Class 9 Science NCERT Solution chapter 6 How Forces Affect Motion

Revise, Reflect, Refine — Complete Solutions

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🟦 QUESTION 1

Using a horizontal force FFF, 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

Since the table moves with constant velocity, its acceleration is zero.

According to Newton’s second law:$$F_{\text{net}}=ma$$Fnet​=ma$$F_{\text{net}}=m\times0=0$$Fnet​=m×0=0

Therefore, the applied force and frictional force are balanced.$$\boxed{\text{Frictional force}=F}$$Frictional force=F​

The frictional force has the same magnitude as the applied force but acts in the opposite direction.

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🟦 QUESTION 2

For a ball moving on a smooth frictionless surface, choose the appropriate option that makes 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

(i) The velocity will remain the same.

If no net force acts on the ball, its acceleration is zero. Therefore, according to Newton’s first law, it continues moving with constant velocity.

(ii) The magnitude of velocity will increase.

A net force acting in the direction of motion produces acceleration in the same direction. Hence, the ball speeds up.

(iii) The magnitude of velocity will decrease.

A net force opposite to the direction of motion produces acceleration opposite to the velocity. Therefore, the ball slows down.

If the opposite force continues even after the ball stops, the ball may begin moving in the opposite direction.

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🟦 QUESTION 3

Two blocks P and Q on a smooth horizontal surface are shown in Fig. 6.36. Two forces of magnitudes 4 N and 5 N act in opposite directions on block P, while block Q moves with 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.

🟩 ANSWER

Correct option: (i) P experiences a net force and Q does not experience a net force.

For block P:$$F_{\text{net}}=5-4=1\text{ N}$$Fnet​=5−4=1 N

Therefore, block P experiences a net force of 1 N in the direction of the 5 N force.

Block Q is moving with constant velocity. Therefore, its acceleration is zero and the net force acting on it is also zero.

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🟦 QUESTION 4

While practising for the snake boat race, 100 oarsmen are rowing a boat together. Out of these, 95 row backwards to propel the boat forward. 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 and air friction.

🟩 ANSWER

Force produced by 95 oarsmen in the forward direction:$$F_1=95\times200$$F1​=95×200$$F_1=19000\text{ N}$$F1​=19000 N

Force produced by 5 oarsmen in the opposite direction:$$F_2=5\times200$$F2​=5×200$$F_2=1000\text{ N}$$F2​=1000 N

Net force:$$F_{\text{net}}=F_1-F_2$$Fnet​=F1​−F2​$$F_{\text{net}}=19000-1000$$Fnet​=19000−1000$$F_{\text{net}}=18000\text{ N}$$Fnet​=18000 N

Final Answer:$$\boxed{\text{Net force}=18000\text{ N in the forward direction}}$$Net force=18000 N in the forward direction​

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🟦 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

Correct option: (iv)

According to Newton’s second law:$$a=\frac{F}{m}$$a=mF​

The acceleration of an object:

• Occurs in the direction of the net force
• Is directly proportional to the net force
• Is inversely proportional to the mass of the object

Therefore, for a fixed mass, a greater force produces greater acceleration.

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🟦 QUESTION 6

The position-time graphs of four objects A, B, C and D moving along a straight line are shown in Fig. 6.37. A net force acts on which object?

(i) Object A
(ii) Object B
(iii) Object C
(iv) Object D

🟩 ANSWER

Correct option: (iii) Object C

The slope of a position-time graph represents velocity.

• Object A: Straight line with a constant positive slope; therefore, it moves with constant velocity and experiences no net force.
• Object B: Horizontal line; therefore, it remains at rest and experiences no net force.
• Object C: Curved graph with an increasing slope; therefore, its velocity changes and it accelerates.
• Object D: Straight line with a constant negative slope; therefore, it moves with constant velocity in the opposite direction and experiences no net force.

Since only Object C is accelerating, a net force acts on it.

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🟦 QUESTION 7

A sailor jumps out from a small boat towards the shore. As the sailor jumps forward, will the boat move? If yes, in which direction and why?

🟩 ANSWER

Yes, the boat will move backwards, away from the shore.

When the sailor jumps, they push the boat backwards. In response, the boat exerts an equal and opposite force on the sailor, pushing the sailor forward.

According to Newton’s third law:

For every action, there is an equal and opposite reaction.

Therefore:

• Action: Sailor pushes the boat backwards.
• Reaction: Boat pushes the sailor forwards.

The boat moves backwards because it receives a force opposite to the sailor’s jump.

