Students can use Exploration Class 9 Science Solutions Chapter 6 How Forces Affect Motion Question Answer NCERT Solutions as a quick reference guide.
Class 9 Science Exploration Chapter 6 Question Answer
Class 9 Science Ch 6 How Forces Affect Motion Question Answer
How Forces Affect Motion Class 9 Questions and Answers (Exercise)
Revise, Reflect, Refine (NCERT Textbook Page No. 112 – 114)
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?
Answer:
The frictional force exerted by the floor on the table is equal in magnitude to the applied force F.
When an object moves at constant velocity, its acceleration is zero. By Newton’s Second Law (F = ma), a zero acceleration means the net force on the object is zero. This means all forces balance out perfectly. The only horizontal forces acting on the table are the applied force F (pushing it forward) and friction / (acting backward, opposing the motion). For the net force to be zero, the frictional force must be exactly equal and opposite to the applied force. Therefore, frictional force = F is acting in the direction opposite to the motion
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 hall will remain the same / increase / decrease.
Answer:
(i) remain the same
This is Newton’s First Law of Motion (Law of Inertia). On a frictionless surface, there is no friction to slow the ball down. If no net force acts on the ball, there is nothing to change its motion. The ball will continue moving in the same direction with the same speed indefinitely. Velocity (which includes both speed and direction) remains constant.
(ii) increase
By Newton’s Second Law, a net force in the direction of motion produces acceleration in that same direction. Acceleration means the speed is increasing. So when a net force pushes the ball forward (same direction as motion), the ball speeds up. The magnitude of velocity (speed) increases.
(iii) decrease
A net force opposite to the direction of motion causes a deceleration (negative acceleration, also called retardation). The ball is slowing down because the force is opposing its motion. The magnitude of velocity (speed) decreases. If the force continues long enough, the ball will eventually come to rest, and if it persists further, the ball will reverse direction.
Question 3.
Two blocks P and Q on a smooth horizontal surface are shown in Fig. 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 option (i): P experiences a net force and Q does not experience a net force.
Explanation for Block P:
Two forces act on block P in opposite directions: 5 N to the right and 4 N to the left. Net force on P = 5 N – 4 N = 1 N to the right. Since the net force is not zero, block P experiences a net force and will accelerate in the direction of the 5 N force.
Explanation for Block Q:
Block Q is moving with constant velocity. As explained in Question 1, constant velocity means zero acceleration, which by Newton’s Second Law means zero net force. Therefore block Q does not experience a net force. The surface is smooth (frictionless), so no friction acts. No other force is mentioned, meaning Q moves freely with whatever velocity it had, unchanged.
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 N, what is the net force on the snake boat? (Ignore drag forces, air friction, etc.)
Answer:
Given:
Total oarsmen: 100
Oarsmen rowing forward (backward stroke): 95
Oarsmen rowing opposite (wrong direction): 5
Force per oarsman: 200 N
Find: Net force on the boat
Force applied by 95 oarsmen (forward direction):
F1 = 95 × 200 N = 19,000 N (forward)
Force applied by 5 oarsmen (backward direction, opposing the boat):
F2 = 5 × 200 N = 1,000 N (backward)
Net force on the boat:
F(net) = 19,000 N – 1,000 N = 18,000 N
The net force on the snake boat = 18,000 N in the forward direction.
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
Answer:
Option (iv) is correct: in the direction of force, with acceleration proportional to the force acting on the object.
This is a direct statement of Newton’s Second Law of Motion: F = ma, where F is the net force, m is the mass of the object, and a is the acceleration. The law tells us two things: First, the direction of acceleration is the same as the direction of the net force (not opposite).
Second, the acceleration is directly proportional to the net force applied, for a fixed mass. The larger the force, the greater the acceleration. This rules out all other options: option (i) and (ii) say acceleration is opposite to force, which is wrong. Option (iii) says acceleration is inversely proportional to force, which is also wrong.
Question 6.
The position-time graph for four objects A, B, C and D moving along a straight line are given in Fig. A net force acts on:
(i) Object A
(ii) Object B
(iii) Object C
(iv) Object D.
Answer:
A net force acts on Object C and Object D (options (iii) and (iv)).
(i) Object A :
The position-time graph for A is a straight line with a positive slope. A straight line on a position-time graph means constant velocity (equal displacement in equal time intervals). Constant velocity means zero acceleration. Zero acceleration means zero net force. No net force acts on A.
(ii) Object B :
The position-time graph for B is a horizontal straight line. This means the position is not changing, so the object is at rest (velocity = zero). Zero velocity and zero acceleration mean zero net force acts on B.
(iii) Object C :
The graph for C is a curved line (like a parabola, curving upward steeply). A curve on a position-time graph means the velocity is changing (the slope of the curve is increasing).
