CBSE Class 8 Science Force and Pressure Chapter Notes

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Force & Pressure

Force and pressure explain why things move, stop, speed up, slow down or change shape. This CBSE Class 8 Science chapter notes page on Force and Pressure covers the effects and types of force, balanced and unbalanced forces, the three equations of uniformly accelerated motion with their derivations, Newton's laws of motion, inertia, linear momentum, impulse, the conservation of momentum, pressure and thrust, density, relative density, buoyancy and Archimedes' principle, followed by the worked examples and the chapter exercise with answers.

Topics Covered

1. Force and its effects
2. Balanced and unbalanced forces, resultant force
3. Classification of forces
4. Equations of uniformly accelerated motion
5. Laws of motion and inertia
6. Linear momentum and Newton's second law of motion
7. Applications of Newton's second law of motion
8. Impulse
9. Newton's third law of motion
10. Principle of conservation of momentum
11. Pressure, thrust, density and buoyancy

FORCE AND ITS EFFECTS

The word force is associated with either a pull or a push, which causes a change in the state of a body. A force cannot be seen. A force can be judged only by the effects it produces on objects around us.

Effects of force:
i) A force can move a stationary object.
ii) A force can stop a moving object.
iii) A force can change the speed of a moving object.
iv) A force can change the direction of a moving object.
v) A force can change the shape (or size) of an object.

Thus, force is that agent which can change the state of rest or of uniform motion, or the shape, or the direction of motion of a body. The S.I. unit of force is the newton (N). Force is a vector quantity, which means it has both magnitude and direction.

Balanced and unbalanced forces
a) Balanced forces cannot change the speed of a body, although they can change the shape of the object. If the resultant of the forces acting on a body is zero, the forces are said to be balanced.
b) Unbalanced forces can cause a change in the speed or the direction of motion of a body. If the resultant of the forces acting on the body is not zero, the forces are said to be unbalanced.

Resultant force: When two or more forces act on a body simultaneously, the single force which produces the same effect as all the forces acting together is known as the resultant force.

Classification of forces: Forces may be classified into two classes.

a) Contact forces: Contact forces arise as a result of physical contact between two bodies, one which exerts the force and the other on which the force is exerted. Examples are:
i) Muscular force: The force exerted by the muscles of our body is called muscular force. The force used to lift a bag or a water bottle is an example of muscular force.
ii) Frictional force: A force that arises when one body moves over the surface of another body, and is directed opposite to the motion, is called frictional force. Friction is a force that slows things down or prevents them from moving.

b) Non-contact forces: Non-contact forces, or action-at-a-distance, are forces which do not involve any physical contact between two bodies but act through the space between them. Examples are:
i) Gravitational force: The force of attraction between two objects because of their mass is called gravitational force.
ii) Magnetic force: A magnet attracts iron filings, nails and other objects made of iron, steel, nickel and cobalt. Magnets exert forces of attraction and repulsion on other magnets. Magnetic force can act from a distance, so it is a non-contact force.
iii) Electrostatic force: When a plastic comb is rubbed with silk, it can pick up small bits of paper. This is because the comb acquires an electric charge, and it can exert a force called electrostatic force. Electrostatic force can also act from a distance, so it is a non-contact force. A body with an electrostatic charge can either attract or repel another charged body. Electrostatic force is used to separate solid pollutant particles from the smoke given out by factories.

[Figure: A plastic comb held near small bits of paper after being rubbed with silk, with the paper bits attracted to the comb]

EQUATIONS OF UNIFORMLY ACCELERATED MOTION

There are three equations for the motion of bodies which travel with a uniform acceleration. These equations give the relationship between initial velocity, final velocity, time taken, acceleration and distance travelled by the bodies.

1. First equation of motion: The first equation of motion is v = u + at. It gives the velocity acquired by a body in time t.

Consider a body having initial velocity u. Suppose it is subjected to a uniform acceleration a, so that after time t its final velocity becomes v. From the definition of acceleration:
Acceleration = (Final velocity - Initial velocity) / Time taken
So, a = (v - u) / t
Therefore, at = v - u, and v = u + at.
where v = final velocity of the body, u = initial velocity of the body, a = acceleration and t = time taken.

This equation has four values in it. If any three values are known, the fourth can be calculated. By paying due attention to the sign of the acceleration, the equation can also be applied to problems of retardation.

2. Second equation of motion: The second equation of motion is s = ut + ½at². It gives the distance travelled by a body in time t.

Suppose a body has initial velocity u and a uniform acceleration a for time t, so that its final velocity becomes v. The distance travelled in this time can be found from the average velocity:
Average velocity = (Initial velocity + Final velocity) / 2 = (u + v) / 2
Distance travelled = Average velocity × Time
So, s = \( \frac{(u + v)}{2} \) × t ... (1)
From the first equation of motion, v = u + at. Putting this value of v in equation (1):
s = \( \frac{(u + u + at)}{2} \) × t = \( \frac{(2u + at)}{2} \) × t
So, s = \( \frac{2ut + at^2}{2} \) = ut + \( \frac{1}{2} \)at²
where s = distance travelled by the body in time t, u = initial velocity of the body and a = acceleration.

3. Third equation of motion: The third equation of motion is v² = u² + 2as. It gives the velocity acquired by a body in travelling a distance s.

It is obtained by eliminating t from the first two equations. From the second equation, s = ut + ½at² ... (1). From the first equation, at = v - u, so t = (v - u) / a ... (2). Putting the value of t from (2) into (1), we get:
s = \( \frac{u(v-u)}{a} + \frac{1}{2}a\left(\frac{v-u}{a}\right)^2 \)
or s = \( \frac{uv-u^2}{a} + \frac{a(v^2+u^2-2uv)}{2a^2} \) [because \( (v-u)^2 = v^2+u^2-2vu \)]
or s = \( \frac{uv-u^2}{a} + \frac{v^2+u^2-2uv}{2a} \)
or s = \( \frac{2uv-2u^2+v^2+u^2-2uv}{2a} \)
or s = \( \frac{v^2-u^2}{2a} \), which on simplifying gives:
s = \( \frac{(v^2 - u^2)}{2a} \)
So, v² = u² + 2as
where v = final velocity, u = initial velocity, a = acceleration and s = distance travelled.