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🟦 QUESTION 8

During a high-jump event, a landing mat or sand bed is placed for the athlete to fall upon. Explain the reason behind it.

🟩 ANSWER

When an athlete lands, their velocity must decrease to zero.

A landing mat or sand bed increases the time taken for the athlete to come to rest.

Acceleration is:$$a=\frac{v-u}{t}$$a=tv−u​

For the same change in velocity, increasing the stopping time decreases the magnitude of acceleration.

According to:$$F=ma$$F=ma

A smaller acceleration produces a smaller force on the athlete’s body.

Therefore, the mat or sand bed:

• Increases the stopping time
• Reduces the impact force
• Prevents serious injuries

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🟦 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 forces of equal magnitude on each other.

🟩 ANSWER

Correct option: (iv)

According to Newton’s third law, when two objects interact, they exert forces on each other that are:

• Equal in magnitude
• Opposite in direction
• Acting on different objects

Therefore, the loaded cart and empty cart exert equal and opposite forces on each other.

However, their accelerations are different because:$$a=\frac{F}{m}$$a=mF​

The empty cart has less mass, so it experiences greater acceleration.

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🟦 QUESTION 10

The acceleration-mass graph for the acceleration produced by a force on objects of different masses is shown in Fig. 6.40. Plot the force-mass graph for this case.

🟩 ANSWER

From the graph, the approximate values are:

Mass (kg)	Acceleration (m s⁻²)	Force F=maF=maF=ma
1	10	$1\times10=10\text{ N}$1×10=10 N
2	5	$2\times5=10\text{ N}$2×5=10 N
4	2.5	$4\times2.5=10\text{ N}$4×2.5=10 N
5	2	$5\times2=10\text{ N}$5×2=10 N

The force remains constant at 10 N for all masses.

Therefore, the force-mass graph will be a horizontal straight line parallel to the mass axis at 10 N.

Points to plot:$$(1,10),\ (2,10),\ (4,10),\ (5,10)$$(1,10), (2,10), (4,10), (5,10)

This shows that the same force produces less acceleration when the mass is increased.

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🟦 QUESTION 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 using the graph.

🟩 ANSWER

From the graph:$$u=10\text{ m s}^{-1}$$u=10 m s−1$$v=30\text{ m s}^{-1}$$v=30 m s−1$$t=8\text{ s}$$t=8 s

Acceleration is equal to the slope of the velocity-time graph:$$a=\frac{v-u}{t}$$a=tv−u​$$a=\frac{30-10}{8}$$a=830−10​$$a=\frac{20}{8}$$a=820​$$a=2.5\text{ m s}^{-2}$$a=2.5 m s−2

Mass of the object:$$m=10\text{ kg}$$m=10 kg

Using Newton’s second law:$$F=ma$$F=ma$$F=10\times2.5$$F=10×2.5$$F=25\text{ N}$$F=25 N

Final Answer:$$\boxed{\text{Force acting on the object}=25\text{ N}}$$Force acting on the object=25 N​

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🟦 QUESTION 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 inside the block.

🟩 ANSWER

Given:$$m=50\text{ g}=0.05\text{ kg}$$m=50 g=0.05 kg$$u=100\text{ m s}^{-1}$$u=100 m s−1$$v=0$$v=0$$s=50\text{ cm}=0.5\text{ m}$$s=50 cm=0.5 m

Using:$$v^2=u^2+2as$$v2=u2+2as$$0=100^2+2(a)(0.5)$$0=1002+2(a)(0.5)$$0=10000+a$$0=10000+a$$a=-10000\text{ m s}^{-2}$$a=−10000 m s−2

Using:$$F=ma$$F=ma$$F=0.05\times(-10000)$$F=0.05×(−10000)$$F=-500\text{ N}$$F=−500 N

The negative sign shows that the force acts opposite to the bullet’s motion.

Final Answer:$$\boxed{\text{Stopping force}=500\text{ N opposite to the motion}}$$Stopping force=500 N opposite to the motion​

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🟦 QUESTION 13

A footballer converted a penalty shot by kicking the football with a speed of 108 km h⁻¹. The estimated force imparted was 800 N. The mass of the football was 0.4 kg. Calculate the time of contact between the footballer’s foot and the ball.

