NCERT Class 9 Science Chapter 6 Solutions – Forces & Motion
Looking for accurate and complete ncert class 9 science chapter 6 solutions for the new CBSE curriculum (2026-27)? You are in the right place! In this guide, we provide step-by-step, direct, and easy-to-understand solutions for all exercise questions from NCERT Exploration Class 9 Science Chapter 6: How Forces Affect Motion.
These answers are crafted specifically to help students write clean, exam-ready answers in their notebooks without extra confusion or lengthy fluff.
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?
The frictional force exerted by the floor on the table is equal in magnitude to $F$ and acts in the opposite direction ($-F$).
Reasoning: Since the table moves at a constant velocity, its acceleration is zero ($a = 0$). According to Newton's First Law of Motion, the net horizontal force acting on the table must be zero ($\text{Net Force} = F - f_{\text{friction}} = 0 \implies f_{\text{friction}} = F$).
Question 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.
- (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 magnitude of velocity will increase.
- (iii) If a net force is applied opposite to the direction of motion, the magnitude of velocity will decrease.
Question 3
Two blocks P and Q on a smooth horizontal surface are shown in Fig. 6.36. Two forces of magnitudes $4\text{ N}$ and $5\text{ N}$ act in opposite directions on block P, while block Q moves with a constant velocity. Which 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.
Correct Option: (i) P experiences a net force and Q does not experience a net force.
Reasoning:
- For Block P: $\text{Net Force} = 5\text{ N} - 4\text{ N} = 1\text{ N}$ (non-zero net force).
- For Block Q: Moving with constant velocity means acceleration $a = 0$, so $\text{Net Force} = 0$.
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?
- Number of oarsmen rowing forward = $95$
- Number of oarsmen rowing backward = $5$
- Effective net number of oarsmen propelling forward = $95 - 5 = 90$
- Force applied by each oarsman = $200\text{ N}$
- $$\text{Net Force on Boat} = 90 \times 200\text{ N} = \mathbf{18,000\text{ N} \text{ (forward)}}$$
Question 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.
Correct Option: (iv) in the direction of force, with acceleration proportional to the force acting on the object.
Reasoning: According to Newton's Second Law of Motion ($F = ma \implies a = \frac{F}{m}$), acceleration is directly proportional to the applied net force ($a \propto F$) and acts in the direction of the force.
Question 6
The position-time graphs 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
Correct Option: (iii) Object C
Reasoning:
- Objects A & D have straight line $s\text{-}t$ graphs (constant velocity $\implies a = 0 \implies \text{Net Force} = 0$).
- Object B is stationary ($v = 0 \implies a = 0 \implies \text{Net Force} = 0$).
- Object C has a curved $s\text{-}t$ graph, indicating changing velocity (acceleration $a \neq 0$), which requires a non-zero 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?
Yes, the boat will move in the backward direction (away from the shore).
Reasoning: According to Newton's Third Law of Motion, every action has an equal and opposite reaction. When the sailor exerts a force on the boat in the backward direction (Action) to push themselves forward onto the shore, the boat exerts an equal reaction force pushing the sailor forward. As a result, the boat moves backwards into 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.
A cushioned landing mat or sand bed increases the time interval ($\Delta t$) during which the athlete's momentum drops to zero upon landing.
According to Newton's Second Law ($\text{Force} = \frac{\Delta p}{\Delta t}$), increasing the impact time reduces the rate of change of momentum, thereby significantly reducing the impact force on the athlete's body and preventing injury.
Question 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.
Correct Option: (iv) the loaded cart and the empty cart, both exert an equal magnitude of force on each other.
Reasoning: According to Newton's Third Law of Motion, action and reaction forces are always equal in magnitude and opposite in direction, regardless of the masses of the colliding bodies.
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.
Finding the Constant Force Value:
From Fig. 6.40, when Mass $m = 1\text{ kg}$, Acceleration $a = 10\text{ m s}^{-2}$.
$$\text{Force } (F) = m \times a = 1\text{ kg} \times 10\text{ m s}^{-2} = \mathbf{10\text{ N}}$$
Since force remains constant ($10\text{ N}$) for all masses, the Force-Mass graph is a horizontal straight line at $F = 10\text{ N}$.
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.