Changing velocity means the object is accelerating. A non-zero acceleration means a net force is acting on C. Net force acts on Object C.
(iv) Object D :
The graph for D is a curved line sloping downward and flattening. This means the object is decelerating (slowing down). The velocity is decreasing, meaning there is a non-zero acceleration (retardation). A non-zero acceleration means a net force is acting on D. Net force acts on Object D.
Question 7.
A sailor jumps out from a small boat to the shore (Fig. below). As the sailor jumps forward, will the boat move? If yes, in which direction and why?
Answer:
Yes, the boat will move. It will move in the direction opposite to the direction in which the sailor jumps (i.e., away from the shore).
This is a direct application of Newton’s Third Law of Motion, which states: For every action, there is an equal and opposite reaction.
When the sailor jumps forward toward the shore, they push backward on the boat with their feet. This backward push on the boat is the action force. By Newton’s Third Law, the boat exerts an equal and opposite force on the sailor (pushing the sailor forward). The reaction to the sailor pushing backward on the boat is that the boat gets pushed backward, away from the shore. Since the boat is on water (very low friction), this reaction force causes the boat to move backward (away from the shore). The sailor moves forward and the boat moves backward, which is why this situation beautifully illustrates Newton’s Third Law.
Question 8.
During a high jump event, a landing mat or sand bed is placed for the athlete to fall upon (Fig. below). Explain the reason behind it.
Answer:
According to Newton’s Second Law (F = ma), the force exerted on an athlete is proportional to their rate of change of momentum. By landing on a soft mat or sand, the time taken for the athlete’s velocity to reach zero is significantly increased. This results in a smaller acceleration (deceleration), thereby reducing the impact force acting on the body to prevent 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.
Answer:
Option (iv) is correct: the loaded cart and the empty cart both exert an equal magnitude of force on each other.
This is a direct consequence of Newton’s Third Law of Motion. The Third Law states that forces always occur in action-reaction pairs. Whenever two objects interact, they exert forces on each other that are equal in magnitude and opposite in direction. This law applies regardless of the masses of the objects, their speeds, or whether one is loaded and the other is empty. The loaded cart exerts a force on the empty cart, and the empty cart exerts an equal and opposite force on the loaded cart. The forces are always equal in magnitude. The difference between the two carts is not in the force they exert on each other, but in how much they accelerate. The lighter, empty cart will accelerate more than the heavier, loaded cart because F = ma, so for the same force, less mass means more acceleration.
Question 10.
The acceleration-mass graph for the acceleration produced by a force on objects of different masses is plotted in Fig. Plot the force-mass graph for this case.
Answer:
From Newton’s Second Law: F = m × a. The force applied is the same in all cases (constant force F).
From the graph in Fig., the force can be calculated at any point. For example, reading from the graph: when mass = 1 kg, acceleration is approximately 10 m/s2, so F = 1 × 10 = 10 N. When mass = 2 kg, acceleration is approximately 5 m/s2, so F = 2 × 5 = 10 N. When mass = 4 kg, acceleration is approximately 2.5 m/s2, so F = 4 × 2.5 = 10 N. This confirms the force is constant at 10 N.
For the force – mass graph:
Since F = m × a and the force is constant (F = 10 N), the relationship is: a = F/m = 10/m. The force does not depend on mass (it is constant). Therefore, the force-mass graph is a horizontal straight line at F = 10 N, parallel to the mass axis. This means no matter what the mass of the object is, the same constant force of 10 N is being applied. The force does not change with mass.
| Mass (kg) | Acceleration (m/s2) | Description | Force F = m × a (N) |
| 1 | 10.0 | F = 1 × 10 = 10 N | 10 N |
| 2 | 5.0 | F = 2 × 5 = 10 N | 10 N |
| 3 | 3.3 | F = 3 × 3.3 = 10 N | 10 N |
| 4 | 2.5 | F = 4 × 2.5 = 10 N | 10 N |
| 5 | 2.0 | F = 5 × 2.0= 10 N | 10 N |
The force – mass graph is a horizontal straight line at F = 10 N, showing that force is constant and independent of mass in this experiment.
Question 11.
The velocity – time graph of an object of mass 10 kg moving along a straight line is shown in Figure. Calculate the force acting on the object by using the graph.
Answer:
Given :
Mass of object : m = 10 kg
From graph :
At time t = 4 s, velocity u = 20 m/s
At time t = 8 s, velocity v = 30 m/s (reading from graph)
Find : Force F acting on the object
Find : Net force on the boat
The acceleration is the slope of the velocity-time graph.
a = \(\frac{v-u}{t}\)
= \(\frac{30-20}{4}\)
= \(\frac{10}{4}\) = 2.5 m/s2
Apply Newton’s Second Law :
F = m × a
= 10 kg × 2.5 m/s2 = 25 N
The graph shows uniform acceleration from 20 to 30 m/s in 4 seconds, giving a constant acceleration of 2.5 m/s2.