LAWS OF MOTION AND INERTIA

Newton's first law of motion: According to Newton's first law of motion, a body continues to be in a state of rest, or in a state of uniform motion along a straight line, unless an external unbalanced force is applied on the body.

Inertia: The inherent property of a body, by virtue of which it is not able to change by itself its state of rest, its state of uniform motion, or its direction of motion, is called inertia of the body.

Inertia of rest: It is the property by virtue of which a body at rest cannot start moving on its own. The body at rest opposes any force that tries to move it.
Example: When a carpet is suddenly jerked, the dust flies off. The carpet comes into motion, but the dust particles tend to remain at rest due to inertia of rest, and hence fall off.

Inertia of motion: It is the property by virtue of which a body in motion cannot stop on its own. The body in motion opposes any force that tries to stop it.
Example: When a running car stops suddenly, the passenger is jerked forward. In a running car, the whole body of the passenger is in a state of motion. When the car stops suddenly, the lower part of the body, in contact with the car, comes to rest, but the upper part remains in motion due to inertia of motion. Thus the passenger is jerked forward.

Inertia of direction: It is the property by virtue of which a body moving along a particular straight line cannot change its direction of motion by itself. The body opposes any force that tries to change its direction of motion.
Example: The mud guard over the wheels of a vehicle is based on inertia of direction.

Note: The mass of a body is a measure of its inertia. The heavier the body, the greater its inertia.

LINEAR MOMENTUM AND NEWTON'S SECOND LAW OF MOTION

Whenever we examine a moving object, we must consider both its mass and its velocity, because the quantity of motion possessed by a body depends on these two factors.

1. A cricket ball is heavier than a tennis ball. If both are thrown with the same speed, more force is required to stop the cricket ball (which has more mass), and less force is required to stop the tennis ball. So, if two bodies of different masses move with the same velocity, the force needed to stop the heavier body is more than that needed for the lighter body.
2. If two cricket balls of the same mass are thrown with different speeds, more force is required to stop the ball moving with the higher velocity, and less force is required to stop the ball moving with the lower velocity. So, the force needed to stop a moving body is directly proportional to its mass and its velocity. This gives us the term known as momentum.

Linear momentum: The linear momentum of a body of mass m travelling with velocity v is defined as the product of its mass and velocity. It is the quantity of motion contained in a body.
Linear momentum, p = mv

Mass is a scalar quantity and velocity is a vector quantity, so their product, momentum, is a vector quantity. The direction of momentum is the direction of velocity.

S.I. unit of linear momentum: The S.I. unit of mass is kg and the S.I. unit of velocity is m/s. So the S.I. unit of linear momentum is kilogram metre per second, written as kg m s^-1 or kg m/s.

Newton's second law of motion: According to Newton's second law of motion, the rate of change of linear momentum of a body is directly proportional to the applied unbalanced force, and the change takes place in the same direction in which the force is applied.

Mathematical formulation of the second law of motion: Consider a body of mass m having an initial velocity u. Suppose a force F acts on this body for time t, and its final velocity becomes v.
Initial linear momentum of the body, Pi = mu
Final linear momentum of the body, Pf = mv
Change in linear momentum = Pf - Pi = mv - mu = m(v - u)
Rate of change of linear momentum = Change in linear momentum / Time taken = m(v - u) / t ... (i)
Since v = u + at, we have (v - u) / t = a ... (ii)
Putting the value from (ii) in (i), the rate of change of linear momentum = ma.
According to Newton's second law, F ∝ rate of change of linear momentum, so F ∝ ma.
Introducing a constant k, F = k × m × a ... (iii)
The S.I. unit of force is the newton (N). One newton is that force which, acting on a mass of 1 kg, produces in it an acceleration of 1 m/s² in its own direction.
So, when m = 1 kg and a = 1 m/s², F = 1 N. Substituting in (iii): 1 N = k × 1 kg × 1 m/s², so k = 1.
Therefore, F = ma. This equation is the mathematical statement of Newton's second law of motion.

The second law gives us the method to measure force. If the mass of a body and the acceleration it produces are known, the force acting on it can be measured.

APPLICATIONS OF NEWTON'S SECOND LAW OF MOTION

Many everyday experiences are applications of Newton's second law. We either increase or decrease the rate of change of momentum by changing the time in which the change in momentum takes place. The force is directly proportional to the rate of change of momentum, so a greater rate of change of momentum means a greater impact force.

1. A karate player breaks a pile of tiles or a slab of ice in a single blow: The karate player moves the hand very fast, which decreases the time of impact and increases the rate of change of momentum. The momentum of the hand is reduced to zero in a very short time, so the force delivered on the tiles or the ice is large enough to break it.
2. In cricket, a fielder moves the arms backwards in the direction of the ball while taking a catch: The fielder increases the time of impact and so decreases the rate of change of momentum. The momentum of the ball is reduced to zero over a long time, so the force on the hands is small and the hands are not hurt. If the ball were stopped suddenly, the rate of change of momentum would be large, and the fielder's palm might be hurt.
3. When a car goes out of control, the driver prefers to hit something soft (such as a hay stack) rather than something hard (such as a concrete wall): A soft object increases the time of stopping, so the rate of change of momentum is small and the reaction force on the car is small. A concrete wall gives a very large reaction force, which may cause a serious accident.
4. Athletes performing the high jump are made to fall on a cushioned bed or on a sand bed: The cushioned bed or sand increases the time over which the body comes to rest. The rate of change of momentum is small, so the impact of the reaction force is small and the athlete is not injured.
5. Shock absorbers are provided in vehicles: They increase the time interval of a jerk, and hence reduce the impact force during the jerks.
6. Athletes are advised to come to a stop slowly after finishing a fast run: Slowing down over a longer time interval reduces the impact force and helps avoid injury.
7. Car passengers are advised to wear seat belts: During a sudden stop, the seat belt stops the passenger from falling forward suddenly. It increases the time over which the passenger's momentum is reduced to zero, so the impact force is small and injury is prevented.
8. China and glassware are packed in soft material for transportation: The soft packing material slows down the rate of change of momentum during jerks. The impact force is reduced, and the items are not broken.
9. Springs are provided in car seats: The springs compress when a passenger sits down suddenly. The compression increases the time for the passenger to come to rest on the seat. Since the rate of change of momentum is small, the passenger applies less force on the seat, and there is no chance of injury.

