Force and Laws of Motion
NCERT Solutions • Class 9 Science • Chapter 8Chapter Exercises
1. An object experiences a net zero external unbalanced force. Is it possible for the object to be travelling with a non-zero velocity? If yes, state the conditions.
Newton’s First Law
Yes, it is possible. According to Newton’s First Law of Motion, if the net unbalanced force is zero, an object in motion will continue to move with a uniform velocity (constant speed in a straight line).
Conditions:
1. The object must already be moving.
2. There must be no change in speed or direction (acceleration is zero).
3. Frictional forces and air resistance must be zero or balanced by an applied force.
Yes, it is possible. According to Newton’s First Law of Motion, if the net unbalanced force is zero, an object in motion will continue to move with a uniform velocity (constant speed in a straight line).
Conditions:
1. The object must already be moving.
2. There must be no change in speed or direction (acceleration is zero).
3. Frictional forces and air resistance must be zero or balanced by an applied force.
2. When a carpet is beaten with a stick, dust comes out of it. Explain.
Inertia of Rest
Initially, both the carpet and the dust particles are at rest. When the carpet is beaten with a stick, the carpet suddenly comes into motion. However, due to the inertia of rest, the dust particles tend to remain stationary. As a result, the dust separates from the moving carpet and falls out.
Initially, both the carpet and the dust particles are at rest. When the carpet is beaten with a stick, the carpet suddenly comes into motion. However, due to the inertia of rest, the dust particles tend to remain stationary. As a result, the dust separates from the moving carpet and falls out.
3. Why is it advised to tie any luggage kept on the roof of a bus with a rope?
This is due to Inertia:
1. Starting: When a stationary bus starts suddenly, the luggage tends to stay at rest (inertia of rest) and may fall backward.
2. Stopping: When a moving bus stops suddenly, the luggage tends to continue moving forward (inertia of motion) and may fall off the roof.
3. Turning: When the bus turns sharply, the luggage tends to continue in a straight line (inertia of direction).
Tying the luggage secures it against these forces.
1. Starting: When a stationary bus starts suddenly, the luggage tends to stay at rest (inertia of rest) and may fall backward.
2. Stopping: When a moving bus stops suddenly, the luggage tends to continue moving forward (inertia of motion) and may fall off the roof.
3. Turning: When the bus turns sharply, the luggage tends to continue in a straight line (inertia of direction).
Tying the luggage secures it against these forces.
4. A batsman hits a cricket ball which then rolls on a level ground. After covering a short distance, the ball comes to rest. The ball slows to a stop because…
(c) there is a force on the ball opposing the motion.
Explanation: This opposing force is the force of friction acting between the ball and the ground, which acts in the direction opposite to the motion of the ball.
Explanation: This opposing force is the force of friction acting between the ball and the ground, which acts in the direction opposite to the motion of the ball.
5. A truck starts from rest and rolls down a hill with a constant acceleration. It travels a distance of 400 m in 20 s. Find its acceleration. Find the force acting on it if its mass is 7 tonnes.
Given: $u = 0$, $s = 400 \text{ m}$, $t = 20 \text{ s}$, $m = 7 \text{ tonnes} = 7000 \text{ kg}$.
Step 1: Find Acceleration ($a$)
Using $s = ut + \frac{1}{2}at^2$:
$400 = 0 + \frac{1}{2} \times a \times (20)^2$
$400 = \frac{1}{2} \times a \times 400 \Rightarrow 400 = 200a$
$a = \frac{400}{200} =$ $2 \text{ m/s}^2$.
Step 2: Find Force ($F$)
$F = ma = 7000 \times 2 =$ $14000 \text{ N}$.
Step 1: Find Acceleration ($a$)
Using $s = ut + \frac{1}{2}at^2$:
$400 = 0 + \frac{1}{2} \times a \times (20)^2$
$400 = \frac{1}{2} \times a \times 400 \Rightarrow 400 = 200a$
$a = \frac{400}{200} =$ $2 \text{ m/s}^2$.
Step 2: Find Force ($F$)
$F = ma = 7000 \times 2 =$ $14000 \text{ N}$.
