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

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. SI unit of force is newton.

Types of forces : 

(a)non contact :

These are the forces in which contact between two objects is not necessary. Gravitational force between two bodies and electrostatic force between two charges are two examples of field forces.  

(b)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 known as 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 then frictional force arise which oppose 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 normal reaction. Thus, the contact force (F) between two objects is made up of two forces.

(i) Normal reaction (N) (ii) Force of friction (f)

 and since these two forces are mutually perpendicular.

 and since these two forces are mutually perpendicula

Ø NEWTON'S LAWS OF MOTION

(A) Newton´s Ist law :

A body can not change its state of motion by itself. If the object is at rest it will remain at rest and if it is in uniform motion, it continues to be in motion unless some external force is applied on it.

 Inertia :

u There is an inherent property of an object by virtue of which it cannot change its state of motion or rest by itself. This property is called 'inertia'.

u Inertia is of two types– inertia of rest and inertia of motion.

(a) Inertia of rest:

If the body is at rest, it will continue to be at rest unless some external force is applied on it. Examples are following.

Ex. When a train at rest starts moving suddenly, a passenger standing inside the compartment tends to fall backward.

Ex. When a carpet is beaten up with a stick, the dust particles are detached.

Ex. When a bullet is fired into a glass pane, it pierces a hole only at the pt where the bullet hits the glass without breaking the entire glass pane into pieces.

(b) Inertia of motion :

When a body is in uniform motion, it will continue to remain in its uniform motion, i.e. it resists any change in its state of motion due to inertia of motion.

Ex. when a person jumps out of a moving bus, he should run in the direction in which bus is moving otherwise he will fall down.

Ex. A train moving with a uniform speed and if a ball is thrown upwards inside the train by a passenger, then the ball comes back to his hand.

u Relation between mass and inertia :

Larger the mass of the body, larger is the inertia.

Ex. eg : it is more difficult to stop a cricket ball than a tennis ball.

(B) Newton's second law of motion

Momentum : The product of mass and velocity is called 'momentum'. i.e. p = m v

(a) Unit : SI unit of momentum is kg-m/s. (b) It is a vector quantity.

Newton's second law states "the rate of change of momentum of a body is directly proportional to force and takes place in the direction of force."

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

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

(b) Unit of force in SI system is newton.

(c) 1N is equivalent to that force which can produce an acceleration of 1m/s2 in a body of mass 1 kg. (d) Unit of force in CGS system is dyne.

1 dyne = 1 gm - cm/s²

1 N = 105 dynes

Ex.1 Calculate the force required to produce an acceleration of 5 m/s2 in a body of mass

2.4 kg.

Sol. We know that force = mass × acceleration

= 2.4 kg × 5 m/s²

= 12.0N

Ex.2 A body of mass 2.5 kg is moving with a velocity of 20 m/s. Calculate its momentum.

Sol. Momentum, p = mass × velocity

Here, mass m = 2.5 kg

Velocity, v = 20 m/s

\ Momentum, p = mv = 2.5 × 20 kg-m/s

Impulse :

= 50 kg-m/s

u If a force F is applied on a body of mass m for a time interval t and if the change in velocity is △v then

\ Impulse = F dt = m △ v

Impulse = change in momentum

 Unit of impulse is newton × second,

Examples of impulse

Ex. While catching a cricket ball a player moves his hands backwards. Cricket ball coming towards fielder has a large momentum. By doing so he increases the time interval to reduce the momentum of the ball. Rate of change of momentum becomes slow.

Ex.3 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 4m in the next one second. Calculate the magnitude of force.

Sol. When the force ceases to act, the body will move with a constant velocity. Since it moves a distance of 4 m in 1 s, therefore, its uniform velocity = 4m/s.

Now, initial velocity, u = 0

Final velocity, v = 4 m/s

Time interval Dt = 0.2 s

\Acceleration, a =

v - u /△t  = 4 - 0 /0.2

= 20m/s²

Force, F = 2.5 × 20 = 50 N

Ex.4 A ball of mass 20 gm is initially moving with a velocity of 100 m/s. On applying a constant force on the ball for

0.5s, it acquires a velocity of 150 m/s. Calculate the following : (i) Initial momentum of the ball

(ii) Final momentum of the ball (iii) Rate of change of momentum (iv) Acceleration of the ball

(v) Magnitude of the force applied

Sol. Given , m = 20 gm = 20/1000

kg = 0.02 kg

Initial velocity, u = 100 m/s Time interval, t = 0.5 s Final velocity, v = 150 m/s

(i) Initial momentum of the ball

= mass × initial velocity or P₁ = mu = 0.02 kg × 100 m/

= 2 kg-ms^–1

(ii) Final momentum of the ball

= mass × final velocity or P₂ = mv = 0.02 kg × 150 m/s

= 3 kg-ms^–1

(iii) Rate of change of momentum

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