Question 12.
A bullet of mass 50 g moving with a speed of 100 ms-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 of bullet : m = 50 g = 0.05 kg
Initial speed : u = 100 m/s
Final speed : v = 0 m/s (bullet stops)
Distance penetrated : s = 50 cm = 0.50 m
Find : Stopping force F on the bullet
Acceleration using the equation of motion
v2 = u2 + 2as
02 = (100)2 + 2 × a × 0.50
0 = 10,000 + a
a = – 10,000 m/s2
(The negative sign shows deceleration, i.e., the bullet is slowing down.)
Apply Newton’s Second Law:
F = m × a
F = 0.05 kg × 10,000 m/s2 = 500 N
The stopping force acting on the bullet = 500 N (directed opposite to the bullet’s motion).
Question 13.
An ace footballer converted a penalty shot by kicking the football with a speed of 108 km/h. 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:
Final speed of ball : v = 108 km/h
Convert : v = 108 × (1000/3600) = 108 × (5/18) = 30 m/s
Initial speed of ball: u = 0 m/s (ball was stationary)
Force applied : F = 800 N
Mass of football : m = 0.4 kg
Find acceleration of the ball using F = m × a
a = \(\frac{F}{m}=\frac{800}{0.4}\)
= 2000 m/s2
From equation of motion v = u + at to find time
30 = 0 + 2000 × t
t = \(\frac {30}{2000}\) = 0.015 s
The time of contact between the foot and the ball = 0.015 s (15 milliseconds)
Alternatively, using Impulse = Change in momentum:
F × t = m × (v – u)
= 0.4 × (30 – 0) = 12 N.s,
so t = \(\frac {12}{800}\) = 0.015 s.
Question 14.
An object of mass 2 kg moving with a constant velocity of 10 ms-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 kg
Initial velocity : u = 10 m/s
Frictional force : f = 7 N (opposing motion)
Additional opposing force: Fext = 3 N (opposing motion)
Final velocity : v = 0 m/s (object comes to rest)
Find : Distance s travelled before stopping
Find total opposing (retarding) force
Total retarding force = friction + additional force
= 7 + 3 = 10 N
Using F = m × a
10 = 2 × a
a = 5 m/s2
deceleration, so a = – 5 m/s
Use equation of motion v2 = u2 + 2as to find distance
v2 = u2 + 2 × (- 5) × s
⇒ 0 = (10)2 – 10s
⇒ 10s = 100
⇒ s = 10 m
The object travels a distance of 10 m before coming to rest.
Question 15.
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:
We know F = ma.
For the harrow alone:
For the trolley alone:
F = m2 x a2
⇒ m2 = \(\frac{F}{a_2}\)
F = m1 × a1
⇒ m1 = \(\frac{F}{a_1}\)
When the trolley carries the harrow, the total mass being pulled is (m1 + m2).
The same force F is applied. Let the new acceleration be a.
F = (m1 + m2) × a
⇒ a = \(\frac{F}{\left(m_1+m_2\right)}\)
Substitute the expressions for m1 and m2:
This shows that the combined acceleration is less than both a1 and a2 individually, because more total mass is being moved by the same force.
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.). Explain why.
Answer:
The compass needle moves while the bar magnet remains stationary because of the huge difference in their masses.
Newton’s Third Law tells us that the force the bar magnet exerts on the compass needle is equal in magnitude to the force the compass needle exerts on the bar magnet. Both forces are identical in size. However, Newton’s Second Law tells us that Force F
acceleration = \(\frac{\text { Force }}{\text { mass }}\) (a = \(\frac {F}{m}\)).
The compass needle is extremely light (mass of just a few grams, typically 0.5 to 2 grams). The bar magnet is much heavier (typically hundreds of grams). Since the force is the same on both, but the bar magnet has a much larger mass, the bar magnet’s acceleration is negligibly small (essentially zero for practical purposes). The tiny compass needle, having a very small mass, experiences the same force but responds with a very large acceleration, causing it to rotate and align with the magnetic field.
Acceleration of needle = \(\frac{F}{m \text { (needle) }}\)
= \(\frac{F}{\text { (very small mass) }}\)
= very large
Acceleration of magnet = \(\frac{F}{m \text { (magnet) }}\)
= \(\frac{F}{\text { (very large mass) }}\)
= nearly zero
This is similar to a cannon firing a cannonball: the explosion exerts equal forces on both the ball and the cannon, but the very light cannonball flies forward with high speed while the heavy cannon barely moves (or recoils only slightly). Equal forces do not mean equal accelerations; they mean equal and opposite changes in momentum.
