## NEWTON’S LAWS OF MOTION

**NEWTON’S LAWS OF MOTION:Newton’s first law (law of inertia) **

This law states that ** “A body continues in its state of rest or uniform motion unless an unbalanced force acts on it”. **The mass of a body is a measure of its inertia. Inertia is the property that keeps an object in its state of motion and resists any efforts to change it.

**Newton’s second law (law of momentum) **

Momentum of a body is defined as the product of its mass and its velocity.

**Momentum ‘p’=mv.** The SI unit for momentum is **kgm/s** or** Ns**. The Newton’s second law states that “*The rate of change of momentum of a body is proportional to the applied force and takes place in the direction in which the force acts”*

Change in momentum= **mv-mu**

Rate of change of momentum= **mv-mu/∆t**

Generally the second law gives rise to the equation of force** F=ma**

Hence **F=mv-mu/∆t **and **F∆t=mv-mu**

The quantity **F∆t** is called impulse and is equal to the change of momentum of the body. The SI unit for impulse is **Ns**.

NEWTON’S LAWS OF MOTION

*Examples *

*A van of mass 3 metric tons is travelling at a velocity of 72 km/h. Calculate the momentum of the vehicle.*

*Solution *

Momentum=mv=72km/h = (20m/s) ×3×10^{3} kg

=6.0×10^{4} kgm/s

*A truck weighs 1.0×10*^{5 }N and is free to move. What force will give it an acceleration of

*1.5 m/s ^{2}? (take g=10N/kg) *

*Solution *

Mass of the truck = (1.0×10^{5})/10=6.0×10^{4}

Using F=ma

=1.5×10×10^{4 }

=1.5×10^{4} N

*A car of mass 1,200 kg travelling at 45 m/s is brought to rest in 9 seconds. Calculate the average retardation of the car and the average force applied by the brakes.*

*Solution *

Since the car comes to rest, v=0, a= (v-u)/t = (0-45)/9=-5m/s (retardation)

F=ma = (1200×-5) N =-6,000 N (braking force)

*A truck of mass 2,000 kg starts from rest on horizontal rails. Find the speed 3 seconds after starting if the tractive force by the engine is 1,000 N.*

*Solution *

Impulse = Ft=1,000×3= 3,000 Ns

Let v be the velocity after 3 seconds. Since the truck was initially at rest then u=0.

Change in momentum=mv-mu

= (2,000×v) – (2,000×0)

=2,000 v

But impulse=change in momentum

2,000 v = 3,000 v = 3/2=1.5 m/s.

*Weight of a body in a lift or elevator *

When a body is in a lift at rest then the weight

**W = mg **

When the lift moves upwards with acceleration ‘a’ then the weight becomes

**W = m (a + g) **

If the lift moves downwards with acceleration ‘a’ then the weight becomes

**W = m (g-a) **

NEWTON’S LAWS OF MOTION

*Example *

*A girl of mass stands inside a lift which is accelerated upwards at a rate of 2 m/s ^{2}. Determine the reaction of the lift at the girls’ feet. *

*Solution *

Let the reaction at the girls’ feet be ‘R’ and the weight ‘W’

The resultant force F= R-W

= (R-500) N

Using F = ma, then R-500= 50×2, R= 100+500 = 600 N.

**Newton’s third law (law of interaction) **

This law states that *“For every action or force there is an equal and opposite force or reaction”**Example *

*A girl of mass 50 Kg stands on roller skates near a wall. She pushes herself against the wall with a force of 30N. If the ground is horizontal and the friction on the roller skates is negligible, determine her acceleration from the wall. *

*Solution *

Action = reaction = 30 N

Force of acceleration from the wall = 30 N

F = ma

a = F/m = 30/50 = 0.6 m/s^{2}

Linear collisions

** Linear collision occurs when two bodies collide head-on and move along the same straight line.** There are two types of collisions;

- Inelastic collision: – this occurs when two bodies collide and stick together i.e. hitting putty on a wall. Momentum is conserved.
- Elastic collision: – occurs when bodies collide and bounce off each other after collision. Both momentum and kinetic energy are conserved.

Collisions bring about a law derived from both Newton’s third law and conservation of momentum. This law is known as **the law of conservation of linear momentum** which states that ** “when no outside forces act on a system of moving objects, the total momentum of the system stays constant”**.

NEWTON’S LAWS OF MOTION

*Examples *

*A bullet of mass 0.005 kg is fired from a gun of mass 0.5 kg. If the muzzle velocity of the bullet is 300 m/s, determine the recoil velocity of the gun.*

*Solution *

Initial momentum of the bullet and the gun is zero since they are at rest.

