CBSE Class 9 Science Force and laws of motion Exam Notes

CBSE Class 9 Science Force and laws of motion Exam Notes. Please refer to the examination notes which you can use for preparing and revising for exams. These notes will help you to revise the concepts quickly and get good marks.

FORCE AND LAWS OF MOTION

FORCE

The external agent which tends to set a body in motion, or which changes the speed and direction of motion of a body, or which can change the shape of a body, is called force. The SI unit of force is the newton.

Types of forces

1. Non-contact forces: the forces in which contact between two objects is not necessary. The gravitational force between two bodies and the electrostatic force between two charges are two examples of field forces.
2. Contact forces: two bodies in contact exert equal and opposite forces on each other. If the contact is frictionless, the contact force is perpendicular to the common surface and is known as the normal reaction. If, however, the objects are in rough contact and move (or have a tendency to move) relative to each other without losing contact, a frictional force arises which opposes such motion. Again each object exerts a frictional force on the other and the two forces are equal and opposite. This force is perpendicular to the normal reaction. Thus the contact force (F) between two objects is made up of two forces: (i) the normal reaction (N) and (ii) the force of friction (f). Since these two forces are mutually perpendicular, \(F=\sqrt{N^2+f^2}\).

NEWTON’S LAWS OF MOTION

(A) Newton’s first law. A body cannot change its state of motion by itself. If the object is at rest it remains at rest, and if it is in uniform motion it continues to be in motion, unless some external force is applied on it.

Inertia. The inherent property of an object by virtue of which it cannot change its state of rest or of motion by itself is called inertia. Inertia is of two types: inertia of rest and inertia of motion.

(a) Inertia of rest. If a body is at rest, it continues to be at rest unless some external force is applied on it. Examples:

• When a train at rest starts moving suddenly, a passenger standing inside the compartment tends to fall backward.
• When a carpet is beaten with a stick, the dust particles are detached.
• When a bullet is fired into a glass pane, it pierces a hole only at the point where the bullet hits, without breaking the entire pane into pieces.

(b) Inertia of motion. When a body is in uniform motion, it continues in its uniform motion, i.e. it resists any change in its state of motion. Examples:

• When a person jumps out of a moving bus, he should run in the direction in which the bus is moving, otherwise he will fall down.
• A train is moving with uniform speed and a passenger throws a ball upwards inside the train; the ball comes back to his hand.

Relation between mass and inertia. The larger the mass of a body, the larger its inertia. For example, it is more difficult to stop a cricket ball than a tennis ball moving at the same speed.

(B) Newton’s second law of motion. Momentum is the product of mass and velocity: \(p=mv\). Its SI unit is kg m/s, and it is a vector quantity.

Newton’s second law states: “the rate of change of momentum of a body is directly proportional to the force and takes place in the direction of the force.”

\[F\propto\dfrac{p_2-p_1}{t}\quad\text{or}\quad F=\dfrac{m(v-u)}{t}=ma\]

where \(p_1=mu\) is the initial momentum and \(p_2=mv\) is the final momentum. The unit of force in the SI system is the newton; 1 N is the force which can produce an acceleration of 1 m/s² in a body of mass 1 kg. The unit of force in the CGS system is the dyne: 1 dyne = 1 g cm/s², and 1 N = \(10^5\) dyne.

Example. Calculate the force required to produce an acceleration of 5 m/s² in a body of mass 2.4 kg.

Solution. Force = mass × acceleration = 2.4 kg × 5 m/s² = 12.0 N.

Example. A body of mass 2.5 kg is moving with a velocity of 20 m/s. Calculate its momentum.

Solution. Momentum \(p=mv=2.5\times20=50\) kg m/s.

Impulse. If a force F is applied on a body of mass m for a time interval t, and the change in velocity is \(\Delta v\), then impulse \(=F\,t=m\,\Delta v\) = change in momentum. The unit of impulse is newton second.

Example of impulse: while catching a cricket ball a player moves his hands backwards. The ball coming towards the fielder has a large momentum. By moving his hands back he increases the time in which the momentum of the ball is reduced to zero, so the rate of change of momentum (the force on the hands) becomes small.

Example. A force acts for 0.2 s on a body of mass 2.5 kg initially at rest. The force then ceases to act and the body moves through 4 m in the next one second. Calculate the magnitude of the force.