6. A stone of 1 kg is thrown with a velocity of $20 \text{ m s}^{-1}$ across the frozen surface of a lake and comes to rest after travelling a distance of 50 m. What is the force of friction between the stone and the ice?
Given: $m = 1 \text{ kg}$, $u = 20 \text{ m/s}$, $v = 0$ (comes to rest), $s = 50 \text{ m}$.
Step 1: Find Acceleration ($a$)
Using $v^2 – u^2 = 2as$:
$0^2 – (20)^2 = 2 \times a \times 50$
$-400 = 100a \Rightarrow a = -4 \text{ m/s}^2$.
Step 2: Find Force ($F$)
$F = ma = 1 \times (-4) = -4 \text{ N}$.
The negative sign indicates opposing force.
Force of Friction = 4 N.
Step 1: Find Acceleration ($a$)
Using $v^2 – u^2 = 2as$:
$0^2 – (20)^2 = 2 \times a \times 50$
$-400 = 100a \Rightarrow a = -4 \text{ m/s}^2$.
Step 2: Find Force ($F$)
$F = ma = 1 \times (-4) = -4 \text{ N}$.
The negative sign indicates opposing force.
Force of Friction = 4 N.
7. A 8000 kg engine pulls a train of 5 wagons, each of 2000 kg, along a horizontal track. If the engine exerts a force of 40000 N and the track offers a friction force of 5000 N, then calculate: (a) the net accelerating force and (b) the acceleration of the train.
(a) Net Accelerating Force ($F_{net}$)
$F_{net} = \text{Force Exerted} – \text{Friction Force}$
$F_{net} = 40000 – 5000 =$ $35000 \text{ N}$.
$F_{net} = \text{Force Exerted} – \text{Friction Force}$
$F_{net} = 40000 – 5000 =$ $35000 \text{ N}$.
(b) Acceleration ($a$)
Mass of train (5 wagons) $= 5 \times 2000 = 10000 \text{ kg}$.
$a = \frac{F_{net}}{m} = \frac{35000}{10000} =$ $3.5 \text{ m/s}^2$.
Mass of train (5 wagons) $= 5 \times 2000 = 10000 \text{ kg}$.
$a = \frac{F_{net}}{m} = \frac{35000}{10000} =$ $3.5 \text{ m/s}^2$.
8. An automobile vehicle has a mass of 1500 kg. What must be the force between the vehicle and road if the vehicle is to be stopped with a negative acceleration of $1.7 \text{ m s}^{-2}$?
Given: $m = 1500 \text{ kg}$, $a = -1.7 \text{ m/s}^2$.
Formula: $F = ma$
$F = 1500 \times (-1.7) = -2550 \text{ N}$.
The force must be 2550 N in the direction opposite to motion.
Formula: $F = ma$
$F = 1500 \times (-1.7) = -2550 \text{ N}$.
The force must be 2550 N in the direction opposite to motion.
9. What is the momentum of an object of mass m, moving with a velocity v?
(d) mv
Momentum ($p$) is defined as the product of mass ($m$) and velocity ($v$).
Momentum ($p$) is defined as the product of mass ($m$) and velocity ($v$).
10. Using a horizontal force of 200 N, we intend to move a wooden cabinet across a floor at a constant velocity. What is the friction force that will be exerted on the cabinet?
Since the cabinet moves with constant velocity, the net force acting on it is zero (Acceleration $a = 0$).
Therefore, the Applied Force must be balanced by the Friction Force.
Friction Force = Applied Force = 200 N (in opposite direction).
Therefore, the Applied Force must be balanced by the Friction Force.
Friction Force = Applied Force = 200 N (in opposite direction).
11. According to the third law… A student justifies that the two opposite and equal forces cancel each other. Comment on this logic and explain why the truck does not move.
Logic Comment: The student’s logic is incorrect. Action and Reaction forces act on two different objects (Student pushes Truck, Truck pushes Student). Since they act on different objects, they cannot cancel each other out on a single object.
Why Truck doesn’t move: The truck does not move because the push applied by the student is balanced by the Force of Static Friction between the truck’s tyres and the road. The applied force is insufficient to overcome this static friction.
