) m s^–2 = 18 ÷ 9 = 2 m s^–2
This is smaller than both a₁ and a₂ — as it must be, since the same force now pulls a larger total mass

Why the reciprocals appear: masses add, but accelerations do not. Since a = F/m, mass is inversely related to acceleration for a fixed force. Adding the masses therefore means adding the reciprocals of the accelerations — which is why the answer has the same form as resistors in parallel or lenses in contact.

Q16.

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

Because equal forces do not produce equal accelerations — the compass needle has a far smaller mass and almost no friction opposing it, so the same force moves it easily.

Newton's third law: |F on needle| = |F on bar magnet| = F

Newton's second law: a = F ÷ m
Needle: very small mass m_needle → large acceleration
Bar magnet: much larger mass m_magnet → very small acceleration

Mass. A compass needle is a tiny sliver of magnetised metal of a fraction of a gram; the bar magnet is many times heavier. Since a = F/m, the same F gives the needle a far bigger acceleration.
Friction. The needle is balanced on a sharp pivot, so almost nothing resists its turning. The bar magnet lies on a table, where friction between it and the surface easily balances the small magnetic pull, keeping its net force at zero.

The general principle: Newton's third law fixes only the forces, never the outcomes. What each body then does depends on its own mass and on the other forces acting on it. Example 6.7 makes the same point on a grand scale — the Earth and a falling fruit pull each other equally, but the Earth's huge mass makes its acceleration unmeasurably small. Example 6.8 makes it with a gun and a bullet: the same 2 N gives the bullet 20 m s^–2 but the gun only 0.4 m s^–2.

Try This: hold the bar magnet loosely on a smooth surface, or float it on a piece of thermocol in water, and bring the compass near. Now you will see the magnet move too — because you have removed the friction that was holding it.

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