Solution. When the force ceases to act, the body moves with constant velocity. Since it moves 4 m in 1 s, its uniform velocity is 4 m/s. Initial velocity u = 0, final velocity v = 4 m/s, time t = 0.2 s.

Acceleration \(a=\dfrac{v-u}t=\dfrac{4-0}{0.2}=20\) m/s².

Force \(F=2.5\times20=50\) N.

Example. A ball of mass 20 g is initially moving with a velocity of 100 m/s. On applying a constant force on the ball for 0.5 s, it acquires a velocity of 150 m/s. Calculate (i) the initial momentum, (ii) the final momentum, (iii) the rate of change of momentum, (iv) the acceleration of the ball, (v) the magnitude of the force applied.

Solution. m = 20 g = 0.02 kg, u = 100 m/s, v = 150 m/s, t = 0.5 s.

(i) Initial momentum \(p_1=mu=0.02\times100=2\) kg m/s.

(ii) Final momentum \(p_2=mv=0.02\times150=3\) kg m/s.

(iii) Rate of change of momentum \(=\dfrac{p_2-p_1}t=\dfrac{3-2}{0.5}=2.0\) kg m/s² \(=2.0\) N.

(iv) Acceleration \(a=\dfrac{v-u}t=\dfrac{150-100}{0.5}=100\) m/s².

(v) Force \(F=ma=0.02\times100=2.0\) N.

Example. A cricket ball of mass 200 g moving with a speed of 40 m/s is brought to rest by a player in 0.04 s. Calculate (i) the change in momentum of the ball, (ii) the average force applied by the player.

Solution. m = 200 g = 0.2 kg, u = 40 m/s, v = 0, t = 0.04 s.

(i) Initial momentum \(p_1=0.2\times40=8.0\) kg m/s. Final momentum \(p_2=0.2\times0=0\). Change in momentum \(=p_2-p_1=-8.0\) kg m/s.

(ii) Average force \(=\dfrac{\text{change in momentum}}{\text{time}}=\dfrac{0-8.0}{0.04}=-200\) N. The negative sign shows that the force is applied in a direction opposite to the direction of motion of the ball.

Example. A motorcycle is moving with a velocity of 108 km/h and it takes 5 s to stop it after the brakes are applied. Calculate the force exerted by the brakes on the motorcycle if its mass along with the rider is 250 kg.

Solution. Initial velocity = 108 km/h = 30 m/s, final velocity = 0, time = 5 s, mass = 250 kg. The change in velocity is 0 – 30 = –30 m/s, so the acceleration is \(a=\dfrac{-30}5=-6\) m/s².

The magnitude of the force applied by the brakes is \(F=ma=250\times6=1500\) N.

(C) Newton’s third law of motion. Newton’s first law gives a qualitative idea of force, while the second law gives a way to measure force. The third law states: “if a body A exerts a force on a body B, the body B will also exert an equal and opposite force on A.” The force exerted by A on B is called the action and the force exerted by B on A is called the reaction. It is also stated as “to every action there is an equal and opposite reaction.” Forces always occur in pairs, and action and reaction always act on different bodies. Example: by hitting a table with the palm we apply a force; the table also exerts a force on the palm.

Applications of Newton’s third law

• Recoil of a gun: when a bullet is fired from a gun, an equal and opposite force is applied on the gun, due to which the gun recoils backward.
• Walking: while moving forward we push the ground backwards (action). An equal and opposite force is applied by the ground on the man (reaction), due to which he moves forward.
• Rowing a boat: when we push the water backward with the oars (a backward force), an equal and opposite force acts on the boat. This reaction moves the boat forward.
• Launching a rocket: in a rocket, gases are produced in large amount by combustion. They come out and move backwards with a force, and an equal and opposite force acts on the rocket and moves it forward.

LAW OF CONSERVATION OF MOMENTUM

According to the law of conservation of momentum, “if there is no external force acting on a system, the momentum of the system remains unchanged.” Generalising: if a group of bodies are exerting forces on each other, their total momentum remains conserved before and after the interaction, provided there is no external force acting on them:

\[m_1u_1+m_2u_2=m_1v_1+m_2v_2\]

Example. A rifle of mass 5 kg fires a bullet of mass 40 g. The bullet leaves the barrel with a velocity of 200 m/s. If the bullet takes 0.004 s to move through the barrel, calculate (i) the recoil velocity of the rifle, (ii) the force experienced by the rifle due to its recoil.