Why Truck doesn’t move: The truck does not move because the push applied by the student is balanced by the Force of Static Friction between the truck’s tyres and the road. The applied force is insufficient to overcome this static friction.
12. A hockey ball of mass 200 g travelling at $10 \text{ m s}^{-1}$ is struck by a hockey stick so as to return it along its original path with a velocity at $5 \text{ m s}^{-1}$. Calculate the magnitude of change of momentum.
Given: $m = 200 \text{ g} = 0.2 \text{ kg}$.
Initial velocity ($u$) = $10 \text{ m/s}$.
Final velocity ($v$) = $-5 \text{ m/s}$ (return path implies opposite direction).
Change in Momentum ($\Delta p$):
$\Delta p = mv – mu = m(v – u)$
$\Delta p = 0.2(-5 – 10) = 0.2 \times (-15) = -3 \text{ kg m/s}$.
Magnitude of change in momentum = $3 \text{ kg m/s}$.
Initial velocity ($u$) = $10 \text{ m/s}$.
Final velocity ($v$) = $-5 \text{ m/s}$ (return path implies opposite direction).
Change in Momentum ($\Delta p$):
$\Delta p = mv – mu = m(v – u)$
$\Delta p = 0.2(-5 – 10) = 0.2 \times (-15) = -3 \text{ kg m/s}$.
Magnitude of change in momentum = $3 \text{ kg m/s}$.
13. A bullet of mass 10 g travelling horizontally with a velocity of $150 \text{ m s}^{-1}$ strikes a stationary wooden block and comes to rest in 0.03 s. Calculate distance of penetration and force exerted.
Given: $m = 10 \text{ g} = 0.01 \text{ kg}$, $u = 150 \text{ m/s}$, $v = 0$, $t = 0.03 \text{ s}$.
Step 1: Acceleration ($a$)
$a = \frac{v – u}{t} = \frac{0 – 150}{0.03} = -\frac{15000}{3} = -5000 \text{ m/s}^2$.
Step 2: Distance ($s$)
$s = ut + \frac{1}{2}at^2 = (150 \times 0.03) + \frac{1}{2}(-5000)(0.03)^2$
$s = 4.5 – 2500(0.0009) = 4.5 – 2.25 =$ $2.25 \text{ m}$.
Step 3: Force ($F$)
$F = ma = 0.01 \times (-5000) = -50 \text{ N}$.
Magnitude of Force = 50 N.
Step 1: Acceleration ($a$)
$a = \frac{v – u}{t} = \frac{0 – 150}{0.03} = -\frac{15000}{3} = -5000 \text{ m/s}^2$.
Step 2: Distance ($s$)
$s = ut + \frac{1}{2}at^2 = (150 \times 0.03) + \frac{1}{2}(-5000)(0.03)^2$
$s = 4.5 – 2500(0.0009) = 4.5 – 2.25 =$ $2.25 \text{ m}$.
Step 3: Force ($F$)
$F = ma = 0.01 \times (-5000) = -50 \text{ N}$.
Magnitude of Force = 50 N.
14. An object of mass 1 kg travelling… at $10 \text{ m s}^{-1}$ collides with, and sticks to, a stationary wooden block of mass 5 kg. Then they both move off together… Calculate total momentum before and after impact, and velocity of combined object.
Given: $m_1 = 1 \text{ kg}, u_1 = 10 \text{ m/s}$. $m_2 = 5 \text{ kg}, u_2 = 0$.
Total Momentum Before Impact:
$P_{initial} = m_1 u_1 + m_2 u_2 = (1 \times 10) + (5 \times 0) =$ $10 \text{ kg m/s}$.
Total Momentum After Impact:
According to Law of Conservation of Momentum, $P_{final} = P_{initial} =$ $10 \text{ kg m/s}$.
Combined Velocity ($v$):
Total Mass $M = m_1 + m_2 = 1 + 5 = 6 \text{ kg}$.
$P_{final} = M \times v \Rightarrow 10 = 6v$
$v = \frac{10}{6} =$ $1.67 \text{ m/s}$.