NEWTON'S FIRST LAW IS CONTAINED IN THE SECOND LAW OF MOTION

Newton's first law of motion is contained in Newton's second law of motion. According to the second law, F = m × a ... (i).

a) If no external force acts on an object at rest, F = 0 ... (ii). From (i) and (ii), m × a = 0. Since m is not zero, a = 0. Thus, a body will not be accelerated, and a body at rest will continue to be at rest if no external force acts on it.

b) If no external force acts on an object in motion, then F = rate of change of momentum = (mv - mu) / t. If F = 0, then (mv - mu) / t = 0, so mv - mu = 0, which gives v = u (since m is not zero). This means the initial and final velocities are equal, and the body continues in uniform motion along a straight line.

From these two points, we conclude that a body continues to be in its state of rest, or of uniform motion in a straight line, if no external force acts on it. This is Newton's first law of motion. Thus, Newton's first law is contained in the second law of motion.

IMPULSE

Forces which act on bodies for a short time are called impulsive forces. Examples are the forces involved in hitting a ball with a bat, firing a gun, and driving a nail into a wooden block with a hammer.

An impulsive force does not remain constant. It changes first from zero to a maximum, and then from the maximum back to zero. So its value is not easy to measure, because it changes with time. In such cases we measure the total effect of the force, called impulse. Impulse is a measure of the total effect of a force applied for a short time. It is given by the product of the average force and the time for which the force acts on the body.
Impulse = Average force × Time, that is, I = F × t

From Newton's second law, F = Change in linear momentum / Time taken. So F × t = Change in linear momentum, and therefore:
Impulse = Change in linear momentum.
So, impulse is also defined as the change in linear momentum of a body.

Units of impulse: Impulse = force × time, so the unit of impulse is the unit of force × the unit of time. In the S.I. system, the unit of impulse is N × s = Ns. Since impulse equals the change in momentum, its unit is also the unit of momentum, that is, kg m/s. So the S.I. unit of impulse is newton second (Ns), or kilogram metre per second (kg m/s).

NEWTON'S THIRD LAW OF MOTION

Newton's first law gives the definition of force. Newton's second law gives a method of measuring force. Newton's third law describes the relationship between the forces of interaction between two objects.

According to Newton's third law of motion, to every action there is an equal and opposite reaction. Action and reaction forces act on different bodies.

In other words, whenever two bodies interact, the force exerted by the first body on the second (called action) is equal and opposite to the force exerted by the second body on the first (called reaction).

For example, a block placed on the ground exerts a downward force equal to its weight W (mg) on the ground, due to the attraction of the Earth. This is the action. The ground exerts an equal and opposite force W (mg) on the block in the upward direction. This is the reaction.

[Figure: A block resting on the ground, with a downward arrow labelled action (weight W = mg, exerted on the ground) and an upward arrow labelled reaction (equal force exerted by the ground on the block)]

Important points about action and reaction:
1. Both action and reaction are forces (in this case, F = mg).
2. Action and reaction act simultaneously, but on different bodies. In this case, the block and the ground are two different bodies.
3. The magnitude of the force F_AB applied by body A on body B is equal to the magnitude of the force F_BA applied by body B on body A, but they are opposite in direction. In this case, the force of the block on the earth is equal in magnitude to the force of the earth on the block, but one acts downward and the other acts upward.
4. Action and reaction act on different bodies, so they cannot cancel each other.
5. Action and reaction forces always occur in pairs. If the block were not placed on the ground, no reaction would develop.
6. An action-reaction pair must satisfy the "A on B and B on A" requirement. Two equal and opposite forces do not necessarily form an action-reaction pair.

Examples of Newton's third law of motion:

1. A gun recoils when a shot is fired: The bullet goes out due to the force applied on it through the trigger (action). According to Newton's third law, the gun recoils backwards due to the reaction acting on it in the opposite direction. This gives a backward jerk to the shoulder of the person shooting.
2. Jet aeroplanes and rockets: In jet engines and rockets, fuel is burnt to produce a large quantity of hot gases. These gases come out of a nozzle (a fine opening) with great force (action). The equal and opposite reaction pushes the jet aeroplane or rocket upward with great speed (reaction).
3. A person swimming: A swimmer pushes the water backwards with the hands (action). As the reaction, the water pushes the person forward with an equal force.
4. How we walk: When we walk, our foot pushes the ground backwards (action). In return, the ground pushes our foot forward with an equal and opposite force (reaction). This forward reaction makes us walk forward.
5. Rowing of a boat: The boatman pushes the water backwards with the oars (action). The water exerts an equal and opposite push on the boat, which moves forward (reaction). The harder the boatman pushes the water backwards, the greater the reaction force, and the faster the boat moves forward.
6. The boatman ties the boat before passengers disembark: When the passengers start getting out of the boat, they push the boat backwards with their feet (action). The boat exerts an equal and opposite force on the passengers in the forward direction, which helps them move forward (reaction). Since the boat is in water, it moves backwards due to the reaction, so it is tied to keep it in place.
7. A rubber ball rebounds when struck against a hard floor: The ball exerts a force on the floor (action). The floor exerts an equal and opposite force on the ball (reaction). The rubber ball, being light, rebounds.
8. We feel pain in the hand while pushing a wall: When we push a wall with the palm (action), the wall exerts an equal and opposite force on the palm (reaction), so we feel pain in our hand.

PRINCIPLE OF CONSERVATION OF MOMENTUM

Suppose two balls A and B of masses m_A and m_B, travelling in the same direction along a straight line with velocities u_A and u_B respectively. Let u_A be greater than u_B. The two balls collide, and the collision lasts for time t. During the collision, ball A exerts a force F_AB on ball B, and ball B exerts a force F_BA on ball A. Let v_A and v_B be the velocities of the two balls after the collision.

[Figure: Two balls A and B moving in the same direction before collision with masses m_A and m_B and velocities u_A and u_B, with forces F_AB and F_BA acting between them; after collision the balls move with velocities v_A and v_B]

The momentum of ball A before and after the collision is m_Au_A and m_Av_A respectively. The rate of change of linear momentum of ball A (force F_AB) during the collision is m_A(v_A - u_A) / t.
Similarly, the rate of change of linear momentum of ball B (force F_BA) is m_B(v_B - u_B) / t.
The force F_AB exerted by ball A on ball B and the force F_BA exerted by ball B on ball A are equal and opposite, so F_AB = -F_BA.
Therefore, m_Au_A + m_Bu_B = m_Av_A + m_Bv_B.
This means that the total momentum before the collision is equal to the total momentum after the collision. The total momentum of the two balls remains constant, or conserved.