Solution. (i) Mass of rifle \(m_1=5\) kg, mass of bullet \(m_2=40\) g = 0.04 kg, \(u_1=u_2=0\), \(v_2=200\) m/s. By conservation of momentum, \(0+0=5v_1+0.04\times200\), so \(v_1=-\dfrac{0.04\times200}{5}=-1.6\) m/s.

(ii) Initial momentum of the rifle = 0; final momentum = 5 × (–1.6) = –8 kg m/s; time = 0.004 s. Force \(=\dfrac{-8}{0.004}=-2000\) N.

Example. A bullet of mass 20 g moving with a velocity of 200 m/s gets embedded into a wooden block of mass 980 g suspended by a string. Calculate the velocity acquired by the combined system.

Solution. Mass of bullet \(m_1=0.02\) kg, \(u_1=200\) m/s, so the momentum of the bullet is \(0.02\times200=4\) kg m/s. The mass of block and bullet together = 980 + 20 = 1000 g = 1 kg. If v is the velocity of the combined system, its momentum is \(1\times v\). By conservation of momentum, \(m_1u_1=(m_1+m_2)v\), i.e. \(4=v\), so \(v=4\) m/s.

Example. A rifleman who together with his rifle has a mass of 100 kg stands on a smooth surface and fires 10 shots horizontally. Each bullet has a mass of 10 g and a muzzle velocity of 800 m/s. What velocity does the rifleman acquire at the end of 10 shots?

Solution. Let \(m_1\) and \(m_2\) be the masses of a bullet and the rifleman, and \(v_1\), \(v_2\) their velocities after the first shot. Initially the system is at rest, so the initial momentum is zero. As the external force is zero, momentum is constant: \(0=m_1v_1+m_2v_2\), so \(v_2=-\dfrac{m_1v_1}{m_2}=-\dfrac{(10\times10^{-3}\text{ kg})(800\text{ m/s})}{100\text{ kg}}=-0.08\) m/s.

Velocity acquired after 10 shots \(=10\times(-0.08)=-0.8\) m/s, i.e. the velocity of the rifleman is 0.8 m/s in a direction opposite to that of the bullets.

Example. A body of mass 1 kg strikes elastically with another body at rest and continues to move in the same direction with one-fourth of its initial velocity. What will be the mass of the other body?

Solution. Initial velocity of the first body = u; its final velocity = \(\dfrac u4\). Let the other body have mass m and final velocity \(v_2\). By conservation of momentum, \(1\times u+0=1\times\dfrac u4+mv_2\), so \(mv_2=\dfrac{3u}4\) …(1).

By conservation of energy (elastic collision), \(\dfrac12\times1\times u^2+0=\dfrac12\times1\times\left(\dfrac u4\right)^2+\dfrac12mv_2^2\), so \(mv_2^2=\dfrac{15}{16}u^2\) …(2).

Dividing the square of (1) by (2): \(\dfrac{(mv_2)^2}{mv_2^2}=m=\dfrac{(9/16)u^2}{(15/16)u^2}=0.6\) kg.

FRICTION

If we switch off the engine of a car, it stops after moving a certain distance. This means some retarding force is acting on the car which stops it. The force opposing the motion of the car is called the force of friction. The frictional force is tangential to the surfaces in contact and always acts in a direction opposite to the direction of motion of the object. Friction is a force opposing the relative motion between two surfaces which are in contact with each other.

Reducing friction. Frictional force can be reduced in the following ways:

• Use of lubricants: in machines, friction can be reduced by applying lubricants between the contact surfaces; they fill the fine pores or depressions in the surfaces and make them smooth.
• Polishing: the unevenness of surfaces can be reduced by polishing.
• Use of ball bearings: in rotating machines, shafts are mounted on ball bearings. Rolling friction is less than sliding friction.
• Streamlining: air friction is reduced by designing streamlined bodies of cars and aeroplanes. Similarly, the bodies of boats and ships are streamlined to reduce the friction of water.