Total Momentum Before Impact:
$P_{initial} = m_1 u_1 + m_2 u_2 = (1 \times 10) + (5 \times 0) =$ $10 \text{ kg m/s}$.
Total Momentum After Impact:
According to Law of Conservation of Momentum, $P_{final} = P_{initial} =$ $10 \text{ kg m/s}$.
Combined Velocity ($v$):
Total Mass $M = m_1 + m_2 = 1 + 5 = 6 \text{ kg}$.
$P_{final} = M \times v \Rightarrow 10 = 6v$
$v = \frac{10}{6} =$ $1.67 \text{ m/s}$.
15. An object of mass 100 kg is accelerated uniformly from a velocity of $5 \text{ m s}^{-1}$ to $8 \text{ m s}^{-1}$ in 6 s. Calculate initial and final momentum, and force.
Given: $m = 100 \text{ kg}, u = 5 \text{ m/s}, v = 8 \text{ m/s}, t = 6 \text{ s}$.
Initial Momentum ($P_i$):
$P_i = mu = 100 \times 5 =$ $500 \text{ kg m/s}$.
Final Momentum ($P_f$):
$P_f = mv = 100 \times 8 =$ $800 \text{ kg m/s}$.
Force ($F$):
$F = \frac{\text{Change in Momentum}}{\text{Time}} = \frac{P_f – P_i}{t} = \frac{800 – 500}{6} = \frac{300}{6} =$ $50 \text{ N}$.
Initial Momentum ($P_i$):
$P_i = mu = 100 \times 5 =$ $500 \text{ kg m/s}$.
Final Momentum ($P_f$):
$P_f = mv = 100 \times 8 =$ $800 \text{ kg m/s}$.
Force ($F$):
$F = \frac{\text{Change in Momentum}}{\text{Time}} = \frac{P_f – P_i}{t} = \frac{800 – 500}{6} = \frac{300}{6} =$ $50 \text{ N}$.
16. Akhtar, Kiran and Rahul were riding in a motorcar… an insect hit the windshield… Comment on their suggestions.
Analysis:
• Rahul is correct. According to the Law of Conservation of Momentum, the change in momentum for both objects (insect and car) is equal and opposite. According to the Third Law of Motion, the force exerted is also equal and opposite.
• Kiran’s suggestion is incorrect regarding momentum change (it’s equal), but the insect suffers a greater change in velocity due to its negligible mass.
• Akhtar’s suggestion is incorrect; the forces are equal. The insect dies because the equal force causes massive acceleration/deformation on the tiny insect mass compared to the car.
• Rahul is correct. According to the Law of Conservation of Momentum, the change in momentum for both objects (insect and car) is equal and opposite. According to the Third Law of Motion, the force exerted is also equal and opposite.
• Kiran’s suggestion is incorrect regarding momentum change (it’s equal), but the insect suffers a greater change in velocity due to its negligible mass.
• Akhtar’s suggestion is incorrect; the forces are equal. The insect dies because the equal force causes massive acceleration/deformation on the tiny insect mass compared to the car.
17. How much momentum will a dumb-bell of mass 10 kg transfer to the floor if it falls from a height of 80 cm? Take its downward acceleration to be $10 \text{ m s}^{-2}$.
Given: $m = 10 \text{ kg}, s = 80 \text{ cm} = 0.8 \text{ m}, a = 10 \text{ m/s}^2, u = 0$.
Step 1: Final Velocity ($v$)
$v^2 – u^2 = 2as \Rightarrow v^2 – 0 = 2 \times 10 \times 0.8 = 16$.
$v = \sqrt{16} = 4 \text{ m/s}$.
Step 2: Momentum Transferred
Momentum just before hitting floor = $mv = 10 \times 4 =$ $40 \text{ kg m/s}$.
Step 1: Final Velocity ($v$)
$v^2 – u^2 = 2as \Rightarrow v^2 – 0 = 2 \times 10 \times 0.8 = 16$.
$v = \sqrt{16} = 4 \text{ m/s}$.
Step 2: Momentum Transferred
Momentum just before hitting floor = $mv = 10 \times 4 =$ $40 \text{ kg m/s}$.