PRESSURE AND THRUST

Pressure: When a force acting on a body is spread over a certain surface area of the body and acts everywhere perpendicular to the surface, the force acting on a unit area of the surface is called pressure. Thus, pressure is the normal force per unit area. The S.I. unit of pressure is N m^-2 or pascal (Pa). Pressure is not a vector quantity.

Thrust: The total force exerted by a body perpendicular to a surface is called thrust. Thrust is a vector quantity, and its S.I. unit is newton (N). For example, a rectangular block of mass 4 kg lying on the ground exerts a thrust on the ground equal to its weight.
Thrust = 4 kg × 9.8 m/s² = 39.2 N

Effects of pressure: The effect of a force on a body is measured by the pressure exerted on it. The higher the pressure, the greater the effect it produces.
Pressure = Thrust / Area
i) Pressure is directly proportional to thrust. The more the thrust, the more the pressure, and vice versa.
ii) Pressure is inversely proportional to the area of contact. The smaller the area of contact, the more the pressure, and vice versa.
So the effect of a force is increased by decreasing the area on which it acts.

Some everyday applications of this concept are:
1. Nails and pins have pointed ends: The tips of nails and pins have a very small area. A small force on the head of a pin or nail falls on a small area, exerting considerable pressure (P ∝ 1/A) to drive it into a surface easily.
2. The cutting edges of knives, blades and axes are sharpened: A sharpened edge has a smaller area of contact, so a small force produces a large pressure, which cuts an object easily.

The effect of a force is decreased by increasing the area on which it acts. Some examples are:
i) Heavy trucks are fitted with six to eight wheels: The large number of wheels increases the area of contact on the ground, which reduces the pressure (P ∝ 1/A) exerted by the truck on the ground. For similar reasons, tractors have broad tyres, a wide steel belt is provided over the wheels of army tanks, and wide wooden or concrete sleepers are placed below railway tracks.
ii) Skiers use long flat skis to slide over snow: Long flat skis increase the area of contact, which reduces the pressure exerted on the snow, so the skier can slide over the snow without sinking. For similar reasons, it is easier to walk on a muddy road in flat shoes than in shoes with sharp heels.
iii) High-rise buildings have wide foundations: The wide foundation increases the area over which the weight of the building acts. This reduces the pressure exerted on the ground (P ∝ 1/A), so high-rise buildings do not sink into the ground.

Pressure in a liquid: The pressure at a point at a certain depth in a liquid at rest under gravity is given by:
P = h d g
that is, pressure = vertical depth × density × acceleration due to gravity.

Pressure at all points in the same horizontal plane in a liquid at rest is the same. Pressure at a point in a liquid at rest is the same in all directions.

Atmosphere: The gaseous envelope surrounding the Earth is called the atmosphere. The atmospheric pressure at any point in air at rest is equal to the weight of a vertical column of air of unit area surrounding that point, the column extending to the top of the atmosphere. Atmospheric pressure at sea level is about 10⁵ N m^-2.

DENSITY, RELATIVE DENSITY AND BUOYANCY

Density: The ratio of the mass of a substance to its volume is defined as the density of the material of the substance. It may also be defined as the mass per unit volume of the substance.
Density (d) = mass (m) / volume (V)
The S.I. unit of density is kg m^-3.

Relative density or specific gravity: Relative density is defined as the ratio of the mass of a certain volume of a substance to the mass of an equal volume of water at 4 °C. Since relative density is the ratio of two similar quantities, it is a number and has no units.

Buoyancy of a liquid: A solid body immersed in a fluid (liquid or gas) experiences no horizontal force. When a body is immersed in a fluid, it experiences an upthrust, or an upward force. This upthrust exerted by a liquid on a body immersed in it is called the buoyancy of the liquid.

Archimedes' principle: According to Archimedes' principle, when a body is wholly or partly immersed in a fluid, it experiences an upthrust equal to the weight of the fluid displaced by it. In other words, a body immersed in a fluid loses its weight equal to the weight of the fluid displaced by it.
The upthrust, equal to the buoyant force, equals the loss in weight, which equals the volume of the object immersed × density of the liquid × gravitational acceleration, that is, F = Vdg.
According to Archimedes' principle, relative density or specific gravity = weight of the object in air / loss of weight in water.

Law of floatation: According to the law of floatation, a body floats in a liquid if the weight of the liquid displaced by the immersed portion of the body is equal to the weight of the body.

Viscous force: A liquid also offers a kind of frictional force called viscous force when any object moves through it. This force also acts between any two adjacent layers of a liquid which have a relative velocity. Unlike friction, this force depends on the area.

EXAMPLES

Example 1: Calculate the linear momentum of a toy car weighing 200g and moving with a velocity of 5 m/s.

Answer:
1. Mass of the toy car, m = 200 g = \( \frac{200}{1000} \) kg = 0.2 kg.
2. Velocity of the toy car, v = 5 m/s.
3. Momentum, p = m × v = 0.2 kg × 5 m/s = 1 kg m/s.
4. Hence, the momentum of the toy car is 1 kg m/s.

Teacher's Note:
a) Convert grams into kilograms before using p = mv.
b) Momentum is always expressed in kg m/s.

Example 2: What is the acceleration produced by a force of 12 N exerted on an object of mass 3 kg.

Answer:
1. Force, F = 12 N and mass, m = 3 kg.
2. Using F = m × a, 12 N = 3 kg × a.
3. a = 12 N / 3 kg = 4 m/s².
4. Hence, the acceleration produced in the object is 4 m/s².

Teacher's Note:
a) Rearrange F = ma to find the acceleration when the force and mass are known.
b) Since 1 N = 1 kg m/s², the unit of acceleration is m/s².

Example 3: What force would be needed to produce an acceleration of 4 m/s2 on a ball 6 kg

Answer:
1. Acceleration, a = 4 m/s² and mass, m = 6 kg.
2. Using F = m × a, F = 6 kg × 4 m/s² = 24 kg m/s².
3. Since 1 kg m/s² = 1 N, F = 24 N.
4. So a force of 24 N is required.

Teacher's Note:
a) The force is found by multiplying mass and acceleration.
b) Check the units: kg multiplied by m/s² gives newton.

