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NCERT Class 9 Science Chapter 6 How Forces Affect Motion
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How Forces Affect Motion
Chapter – 6
| Pause And Ponder |
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?
Ans: The two forces acting on the barbell are the upward force applied by the weightlifter and the downward gravitational force (weight). If the weightlifter keeps 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?
Ans: No, the forces are not balanced. The player in whose direction the arm moves has exerted the larger force.
| Pause And Ponder |
1. An object is moving with a constant velocity. Is there a net force acting upon it?
Ans: No. If an object is moving with a constant velocity, the net force acting on it is zero.
2. Suppose, no net force is acting on an object. Which of the following situations are possible?
(i) Object remains at rest if at rest.
Ans: Possible.
(ii) Object keeps moving with a constant velocity if already moving.
Ans: Possible.
(iii) Object is moving with a constant acceleration.
Ans: Not possible, because constant acceleration requires a net force.
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.
Ans: A book resting on a table is an example. The downward gravitational force is balanced by the upward normal force of the table, so the net force is zero.
| Pause And Ponder |
1. A toy car of mass 100 g is moving with a constant velocity of 0.5 m s. What is the net force acting on the toy car?
Ans: Since the toy car is moving with constant velocity, its acceleration is zero. Therefore, the net force acting on it is 0 N.
2. 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.
Ans: A larger force is required for the heavier child because, according to Newton’s Second Law, greater mass requires greater force to produce the same acceleration.
3. How are glass items packed for transportation using a bubble wrap or hay protected from damage?
Ans: Bubble wrap or hay increases the time of impact and absorbs the shock, thereby reducing the force on the glass items and protecting them from breaking.
| Pause And Ponder |
1. Why does a fireperson sometimes struggle when holding the pipe issuing water?
Ans: Water comes out of the pipe with high speed, producing a reaction force on the pipe. This backward force makes it difficult for the fireperson to hold the pipe steadily.
2. 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.
Ans: The spacecraft can change its velocity by firing its engines (thrusters). The expelled gases produce a reaction force that changes the spacecraft’s velocity according to Newton’s Third Law.
| Revise, Reflect, Refine |
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?
Ans: The frictional force exerted by the floor is equal in magnitude to the applied force F, but acts in the opposite direction. According to Newton’s first law of motion, if an object moves with a constant velocity, the net force acting on it must be zero. Therefore, the applied force and the frictional force must completely balance each other.
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.
Ans: 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 on the ball in the direction of its motion, the magnitude of the velocity of the ball will remain the same/ increase/decrease.
Ans: If a net force is applied on the ball in the direction of its motion, the magnitude of the velocity of the ball will increase.
(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.
Ans: 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 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.
Ans: (i) P experiences a net force and Q does not experience a net force.
Explanation: Block P has two unbalanced forces acting on it (4 N and 5 N in opposite directions), resulting in a net force of 1 N in the direction of the 5 N force. Since Block Q is moving with a constant velocity, Newton’s first law states that the net force acting on it must be zero.
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.)
Ans: The net force is 18,000 N in the forward direction.
Explanation: The 95 oarsmen pushing water backward propel the boat forward:
95×200 N=19,000 N (Forward force).
The 5 oarsmen rowing forward push the boat backward:
5×200 N=1,000 N (Backward force).
The net force is the difference between these opposing forces: 19,000 N−1,000 N=18,000 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.
Ans: (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.
Ans: (iii) Object C.
Explanation:
Object A: Straight line → Constant velocity → No net force.
Object B: Horizontal line → At rest → No net force.
Object C: Curved line → Changing velocity (acceleration) → Net force acts.
Object D: Straight line with negative slope → Constant velocity in the opposite direction → No net force.
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.
Ans: Yes, the boat will move backwards (away from the shore). Explanation: According to Newton’s third law of motion, to jump forward, the sailor must push the boat backwards with their feet. Simultaneously, the boat exerts an equal and opposite force on the sailor, pushing them forward to the shore.
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.
Ans: The soft landing mat or sand bed increases the time duration over which the athlete’s high velocity reduces to zero upon impact. By increasing this time, the acceleration is significantly decreased. According to Newton’s second law (F=ma), a smaller acceleration results in a much smaller force exerted on the athlete’s body, thereby minimizing the risk of serious injury.
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.
Ans: (iv) The loaded cart and the empty cart, both exert an equal magnitude of force on each other.
Explanation: Newton’s third law of motion states that whenever one object exerts a force on a second object, the second object simultaneously exerts an equal and opposite force on the first object, regardless of their masses.
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.
Ans: Because the acceleration is produced by a specific “given force”, we assume the applied force is constant. Since Force = mass × acceleration (F=ma), and the force is constant across different masses, the force-mass graph will be a straight horizontal line parallel to the mass axis (X-axis).
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.
Ans: From the velocity-time graph:
Initial velocity, u = 10 m/s
Final velocity, v = 30 m/s
Time taken, t = 8 s
Mass of the object, m = 10 kg
First, calculate the acceleration:
First, calculate the acceleration using:
a = v – u / t
a = 30 – 10 / 8
= 20 /8
= 2.5 m/s2
Now, using Newton’s Second Law of Motion:
F = ma
F = 10 × 2.5
= 25N
Therefore, the force acting on the object is: 25 N
12. 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).
Ans: Given:
m = 50 g = 0.05 kg;
u = 100 m/s
v = 0 m/s (since it stops);
s = 50 cm 0.5 m.
Find acceleration using the kinematic equation:
v2 = u2 + 2as
0 = (100)2 + 2 a (0.5)
0 = 10000 m/s2
Find force using F = ma
F = 0.05 (-10000)
= – 500N
The stopping force is 500 N (the negative sign indicates it opposes motion).
13. 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.
Ans: Given:
v=108 km/h=30 m/s
u=0 m/s
m=0.4 kg
F=800 N
Using F=ma:
800=0.4a
a=2000 m/s2
Using v = u + at
30 = 2000t
t = 30/2000 = 0.015 s
t = 0.015 s
14. 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?
Ans: Total opposing (net) force F=−7 N+(−3 N)=−10 N
Find acceleration (F=ma): −10=2×a⇒a=−5 m/s2
Given u=10 m/s and v=0 m/s.
Find distance (s) using v2 =u2 + 2as: 0
= (10)2 +2(−5)s 0=100−10s⇒10s=100⇒s= 10 m.
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.
Ans: Using F=ma:
F = m1a1
m1 = F/a1
m2 = F/a2
Total mass:
m = m1 + m2 = F/a1 + F/a2
New acceleration:
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.
Ans: Even though the magnetic forces acting on both the compass needle and the bar magnet are completely equal in magnitude, their masses are significantly different. According to Newton’s second law (a=F/m), acceleration is inversely proportional to mass. The compass needle has an extremely small mass, so it experiences a large, noticeable acceleration and moves. The bar magnet has a much larger mass, so the acceleration produced by the exact same force is negligibly small, and its movement is not noticed.
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