Disadvantages of friction

• Wear and tear: machine parts like gears and brakes wear out gradually when they are in continuous contact, and have to be replaced from time to time.
• Friction reduces the efficiency of a machine.
• Friction in machines produces heat and undesirable noise which can damage the machine. To avoid excessive heating, water is generally circulated in machines.

Applications of frictional force

• We would not be able to walk if there were no friction between the soles of our shoes and the ground.
• If there were no friction, the wheels of a car would slip instead of rotating. To increase the grip of the tyres on the road, the tyres are made corrugated.
• When brakes are applied, the vehicle stops due to the force of friction between the brake lining and the drum.
• In the absence of friction we could not write on a blackboard with a chalk stick, because the chalk would slide off the board without leaving any mark.

Example. Three blocks are tied together with strings in a row on a horizontal frictionless table (\(m_1\), \(T_1\), \(m_2\), \(T_2\), \(m_3\)). They are pulled to the right with a force \(T_3=12\) N. If \(m_1=1\) kg, \(m_2=2\) kg and \(m_3=3\) kg, calculate the tension \(T_1\).

Solution. All the blocks move with the same acceleration a. For \(m_3\): \(3a=12-T_2\) …(1). For \(m_2\): \(2a=T_2-T_1\) …(2). Adding (1) and (2): \(5a=12-T_1\) …(3). For \(m_1\): \(1\times a=T_1\), so \(a=T_1\) …(4). From (3) and (4): \(5T_1=12-T_1\), so \(T_1=2\) N.

THRUST AND PRESSURE

(A) Thrust. The force acting normally on a surface is called thrust. It is a vector quantity and is measured in newton (N).

(B) Pressure. The thrust on a unit area of a surface is called pressure.

\[P=\dfrac{\text{thrust}}{\text{area}}=\dfrac FA\]

The SI unit of pressure is newton per square metre (N/m²); other units are pascal and bar. One pascal is the pressure exerted on a surface area of 1 m² by a thrust of 1 newton, i.e. 1 pascal = 1 N/m².

Some examples based on pressure

• Inserting a pointed nail in a wooden block is easier than inserting a rod with the same force, because the nail has a smaller area and therefore experiences more pressure for the same force.
• A sharp knife cuts better than a blunt knife.
• While walking, a man exerts more pressure on the ground than when he is standing.

Pressure in fluids. A substance that can flow is called a fluid. Liquids and gases are considered fluids.

Laws of pressure

• The pressure exerted by a liquid is the same in all directions about a point.
• In a stationary liquid, the pressure is the same at all points in a horizontal plane.
• The pressure at a particular depth is \(P=hdg\), where h = height of the liquid column (depth), d = density of the liquid, g = acceleration due to gravity.

Different types of pressure

• Hydrostatic pressure: the normal force (thrust) exerted by a liquid at rest per unit area of the surface in contact with it is called the pressure of the liquid or hydrostatic pressure.
• Atmospheric pressure: the pressure exerted by the atmosphere. At sea level, atmospheric pressure is the pressure exerted by a 0.76 m column of mercury, i.e. about \(10^5\) pascal. 1 atm \(\approx10^5\) Pa.

Example. What will be the pressure in N/m² at a depth of 1.5 m in brine of density 120 kg/m³?

Solution. \(P=hdg=1.5\times120\times10=1800\) N/m².

Example. Calculate the density of a liquid if the pressure at a point 30 m below its surface is \(32\times10^4\) N/m².

Solution. \(P=hdg\), so \(d=\dfrac P{hg}=\dfrac{32\times10^4}{30\times10}=1066.6\) kg/m³.

Example. A force of 150 N is applied on an area of 1.5 m². Calculate the pressure exerted.

Solution. \(P=\dfrac FA=\dfrac{150}{1.5}=100\) N/m².

Example. A force of 500 dynes is applied on an area of 20 cm². Calculate the pressure exerted.

Solution. F = 500 dyne = \(500\times10^{-5}\) N; A = 20 cm² = \(20\times10^{-4}\) m². \(P=\dfrac{500\times10^{-5}}{20\times10^{-4}}=2.5\) N/m².

BUOYANCY

When a body is immersed in a liquid, the liquid exerts an upward force on the body called upthrust or buoyant force.

Factors affecting upthrust: the larger the volume of the body submerged in the liquid, the greater the upthrust; the larger the density of the liquid, the greater the upthrust.