Example 4: A car weighing 1600 kg moving with a velocity of 30 m/s retards uniformly coming to rest in 20 s. Calculate the a) initial momentum of the car b) final momentum of the car c) rate of change of momentum d) acceleration of the car e) magnitude of the force applied

Answer:
1. Mass of the car, m = 1600 kg; initial velocity, u = 30 m/s; final velocity, v = 0 m/s (the car stops); time taken, t = 20 s.
2. a) Initial momentum = mass × initial velocity = 1600 kg × 30 m/s = 48,000 kg m/s.
3. b) Final momentum = mass × final velocity = 1600 kg × 0 m/s = 0 kg m/s.
4. c) Rate of change of momentum = (Final momentum - Initial momentum) / Time taken = (0 - 48,000 kg m/s) / 20 s = -2400 N.
5. d) Acceleration, a = (v - u) / t = (0 - 30 m/s) / 20 s = -1.5 m/s².
6. e) Force, F = mass × acceleration = 1600 kg × (-1.5 m/s²) = -2400 N.
7. The magnitude of the force is 2400 N. The negative sign shows that the force acts in the direction opposite to the motion of the car.

Teacher's Note:
a) Use a negative velocity change for a body that slows down, to get the correct sign.
b) The magnitude of the force is 2400 N, and the sign shows only its direction.
c) The rate of change of momentum and the force have the same value, as the second law says.

EXERCISES

Q.1 Impulse is
(A) a scalar quantity
(B) a vector quantity
(C) neither a scalar nor a vector
(D) sometimes a scalar and sometimes a vector

Answer: (B) a vector quantity

1. Impulse is equal to the change in linear momentum.
2. Momentum is a vector quantity, so impulse is also a vector quantity.
3. Hence, the correct option is (B).

Teacher's Note:
a) Impulse has the same direction as the change in momentum.
b) Its unit is newton second, which is the same as kg m/s.

Q.2 Choose the wrong statement
(A) kg wt = 9.8 N
(B) Momentum is a vector quantity
(C) Force is always conserved
(D) Momentum is conserved in the absence of an external force.

Answer: (C) Force is always conserved

1. 1 kg wt = 9.8 N, so option (A) is correct.
2. Momentum is a vector, so option (B) is correct.
3. Momentum is conserved when no external force acts, so option (D) is correct.
4. Force is not a conserved quantity, so option (C) is wrong.
5. Hence, the correct option is (C).

Teacher's Note:
a) Momentum is conserved, but force is not.
b) Remember 1 kg wt = 9.8 N for conversions in older numericals.

Q.3 A long-jumper runs before jumping because
(A) he covers a greater distance
(B) he maintains momentum conservation
(C) he gains energy by running
(D) he gains momentum

Answer: (D) he gains momentum

1. Running gives the long-jumper a large velocity before take-off.
2. Since momentum = mass × velocity, the jumper gains momentum from the run.
3. Hence, the correct option is (D).

Teacher's Note:
a) The run builds up momentum that carries the body forward in the jump.
b) Energy is not gained by running in the sense of the question.

Q.4 If a rock is brought from the surface of the moon
(A) its mass will change
(B) its weight will change, but not mass
(C) both mass and weight will change
(D) its mass and weight will remain the same

Answer: (B) its weight will change, but not mass

1. Mass is the amount of matter in the rock, which stays the same everywhere.
2. Weight depends on gravity, which is weaker on the Moon, so the weight changes.
3. Hence, the correct option is (B).

Teacher's Note:
a) Mass is a measure of inertia and does not change with place.
b) Weight is the gravitational force on the body, so it changes from place to place.

Q.5 If 1 kg wt = 10 N, the value of gravitational intensity
(A) 10 m/s²
(B) \( \frac{1}{10} \) m/s²
(C) 1 m/s²
(D) \( \frac{1}{100} \) m/s²

Answer: (A) 10 m/s²

1. Weight of 1 kg = 1 kg × g = 10 N.
2. So g = 10 N / 1 kg = 10 m/s².
3. Hence, the correct option is (A).

Teacher's Note:
a) Gravitational intensity is the same as acceleration due to gravity.
b) Using the value 9.8 N for 1 kg gives g = 9.8 m/s².

Q.6 The force acting on mass of 1 g due to the gravitational pull of the earth is called 1 gwt. One gwt equals
(A) 1 N
(B) 9.8 N
(C) 980 dyne
(D) none of these

Answer: (C) 980 dyne

1. Mass = 1 g = 0.001 kg, and g = 9.8 m/s² = 980 cm/s².
2. Force = mass × g = 1 g × 980 cm/s² = 980 dyne.
3. Hence, the correct option is (C).

Teacher's Note:
a) 1 dyne = 1 g × 1 cm/s², so the CGS value is 980 dyne.
b) 1 N = 10⁵ dyne, which is why option (A) is wrong.

Q.7 The weight of a body would not be zero
(A) at the centre of the earth
(B) during a free fall
(C) in interplanetary space
(D) on a frictionless surface

Answer: (D) on a frictionless surface

1. At the centre of the Earth, gravity is zero, so the weight is zero.
2. During free fall, the apparent weight is zero, and in interplanetary space the gravity is negligible.
3. On a frictionless surface, the body still has its weight, which is equal to mg, so the weight is not zero.
4. Hence, the correct option is (D).

Teacher's Note:
a) Friction has no effect on weight, which depends only on mass and gravity.
b) Options (A), (B) and (C) describe situations where the weight is zero or close to zero.

Q.8 An iron ball and wooden ball of the same radius are released from a height H in vacuum. The times taken by both of them to reach the ground are
(A) roughly equal
(B) unequal
(C) exactly equal
(D) in the inverse ratio of their diameters

Answer: (C) exactly equal

1. In vacuum there is no air resistance.
2. Both balls fall with the same acceleration g, from the same height H.
3. Hence, the times taken are exactly equal, and the correct option is (C).

Teacher's Note:
a) Air resistance is the reason that a feather falls more slowly than a stone in air.
b) In vacuum, all bodies fall with the same acceleration, whatever their mass.

Q.9 A man is standing on a boat in still water. If he walks towards the shore the boat will
(A) move away from the shore
(B) remain stationary
(C) move towards the shore
(D) sink

Answer: (A) move away from the shore

1. The man pushes the boat backwards while walking forward, which is the action.
2. By Newton's third law, the boat moves in the opposite direction, away from the shore.
3. Hence, the correct option is (A).