Additional Exercises
A1. The following is the distance-time table of an object in motion… (a) Conclusion about acceleration? (b) Inference about forces?
(a) Acceleration
Distances are $0, 1, 8, 27, 64…$ which corresponds to $t^3$.
If $s \propto t^3$, then velocity $v \propto t^2$, and acceleration $a \propto t$.
Since acceleration depends on time ($t$), it is increasing (non-uniform).
Distances are $0, 1, 8, 27, 64…$ which corresponds to $t^3$.
If $s \propto t^3$, then velocity $v \propto t^2$, and acceleration $a \propto t$.
Since acceleration depends on time ($t$), it is increasing (non-uniform).
(b) Forces
Since acceleration is non-zero and increasing, the net unbalanced force acting on the object is also increasing with time ($F = ma$).
Since acceleration is non-zero and increasing, the net unbalanced force acting on the object is also increasing with time ($F = ma$).
A2. Two persons push a motorcar of mass 1200 kg at uniform velocity… Three persons produce acceleration of $0.2 \text{ m s}^{-2}$. With what force does each person push?
Let force of one person be $F$ and Friction be $f$.
Case 1 (Uniform Velocity): 2 Persons push $\Rightarrow 2F$. Since velocity is uniform ($a=0$), Applied Force = Friction.
$\therefore f = 2F$.
Case 2 (Acceleration): 3 Persons push $\Rightarrow 3F$.
Net Force = $3F – f = ma$.
Substitute $f = 2F$:
$3F – 2F = 1200 \times 0.2$
$F = 240 \text{ N}$.
Each person pushes with 240 N.
Case 1 (Uniform Velocity): 2 Persons push $\Rightarrow 2F$. Since velocity is uniform ($a=0$), Applied Force = Friction.
$\therefore f = 2F$.
Case 2 (Acceleration): 3 Persons push $\Rightarrow 3F$.
Net Force = $3F – f = ma$.
Substitute $f = 2F$:
$3F – 2F = 1200 \times 0.2$
$F = 240 \text{ N}$.
Each person pushes with 240 N.
A3. A hammer of mass 500 g, moving at $50 \text{ m s}^{-1}$, strikes a nail. The nail stops the hammer in 0.01 s. What is the force of the nail on the hammer?
Given: $m = 500 \text{ g} = 0.5 \text{ kg}, u = 50 \text{ m/s}, v = 0, t = 0.01 \text{ s}$.
Force ($F$):
$F = \frac{m(v – u)}{t} = \frac{0.5(0 – 50)}{0.01} = \frac{-25}{0.01} =$ $-2500 \text{ N}$.
The nail exerts a force of 2500 N opposing the hammer’s motion.
Force ($F$):
$F = \frac{m(v – u)}{t} = \frac{0.5(0 – 50)}{0.01} = \frac{-25}{0.01} =$ $-2500 \text{ N}$.
The nail exerts a force of 2500 N opposing the hammer’s motion.
A4. A motorcar of mass 1200 kg is moving… velocity $90 \text{ km/h}$. Slowed down to $18 \text{ km/h}$ in 4 s… Calculate acceleration, change in momentum, and force.
Given: $m=1200 \text{ kg}, u=90 \text{ km/h} = 25 \text{ m/s}, v=18 \text{ km/h} = 5 \text{ m/s}, t=4 \text{ s}$.
Acceleration
$a = \frac{v-u}{t} = \frac{5-25}{4} = -\frac{20}{4} =$ $-5 \text{ m/s}^2$.
Change in Momentum
$\Delta p = m(v-u) = 1200(5-25) = 1200(-20) =$ $-24000 \text{ kg m/s}$.
Force
$F = ma = 1200 \times (-5) =$ $-6000 \text{ N}$.
Acceleration
$a = \frac{v-u}{t} = \frac{5-25}{4} = -\frac{20}{4} =$ $-5 \text{ m/s}^2$.
Change in Momentum
$\Delta p = m(v-u) = 1200(5-25) = 1200(-20) =$ $-24000 \text{ kg m/s}$.
Force
$F = ma = 1200 \times (-5) =$ $-6000 \text{ N}$.