Momentum of the bullet after firing = (0.005×350) = 1.75 kgm/s

But momentum before firing = momentum after firing hence

0 = 1.75 + 0.5 v where ‘v’ = recoil velocity

0.5 v = -1.75

v =-1.75/0.5 = – 3.5 m/s (recoil velocity)

*A resultant force of 12 N acts on a body of mass 2 kg for 10 seconds. What is the change in momentum of the body?*

*Solution *

Change in momentum = ∆P = mv – mu= Ft

= 12×10 = 12 Ns

*A minibus of mass 1,500 kg travelling at a constant velocity of 72 km/h collides head-on with a stationary car of mass 900 kg. The impact takes 2 seconds before the two move together at a constant velocity for 20 seconds. Calculate a)**The common velocity**The distance moved after the impact**The impulsive force**The change in kinetic energy*

*Solution *

- a) Let the common velocity be ‘v’

Momentum before collision = momentum after collision

(1500×20) + (900×0) = (1500 +900) v

30,000 = 2,400v

v = 30,000/2,400 = 12.5 m/s (common velocity)

- After impact, the two bodies move together as one with a velocity of 12.5 m/s

Distance = velocity × time = 12.5× 20

= 250m

- Impulse = change in momentum

= 1500 (20-12.5) for minibus or

=900 (12.5 – 0) for the car

= 11,250 Ns

Impulse force F = impulse/time = 11,250/2 = 5,625 N d) K.E before collision = ½ × 1,500 × 20^{2} = 3 × 10^{5} J

K.E after collision = ½ × 2400 × 12.5^{2} = 1.875× 10^{5} J

Therefore, change in K.E = (3.00 – 1.875) × 10^{5} = 1.25× 10^{5} J

**Some of the applications of the law of conservation of momentum**

- Rocket and jet propulsion: – rocket propels itself forward by forcing out its exhaust gases. The hot gases are pushed through exhaust nozzle at high velocity therefore gaining momentum to move forward.
- The garden sprinkler: – as water passes through the nozzle at high pressure it forces the sprinkler to rotate.

**Solid friction**

** Friction is a force which opposes or tends to oppose the relative motion of two surfaces in contact with each other**.

*Measuring frictional forces*

We can relate weight of bodies in contact and the force between them. This relationship is called coefficient of friction. ** Coefficient of friction is defined as the ratio of the force needed to overcome friction F_{f} to the perpendicular force between the surfaces F_{n}. **Hence

**µ = F**

_{f }/ F_{n }*Examples*

*A box of mass 50 kg is dragged on a horizontal floor by means of a rope tied to its front. If the coefficient of kinetic friction between the floor and the box is 0.30, what is the force required to move the box at uniform speed?*

*Solution *

F_{f} = µF_{n}

F_{n}= weight = 50×10 = 500 N

F_{f} = 0.30 × 500 = 150 N

*A block of metal with a mass of 20 kg requires a horizontal force of 50 N to pull it with uniform velocity along a horizontal surface. Calculate the coefficient of friction between the surface and the block. (take g = 10 m/s)*

*Solution *

Since motion is uniform, the applied force is equal to the frictional force F_{n} = normal reaction = weight = 20 ×10 = 200 N Therefore, µ =F_{f }/ F_{n} = 50/ 200 = 0.25.

*Laws of friction *

It is difficult to perform experiments involving friction and thus the following statements should therefore be taken merely as approximate descriptions: –

- Friction is always parallel to the contact surface and in the opposite direction to the force tending to produce or producing motion.
- Friction depends on the nature of the surfaces and materials in contact with each other.
- Sliding (kinetic) friction is less than static friction (friction before the body starts to slide).
- Kinetic friction is independent of speed.
- Friction is independent of the area of contact.
- Friction is proportional to the force pressing the two surfaces together.

NEWTON’S LAWS OF MOTION

*Applications of friction *

- Match stick
- Chewing food
- Brakes
- Motion of motor vehicles
- Walking

*Methods of reducing friction *

- Rollers
- Ball bearings in vehicles and machines
- Lubrication / oiling
- Air cushioning in hovercrafts

*Example *

*A wooden box of mass 30 kg rests on a rough floor. The coefficient of friction between the floor and the box is 0.6. Calculate *

*The force required to just move the box**If a force of 200 N is applied the box with what acceleration will it move?*

*Solution *

- Frictional force F
_{f }= µF_{n}= µ(mg)

= 0.6×30×10 = 180 N

- The resultant force = 200 – 180 = 20 N From F =ma, then 20 = 30 a

a = 20 / 30 = 0.67 m/s^{2}

Viscosity

** This is the internal friction of a fluid**. Viscosity of a liquid decreases as temperature increases. When a body is released in a viscous fluid it accelerates at first then soon attains a steady velocity called terminal velocity. Terminal velocity is attained when

**where**

*F + U = mg***F**is viscous force,

**U**is upthrust and

**mg**is weight.

NEWTON’S LAWS OF MOTION

ALL PHYSICS NOTES FORM 1-4 WITH TOPICAL QUESTIONS & ANSWERS

PRIMARY NOTES, SCHEMES OF WORK AND EXAMINATIONS