Teacher's Note:
a) This is the same principle as walking on the ground.
b) The total momentum of man and boat remains zero, so the boat moves opposite to the man.

Q.10 During a planned manoeuvre in a space flight, a free-floating astronaut A pushes another free-floating astronaut B, the mass of A being greater than that of B. Then, the magnitude of the force exerted by astronaut A on astronaut B is
(A) equal to zero
(B) equal to the force exerted by B on A
(C) greater than the force exerted by B on A
(D) less than the force exerted by B on A

Answer: (B) equal to the force exerted by B on A

1. A pushes B, and B pushes A with an equal and opposite force.
2. By Newton's third law, the two forces are equal in magnitude, whatever the masses.
3. Hence, the correct option is (B).

Teacher's Note:
a) The mass of the astronauts does not change the size of the forces.
b) Students often think the heavier body exerts the bigger force; this is wrong.

Q.11 In the above problem, during push
(A) the acceleration of A is greater than that of B
(B) the acceleration of A is less than that of B
(C) neither is accelerated
(D) their accelerations are equal in magnitude but opposite indirection

Answer: (B) the acceleration of A is less than that of B

1. The forces on A and B are equal in magnitude.
2. By F = ma, a = F / m, so the body with the larger mass has the smaller acceleration.
3. Since A is heavier, its acceleration is less than that of B. Hence, the correct option is (B).

Teacher's Note:
a) Equal forces on unequal masses give unequal accelerations.
b) The accelerations are opposite in direction, but they are not equal in magnitude.

Q.12 A bullet of mass A and velocity B is fired into a block of wood of mass C. If loss of any mass and friction by neglected, the velocity of the system must be
(A) \( \frac{AB}{A+C} \)
(B) \( \frac{A+C}{B+C} \)
(C) \( \frac{AC}{B+C} \)
(D) \( \frac{A+B}{AC} \)

Answer: (A) \( \frac{AB}{A+C} \)

1. Momentum before impact = A × B, as only the bullet is moving.
2. After the bullet gets embedded, the mass of the system is (A + C) and it moves with velocity V.
3. By conservation of momentum, (A + C) × V = A × B, so V = \( \frac{AB}{A+C} \).
4. Hence, the correct option is (A).

Teacher's Note:
a) The bullet and the block move together after the collision, so the final mass is A + C.
b) Momentum is conserved even though kinetic energy is not.

Q.13 A driver accelerates this car first at the rate of 1.8 m/s² and then at the rate of 1.2 m/s². The ratio of the forces exerted by the engines will be respectively equal to
(A) 2 : 3
(B) 1 : 2
(C) 2 : 1
(D) 3 : 2

Answer: (D) 3 : 2

1. The mass of the car is the same in both cases, so F ∝ a.
2. Ratio of forces = 1.8 : 1.2 = 3 : 2.
3. Hence, the correct option is (D).

Teacher's Note:
a) Since the mass is constant, the forces are in the same ratio as the accelerations.
b) Writing the ratio as 1.8 : 1.2 and simplifying gives 3 : 2.

Q.14 A body of mass 5 kg undergoes a change in speed from 30 to 40 m/s. Its momentum would increase by
(A) 50 kg m/s
(B) 75 kg m/s
(C) 150 kg m/s
(D) 350 kg m/s

Answer: (A) 50 kg m/s

1. Initial momentum = 5 × 30 = 150 kg m/s.
2. Final momentum = 5 × 40 = 200 kg m/s.
3. Increase in momentum = 200 - 150 = 50 kg m/s.
4. Hence, the correct option is (A).

Teacher's Note:
a) Find the change in momentum, not the momentum itself.
b) Change = m × (change in speed) when the motion is in the same direction.

Q.15 The force needed to produce an acceleration of 6 m/s² in a ball of mass 4 kg will be
(A) 24 N
(B) 30 N
(C) 32 N
(D) 36 N

Answer: (A) 24 N

1. Using F = m × a, F = 4 kg × 6 m/s².
2. F = 24 N.
3. Hence, the correct option is (A).

Teacher's Note:
a) This is a direct use of Newton's second law.
b) Multiplying mass by acceleration gives the force in newtons.

Q.16 A body of mass 5 kg undergoes a change in speed from 20 to 0.20 m/s. The momentum of the body would
(A) increase by 99 kg m/s
(B) decrease by 99 kg m/s
(C) increase by 101 kg m/s
(D) decrease by 101 kg m/s

Answer: (B) decrease by 99 kg m/s

1. Initial momentum = 5 × 20 = 100 kg m/s.
2. Final momentum = 5 × 0.20 = 1 kg m/s.
3. Change in momentum = 1 - 100 = -99 kg m/s, so the momentum decreases by 99 kg m/s.
4. Hence, the correct option is (B).

Teacher's Note:
a) A negative change in momentum means a decrease.
b) Do not confuse the decrease of 99 kg m/s with the final momentum of 1 kg m/s.

Q.17 A bullet of mass 0.01 kg is fired from a gun weighing 5.0 kg. If the initial speed of the bullet is 250 m/s, calculate the speed with which the gun recoils.
(A) -0.50 m/s
(B) - 0.25 m/s
(C) +0.05 m/s
(D) +0.25 m/s

Answer: (A) -0.50 m/s

1. Momentum before firing = 0.

2. Momentum of bullet = 0.01 kg × 250 m/s = 2.5 kg m/s, forward.
3. Momentum of gun = 5.0 kg × V, where V is its recoil velocity.
4. By conservation of momentum, 5.0 × V + 2.5 = 0, so V = -0.50 m/s.
5. The negative sign shows that the gun recoils backwards. Hence, the correct option is (A).

Teacher's Note:
a) Total momentum is zero before firing, so it must stay zero after.
b) The negative sign shows the direction opposite to the bullet.

Q.18 A body of mass 100g is moving with a velocity of 15 m/s. The momentum associated with the ball will be
(A) 0.5 kg m/s
(B) 1.5 kg m/s
(C) 2.5 kg m/s
(D) 3.2 N s

Answer: (B) 1.5 kg m/s

1. Mass = 100 g = 0.1 kg.
2. Momentum = 0.1 kg × 15 m/s = 1.5 kg m/s.
3. Hence, the correct option is (B).

Teacher's Note:
a) Always convert grams into kilograms before multiplying.
b) Option (D) has the unit of impulse, not momentum.