Archimedes’ principle states that when a body is immersed in a liquid, partially or completely, it experiences an upthrust equal to the weight of the liquid displaced. The loss in weight of the block, i.e. the buoyant force acting on the block, is equal to the weight of the liquid displaced:

\[F=Vdg\]

where V = volume of the body (immersed part), d = density of the liquid, g = acceleration due to gravity.

Example. A body weighs 300 gf in air and 260 gf when completely immersed in water. Calculate (i) the loss in weight of the body, (ii) the upthrust on the body.

Solution. Loss in weight = 300 – 260 = 40 gf. Upthrust = loss in weight = 40 gf.

Example. A solid block of volume 2 litres has a weight of 80 N. What will be its weight when immersed completely in water?

Solution. Volume of the block = 2 litres = 2000 cc, so the volume of water displaced is 2000 cc. Weight of water displaced = 2000 g = 2.0 kgf = 2.0 × 9.8 N = 19.6 N. Upthrust = 19.6 N. Hence the weight of the body fully immersed in water = 80 – 19.6 = 60.4 N.

Example. A solid block of density D has a weight W in air. It is fully immersed in a liquid of density d. Calculate its apparent weight when fully immersed in the liquid.

Solution. The weight of the block is W and its density is D, so its volume is \(\dfrac WD\) (in gravitational units, with the same g). The upthrust \(=\dfrac WD\cdot d\), which is also the loss in weight. Hence the apparent weight when fully immersed \(=W-\dfrac WD d=W\left(1-\dfrac dD\right)\).

DENSITY

\[\text{Density}=\dfrac{\text{mass}}{\text{volume}}\quad\text{or}\quad d=\dfrac MV\]

The SI unit of density is kg/m³ and the CGS unit is g/cm³. 1 g/cm³ = 1000 kg/m³. Different substances have different densities, which are affected by temperature. The density of a substance decreases on heating due to expansion of the substance. Water has an anomalous expansion: when water is cooled from room temperature down to 4 °C its volume decreases, but on cooling further below 4 °C its volume starts increasing. The density of water is maximum at 4 °C.

Relative density

\[\text{Relative density}=\dfrac{\text{density of substance}}{\text{density of water at }4^\circ\text{C}}=\dfrac{\text{weight of a certain volume of substance}}{\text{weight of the same volume of water at }4^\circ\text{C}}\]

It is also the ratio of the mass of a unit volume of the substance to the mass of a unit volume of water at 4 °C. Relative density has no unit. The density of water in the CGS system is 1 g/cm³.

Relative density for solids: \(\text{R.D.}=\dfrac{W_1}{W_1-W_2}\), where \(W_1\) = weight of the solid body in air and \(W_2\) = weight of the solid body in water.

Relative density for liquids: \(\text{R.D.}=\dfrac{W-W''}{W-W'}\), where W = weight of the body in air, W′ = weight of the body fully immersed in water, W″ = weight of the body fully immersed in the liquid.

Example. A body weighs 30 N in air and 26 N when fully immersed in water. Calculate its relative density.

Solution. \(W_1=30\) N, \(W_2=26\) N. Relative density \(=\dfrac{W_1}{W_1-W_2}=\dfrac{30}{30-26}=7.5\).

Example. The relative density of copper is 8.8. What is its density in the SI system?

Solution. R.D. = density of copper / density of water, so density of copper = R.D. × density of water = \(8.8\times10^3\) kg/m³.

Example. A solid weighs 60 gf in air and 52 gf when completely immersed in water. Calculate (i) the upthrust, (ii) the volume of the solid, (iii) the relative density of the solid.

Solution. Loss of weight in water = 60 – 52 = 8 gf. (i) Upthrust = loss of weight in water = 8 gf. (ii) Since the density of water is 1 g/cm³, the weight of an equal volume of water is 8 gf, so the volume of the solid = 8 cm³. (iii) Relative density \(=\dfrac{W_1}{W_1-W_2}=\dfrac{60}{8}=7.5\).

(a) i.e. F = P2 - P1 /t or F = m( v - u) ÷ = m a

where p₁ = initial momentum = mu p₂ = final momentum = mv

Sol. Momentum, p = mass × velocity

Here, mass m = 2.5 kg