- Mass of object ($m$) = $10\text{ kg}$
- From Fig. 6.41 ($v\text{-}t$ graph): Initial velocity $u = 0$, velocity $v = 30\text{ m s}^{-1}$ at $t = 8\text{ s}$.
- Acceleration ($a$): $$a = \text{Slope of } v\text{-}t \text{ graph} = \frac{30 - 0}{8 - 0} = 3.75\text{ m s}^{-2}$$
- Force ($F$): $$F = m \times a = 10\text{ kg} \times 3.75\text{ m s}^{-2} = \mathbf{37.5\text{ N}}$$
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.
- Mass of bullet ($m$) = $50\text{ g} = 0.05\text{ kg}$
- Initial velocity ($u$) = $100\text{ m s}^{-1}$
- Final velocity ($v$) = $0\text{ m s}^{-1}$
- Penetration distance ($s$) = $50\text{ cm} = 0.5\text{ m}$
- Acceleration ($a$): Using $v^2 = u^2 + 2as$ $$0 = (100)^2 + 2(a)(0.5) \implies a = -10,000\text{ m s}^{-2}$$
- Stopping Force ($F$): $$F = m \times a = 0.05\text{ kg} \times (-10,000\text{ m s}^{-2}) = \mathbf{-500\text{ N}}$$ (Magnitude of stopping force = $500\text{ N}$)
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.
- Final velocity ($v$) = $108\text{ km h}^{-1} = 108 \times \frac{5}{18} = 30\text{ m s}^{-1}$
- Initial velocity ($u$) = $0\text{ m s}^{-1}$
- Applied force ($F$) = $800\text{ N}$
- Mass of football ($m$) = $0.4\text{ kg}$
- Time of contact ($t$): Using $F = \frac{m(v - u)}{t}$ $$800 = \frac{0.4 \times (30 - 0)}{t} \implies 800t = 12 \implies t = \frac{12}{800} = \mathbf{0.015\text{ s}}$$
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?
- Mass ($m$) = $2\text{ kg}$, Initial velocity ($u$) = $10\text{ m s}^{-1}$, Final velocity ($v$) = $0\text{ m s}^{-1}$
- Total opposing force ($F_{\text{opp}}$) = $7\text{ N (friction)} + 3\text{ N (applied)} = 10\text{ N}$
- Deceleration ($a$): $$a = \frac{-F_{\text{opp}}}{m} = \frac{-10}{2} = -5\text{ m s}^{-2}$$
- Distance travelled ($s$): Using $v^2 = u^2 + 2as$ $$0 = (10)^2 + 2(-5)s \implies 10s = 100 \implies s = \mathbf{10\text{ m}}$$
Question 15
A tractor pulls a harrow 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 $F$), obtain an expression for the resulting acceleration in terms of $a_1$ and $a_2$.
- From first case: $F = m_1 a_1 \implies m_1 = \frac{F}{a_1}$
- From second case: $F = m_2 a_2 \implies m_2 = \frac{F}{a_2}$
- Combined mass ($M$) = $m_1 + m_2 = \frac{F}{a_1} + \frac{F}{a_2} = F \left(\frac{a_1 + a_2}{a_1 a_2}\right)$
- Resulting Acceleration ($a_{\text{net}}$): $$a_{\text{net}} = \frac{F}{M} = \frac{F}{F \left(\frac{a_1 + a_2}{a_1 a_2}\right)} = \mathbf{\frac{a_1 a_2}{a_1 + a_2}}$$
Question 16
When the pole of a bar magnet is brought close to a magnetic compass, the bar magnet and the compass needle exert a magnetic force on each other. As per Newton's third law, both 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.
Although equal forces act on both objects ($F_{\text{magnet}} = F_{\text{needle}}$):
- Mass difference: The mass of the bar magnet ($M$) is extremely large compared to the tiny mass of the compass needle ($m$).
- Acceleration produced ($a = \frac{F}{m}$): Since $a \propto \frac{1}{m}$, the acceleration of the light compass needle is very high and visible, while the acceleration of the heavy bar magnet is negligible and unnoticeable.
- Friction: The static friction between the heavy bar magnet and the table easily overcomes the small magnetic force.
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Based on latest NCERT class 9 book.