🟩 ANSWER

Convert the speed into m s⁻¹:$$v=108\times\frac{5}{18}$$v=108×185​$$v=30\text{ m s}^{-1}$$v=30 m s−1

Assuming that the football was initially at rest:$$u=0$$u=0

Mass:$$m=0.4\text{ kg}$$m=0.4 kg

Force:$$F=800\text{ N}$$F=800 N

Using:$$F=ma$$F=ma$$a=\frac{F}{m}$$a=mF​$$a=\frac{800}{0.4}$$a=0.4800​$$a=2000\text{ m s}^{-2}$$a=2000 m s−2

Using:$$v=u+at$$v=u+at$$30=0+2000t$$30=0+2000t$$t=\frac{30}{2000}$$t=200030​$$t=0.015\text{ s}$$t=0.015 s

Final Answer:$$\boxed{\text{Time of contact}=0.015\text{ s}}$$Time of contact=0.015 s​

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🟦 QUESTION 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. After entering the rough patch, how much distance does the object travel before coming to rest?

🟩 ANSWER

Frictional force opposing motion:$$F_1=7\text{ N}$$F1​=7 N

Additional opposing force:$$F_2=3\text{ N}$$F2​=3 N

Total opposing force:$$F=7+3=10\text{ N}$$F=7+3=10 N

Mass:$$m=2\text{ kg}$$m=2 kg

Acceleration:$$a=\frac{F}{m}$$a=mF​

Since the force opposes motion:$$a=-\frac{10}{2}$$a=−210​$$a=-5\text{ m s}^{-2}$$a=−5 m s−2

Given:$$u=10\text{ m s}^{-1}$$u=10 m s−1$$v=0$$v=0

Using:$$v^2=u^2+2as$$v2=u2+2as$$0=10^2+2(-5)s$$0=102+2(−5)s$$0=100-10s$$0=100−10s$$10s=100$$10s=100$$s=10\text{ m}$$s=10 m

Final Answer:$$\boxed{\text{Distance travelled before stopping}=10\text{ m}}$$Distance travelled before stopping=10 m​

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🟦 QUESTION 15

A tractor pulls a harrow of mass m1m_1m1​ with a net force FFF, resulting in an acceleration a1a_1a1​. The same tractor pulls a trolley of mass m2m_2m2​ with the same force FFF, producing an acceleration a2a_2a2​. If the tractor now pulls the trolley with the harrow placed on it using the same force FFF, obtain an expression for the resulting acceleration in terms of a1a_1a1​ and a2a_2a2​. Ignore friction.

🟩 ANSWER

For the harrow:$$F=m_1a_1$$F=m1​a1​

Therefore:$$m_1=\frac{F}{a_1}$$m1​=a1​F​

For the trolley:$$F=m_2a_2$$F=m2​a2​

Therefore:$$m_2=\frac{F}{a_2}$$m2​=a2​F​

When the harrow is placed on the trolley, total mass is:$$m=m_1+m_2$$m=m1​+m2​$$m=\frac{F}{a_1}+\frac{F}{a_2}$$m=a1​F​+a2​F​$$m=F\left(\frac{1}{a_1}+\frac{1}{a_2}\right)$$m=F(a1​1​+a2​1​)$$m=F\left(\frac{a_1+a_2}{a_1a_2}\right)$$m=F(a1​a2​a1​+a2​​)

Let the combined acceleration be $a$a.$$a=\frac{F}{m}$$a=mF​$$a= \frac{F} {F\left(\frac{a_1+a_2}{a_1a_2}\right)}$$a=F(a1​a2​a1​+a2​​)F​$$\boxed{a=\frac{a_1a_2}{a_1+a_2}}$$a=a1​+a2​a1​a2​​​

Final Answer:$$\boxed{\text{Resulting acceleration}=\frac{a_1a_2}{a_1+a_2}}$$Resulting acceleration=a1​+a2​a1​a2​​​

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🟦 QUESTION 16

When the pole of a bar magnet is brought close to a magnetic compass, the bar magnet and compass needle exert magnetic forces on each other. According to 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. Explain why.

🟩 ANSWER

The bar magnet and compass needle exert equal and opposite forces on each other, as stated by Newton’s third law.

However, equal forces do not always produce equal accelerations.

According to Newton’s second law:$$a=\frac{F}{m}$$a=mF​

The compass needle:

• Has a very small mass
• Is freely mounted and can rotate easily
• Experiences very little resistance at its pivot

Therefore, the magnetic force produces a noticeable acceleration and the needle moves.

The bar magnet:

• Has a much greater mass
• May be held or supported on a surface
• Experiences friction or external support

Therefore, the same force produces an extremely small or unnoticeable acceleration in the magnet.

Thus, the forces are equal, but the accelerations are different because the masses and conditions of motion are different.