Class 9 Science Chapter 6 How Forces Affect Motion Question Answer (InText)
Think It Over (NCERT Textbook Page No. 94)
Question 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:
According to Newton’s Third Law of Motion, when the canoeist pushes water backwards with the paddle, the water exerts an equal and opposite force on the paddle in the forward direction. This pushes both the paddle and the canoe forward. When the canoeist pushes harder, the force on the water is larger, and so by Newton’s Third Law, the water exerts a larger force on the canoe. By Newton’s Second Law (F = ma), a larger force on the same mass produces greater acceleration, so the canoe moves faster.
Question 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 empty canoe will move faster. According to Newton’s Second Law of Motion (ia = F/m), for the same force, a larger mass produces smaller acceleration. The canoe carrying a passenger has greater total mass. Therefore, the same force produces less acceleration in the loaded canoe than in the empty one, meaning the empty canoe will move faster.
Pause and Ponder (NCERT Textbook Page No. 97)
Question 1.
A weightlifter lifts a barbell (Fig.). List two forces that are acting on the barbell. Are these forces balanced if the weightlifter keeps the barbell steady?
Answer:
The two forces acting on the barbell are:
(i) Gravitational force (weight) acting downward due to the Earth, and
(ii) The upward force applied by the weightlifter.
Yes, if the weightlifter keeps the barbell steady (not moving), these two forces are balanced. The net force is zero, which is why the barbell remains at rest.
Question 2.
Two players R and S are participating in an arm-wrestling match (Fig.). 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. Since the arm is moving outward towards the viewer (Player S’s side), Player S has exerted the larger force. When forces are unbalanced, the net force acts in the direction of the larger force, which is why the arm moves in that direction.
Pause and Ponder (NCERT Textbook Page No. 101)
Question 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 it. According to Newton’s First Law of Motion, an object in motion continues to move with the same velocity unless a net force acts on it. Since the velocity is constant (no change in speed or direction), the net force must be zero.
Question 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:
Situations (i) and (ii) are possible. When net force is zero, Newton’s First Law tells us the object either stays at rest or keeps moving with constant velocity. Situation (iii) is NOT possible because acceleration requires a non-zero net force (from Newton’s Second Law, F = ma; if F = 0 then a = 0).
Question 5.
In the real world, it is difficult to find a situation where no forces are acting on an object. But by applying additional forces, a condition can be achieved where the net force on the object is zero. Explain with the help of an example.
Answer:
Yes! A book lying on a table is a perfect example. The gravitational force pulls the book downward, but the table exerts an equal upward normal force on the book. These two forces are equal and opposite, so the net force is zero. Even though two forces are acting, the book remains at rest because the forces are balanced. Similarly, a car moving at constant velocity on a road has balanced forward thrust and backward friction, giving zero net force.
Pause and Ponder (NCERT Textbook Page No. 106)
Question 6.
A toy car of mass 100 g is moving with a constant velocity of 0.5 ms-1. What is the net force acting on the toy car?
Answer:
Since the toy car is moving with constant velocity (not changing), by Newton’s First Law, the net force acting on it is zero. The net force = 0 N.
Question 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:
We would need to apply a larger force to the heavier child. From Newton’s Second Law, F = ma. If the acceleration (a) is to be the same for both children, then the force (F) needed is directly proportional to the mass (m). The heavier child has more mass, so a greater force is required to produce the same acceleration.
Question 8.
How are glass items packed for transportation using a bubble wrap or hay protected from damage?
Answer:
Glass items packed in bubble wrap or hay are protected because these soft materials increase the time over which the glass comes to a stop if dropped. From Newton’s Second Law, force = change in velocity / time. When the stopping time increases, the deceleration decreases, and therefore the force experienced by the glass item is reduced. This prevents breakage.
Question 9.
Why does a fire person sometimes struggle when holding the pipe issuing water?
Answer:
Water is expelled from the pipe in the forward direction at high speed. By Newton’s Third Law, the water exerts an equal and opposite reaction force on the pipe (and the person holding it) in the backward direction. This backward force is quite large at high water flow rates, which is why the fireperson struggles to hold the pipe steady.
Pause and Ponder (NCERT Textbook Page No. 110)
Question 10.
Suppose a spacecraft is moving in a region of space where the gravitational force acting upon it is negligible. Suggest how it can change its velocity.
Answer:
In the absence of gravity, the spacecraft can change its velocity by firing its engines. The rocket engine expels gases in one direction, and by Newton’s Third Law, the gases exert an equal and opposite force on the spacecraft in the other direction. This net force changes the spacecraft’s velocity. This is exactly how rockets and spacecraft manoeuvre in space.