Q.19 A number of discs, each of momentum M kg m/s are striking a wall at the rate of n discs per minute. The force associated with these discs, in newtons, would be
(A) \( \frac{Mn}{60} \)
(B) 60 Mn
(C) \( \frac{M}{60n} \)
(D) \( \frac{n}{60M} \)

Answer: (A) \( \frac{Mn}{60} \)

1. Each disc stops on hitting the wall, so the change in momentum of each disc is M.
2. Discs striking per second = \( \frac{n}{60} \).
3. Force = rate of change of momentum = \( \frac{Mn}{60} \) N.
4. Hence, the correct option is (A).

Teacher's Note:
a) Force is the number of discs per second multiplied by the momentum of each disc.
b) Convert the rate from per minute to per second by dividing by 60.

Q.20 If action and reaction were to act on the same body
(A) the resultant would be zero
(B) the body would not move at all
(C) both (A) and (B) is correct
(D) neither (A) nor (B) is correct

Answer: (C) both (A) and (B) is correct

1. Equal and opposite forces acting on the same body add up to zero.
2. With zero resultant force, the body would not move.
3. Hence, both (A) and (B) are correct, and the correct option is (C).

Teacher's Note:
a) Action and reaction act on different bodies, which is why they do not cancel.
b) If they acted on the same body, no motion would be possible.

Q.21 A stationary ball weighing 0.25 kg acquires a speed of 10 m/s when hit by a hockey stick The impulse imparted to the ball is
(A) 2.5 N s
(B) 2.0 N s
(C) 1.5 N s
(D) 0.5 N s

Answer: (A) 2.5 N s

1. Impulse = change in momentum = m(v - u).
2. Impulse = 0.25 kg × (10 m/s - 0) = 2.5 kg m/s = 2.5 N s.
3. Hence, the correct option is (A).

Teacher's Note:
a) Impulse equals the change in momentum, not the momentum at the end.
b) The unit N s is the same as kg m/s.

Q.22 The combined effect of mass and velocity is taken into account by a physical quantity called
(A) torque
(B) moment of force
(C) momentum
(D) moment of momentum

Answer: (C) momentum

1. Momentum is the product of mass and velocity.
2. So it takes into account both mass and velocity together.
3. Hence, the correct option is (C).

Teacher's Note:
a) Torque and moment of force are rotational quantities.
b) Momentum is a linear quantity.

Q.23 Momentum is a measure of
(A) weight
(B) mass
(C) quantity of motion
(D) velocity

Answer: (C) quantity of motion

1. Momentum is the quantity of motion contained in a body.
2. Hence, the correct option is (C).

Teacher's Note:
a) Mass alone or velocity alone does not give the quantity of motion.
b) Momentum is the product of the two.

Q.24 Momentum has the same unit as that of
(A) impulse
(B) torque
(C) moment of momentum
(D) couple

Answer: (A) impulse

1. Impulse equals the change in momentum, so it has the same unit.
2. Both have the unit kg m/s, or N s.
3. Hence, the correct option is (A).

Teacher's Note:
a) Torque has the unit N m, which is different.
b) Use this link between impulse and momentum to remember the unit.

Q.25 A hammer weighing 3 kg moving with a velocity of 10 m/s, strikes against the head of a spike and drives it into a block of wood. If the hammer comes to rest in 0.025 s, the impulse associated with the ball will be
(A) 30 N s
(B) - 30 N s
(C) 15 N s
(D) -15 N s

Answer: (B) - 30 N s

1. Initial momentum = 3 kg × 10 m/s = 30 kg m/s.
2. Final momentum = 0, as the hammer comes to rest.
3. Impulse = final momentum - initial momentum = 0 - 30 = -30 N s.
4. The negative sign shows that the impulse acts opposite to the initial motion. Hence, the correct option is (B).

Teacher's Note:
a) The time of 0.025 s is not needed to find the impulse, only to find the force.
b) The question's word "ball" refers to the hammer, as printed.

Q.26 Consider two spring balances hooked as shown in Figure. We pull them in opposite directions with equal force. If the reading shown by A is 1.5 N, the reading shown by B will be

[Figure: Two spring balances A and B hooked end to end, with a hand pulling each outer end in opposite directions]

(A) 1.5N
(B) 2.5 N
(C) 3.0 N
(D) zero

Answer: (A) 1.5N

1. The two spring balances are pulled with equal and opposite forces.
2. Each spring balance records the same tension, because the forces are equal.
3. So the reading of B is equal to the reading of A, which is 1.5 N. Hence, the correct option is (A).

Teacher's Note:
a) A spring balance reads the force pulling on it, not zero when both ends are pulled.
b) This follows from Newton's third law applied along the spring.

Q.27 In the problem no. 25, the average (retarding) force acting on the spike will be
(A) 600 N
(B) -600 N
(C) 1200 N
(D) -1200 N

Answer: (D) -1200 N

1. Change in momentum = -30 kg m/s (from Q.25).
2. Time taken = 0.025 s.
3. Average force = \( \frac{-30}{0.025} \) = -1200 N.
4. The magnitude is 1200 N, and the negative sign shows that the force is retarding. Hence, the correct option is (D).

Teacher's Note:
a) The retarding force is the force that brings the hammer to rest.
b) Divide the change in momentum by the time to get the force.

Q.28 In a tug-of-war between the teams A and B, the rope breaks at a point which is nearer to A. Then
(A) A has applied more force
(B) B has applied more force
(C) A and B both have applied same force
(D) none has applied any force

Answer: (C) A and B both have applied same force

1. The two teams pull the rope with equal and opposite forces, by Newton's third law.
2. The rope breaks at the weaker point, which is nearer to A, and this does not mean A pulled harder.
3. Hence, the correct option is (C).

Teacher's Note:
a) The printed key shows option (A); the correct answer is (C), because the two teams apply equal forces on the rope.
b) The break point depends on the strength of the rope at that point, not on the pull of the teams.

Q.29 A rocket works on the
(A) first law of motion
(B) second law of motion
(C) third law of motion
(D) law of conservation of energy

Answer: (C) third law of motion

1. The rocket pushes hot gases backwards, which is the action.
2. The gases push the rocket forward with an equal and opposite force, which is the reaction.
3. Hence, the correct option is (C).

Teacher's Note:
a) A rocket works in empty space, so it does not need air to push against.
b) The second law explains the size of the force, but the principle is the third law.

Q.30 A metallic ball strikes a wall and falls down whereas a tennis ball having the same mass and velocity bounces back. The reason for this is that
(A) both suffer equal change in momentum
(B) the tennis ball suffers a greater change in momentum
(C) metallic ball suffers a greater change in momentum
(D) the momentum of the tennis ball is less than that of the metallic ball

Answer: (B) the tennis ball suffers a greater change in momentum

1. The metallic ball stops, so its change in momentum is m × u.
2. The tennis ball bounces back, so its velocity changes from u to -u, and its change in momentum is 2mu, which is greater.
3. Hence, the correct option is (B).

Teacher's Note:
a) A bounce means a reversal of velocity, which doubles the change in momentum.
b) Both balls have the same initial momentum, so option (D) is wrong.

Q.31 When a bicycle travels on a rough surface, its speed
(A) increases
(B) decreases
(C) remains the same
(D) none of these

Answer: (B) decreases

1. A rough surface produces a larger frictional force opposite to the motion.
2. This frictional force acts as a retarding force, so the speed decreases unless the rider keeps pedalling.
3. Hence, the correct option is (B).

Teacher's Note:
a) The printed key shows option (A); the correct answer is (B), because friction always opposes the motion and slows the bicycle.
b) Friction is larger on a rough surface than on a smooth one.

Q.32 If you are asked to push an object so that the acceleration produced in it is now twice as before, then the force required will be
(A) twice as before
(B) half as before
(C) same as before
(D) four times as before

Answer: (A) twice as before

1. The mass of the object is the same, so F = ma gives F ∝ a.
2. If the acceleration is doubled, the force must also be doubled.
3. Hence, the correct option is (A).

Teacher's Note:
a) The force is directly proportional to the acceleration for a fixed mass.
b) Doubling the mass instead would also need twice the force for the same acceleration.

Q.33 It is difficult to walk on ice because of
(A) absence of friction
(B) absence of inertia
(C) more inertia
(D) more friction

Answer: (A) absence of friction

1. Walking needs friction between the shoe and the ground to push forward.
2. On ice the friction is very small, so the foot slips.
3. Hence, the correct option is (A).

Teacher's Note:
a) Friction helps us walk by pushing the ground backwards, so the ground pushes us forward.
b) Inertia is not the reason, since it does not change with the surface.

Q.34 The law which defines force is
(A) Newton's third law of motion
(B) Newton's first law of motion
(C) Newton's second law of motion
(D) Newton's law of gravitation

Answer: (B) Newton's first law of motion

1. The first law says that a body stays at rest or in uniform motion unless an unbalanced force acts on it.
2. This statement defines force as the agent that changes the state of motion.
3. Hence, the correct option is (B).

Teacher's Note:
a) The second law gives a method of measuring force, not its definition.
b) The third law describes pairs of forces.

Q.35 The law which gives a quantitative measurement of force is
(A) Newton's third law of motion
(B) Newton's first law of motion
(C) Newton's second law of motion
(D) Newton's law of gravitation

Answer: (C) Newton's second law of motion

1. The second law gives F = ma, which measures force quantitatively.
2. Hence, the correct option is (C).

Teacher's Note:
a) The first law defines force, and the second law measures it.
b) The newton is defined using the second law.

Q.36 Internal forces
(A) are always balanced
(B) are never balanced
(C) may or may not be balanced
(D) none of these

Answer: (A) are always balanced

1. Internal forces act between the parts of a system and occur in equal and opposite pairs.
2. Within the system, these pairs cancel out, so they are always balanced.
3. Hence, the correct option is (A).

Teacher's Note:
a) Internal forces cannot change the motion of the whole system.
b) Only external unbalanced forces can change the motion of a system.

Q.37 External forces
(A) are always balanced
(B) never balanced
(C) may or may not be balanced
(D) none of these

Answer: (C) may or may not be balanced

1. External forces come from outside the system.
2. They may add up to zero, when the system is balanced, or they may not, when the system is unbalanced.
3. Hence, the correct option is (C).

Teacher's Note:
a) A body at rest with balanced external forces stays at rest.
b) An unbalanced external force changes the motion of the body.

Q.38 If the mass of a body is 12.1 gm and the density is 2.2 g/cc, its volume is
(A) 5.5 cm³
(B) 8 cc
(C) 11 cc
(D) 55 cc

Answer: (A) 5.5 cm³

1. Density = mass / volume, so volume = mass / density.
2. Volume = 12.1 g / 2.2 g/cm³ = 5.5 cm³.
3. Hence, the correct option is (A).

Teacher's Note:
a) 1 cc is the same as 1 cm³.
b) Dividing the mass by the density gives the volume.

Q.39 Brine has a density of 1.2 g/cc of it are mixed with 30 cc of water. The density of solution is
(A) 2.11 g/cc
(B) 1.11 g/cc
(C) 12.2 g/cc
(D) 20.4 g/cc

Answer: (B) 1.11 g/cc

1. The question as printed does not state the volume of brine, so the most sensible reading is 40 cc of brine of density 1.2 g/cc.
2. Mass of brine = 40 cc × 1.2 g/cc = 48 g. Mass of water = 30 cc × 1 g/cc = 30 g.
3. Total mass = 78 g and total volume = 70 cc.
4. Density of solution = 78 / 70 = 1.11 g/cc.
5. Hence, the correct option is (B).

Teacher's Note:
a) The printed question omits the volume of brine, which is needed to solve it; the reading of 40 cc is used here.
b) Density of a mixture is total mass divided by total volume, not the average of the two densities.

Q.40 If a force of 10 N acts on surfaces of areas in the ratio 1 : 2 then the ratio of thrusts is
(A) 1 : 2
(B) 2 : 1
(C) 3 : 1
(D) 1 : 1

Answer: (D) 1 : 1

1. Thrust is the total force perpendicular to the surface.
2. The same force of 10 N acts in both cases, so the thrusts are equal.
3. Hence, the ratio of thrusts is 1 : 1, and the correct option is (D).

Teacher's Note:
a) Thrust does not depend on the area; pressure does.
b) If the question asked for the ratio of pressures, the answer would be 2 : 1.

Please click the link below to download CBSE Class 8 Science Force and Pressure Chapter Notes