Physics Form 3 Notes : CHAPTER THREE – NEWTON’S LAWS OF MOTION

CHAPTER THREE

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 kg·m/s or Ns. 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.

Examples

1. 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 = 3,000 kg × 20 m/s = 6.0 × 10⁴ kg·m/s

2. A truck weighs 1.0 × 10⁵ N and is free to move. What force will give it an acceleration of 1.5 m/s²? (Take g = 10 N/kg)

   Solution

   Mass of the truck = (1.0 × 10⁵) / 10 = 1.0 × 10⁴ kg

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   Using F = ma

   F = 1.5 × 1.0 × 10⁴ = 1.5 × 10⁴ N

3. 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 = -5 m/s² (retardation)

   F = ma = 1,200 × (-5) = -6,000 N (braking force)

4. 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, u = 0.

   Change in momentum = mv – mu = 2,000 v – 0 = 2,000 v

   But impulse = change in momentum, so 2,000 v = 3,000

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

Weight of a body in a lift or elevator

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

W = mg

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

W = m (g + a)

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

W = m (g – a)

Example

A girl of mass 50 kg stands inside a lift which is accelerated upwards at a rate of 2 m/s². Determine the reaction of the lift at the girl’s feet.

Solution

Let the reaction at the girl’s feet be ‘R’ and the weight ‘W’.

The resultant force F = R – W = R – 500 N

Using F = ma, then R – 500 = 50 × 2, so 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 30 N. 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/s²

Linear collisions

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

1. Inelastic collision: This occurs when two bodies collide and stick together, e.g., hitting putty on a wall. Momentum is conserved.
2. 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.”

Examples

1. 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 × 300 = 1.5 kg·m/s

   But momentum before firing = momentum after firing, hence

   0 = 1.5 + 0.5 v, where ‘v’ = recoil velocity

   0.5 v = -1.5

   v = -1.5 / 0.5 = -3 m/s (recoil velocity)

2. 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 = 120 Ns

3. 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:

   1. The common velocity
   2. The distance moved after the impact
   3. The impulsive force
   4. The change in kinetic energy

   Solution

   1. Let the common velocity be ‘v’.

   Momentum before collision = momentum after collision

   (1500 × 20) + (900 × 0) = (1500 + 900) v

   30,000 = 2,400 v

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

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

   Distance = velocity × time = 12.5 × 20 = 250 m

   1. 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

   1. Kinetic energy before collision = ½ × 1,500 × 20² = 3 × 10⁵ J

   Kinetic energy after collision = ½ × 2,400 × 12.5² = 1.875 × 10⁵ J

   Therefore, change in kinetic energy = (3.00 – 1.875) × 10⁵ = 1.25 × 10⁵ J

Some of the applications of the law of conservation of momentum

1. Rocket and jet propulsion: A rocket propels itself forward by forcing out its exhaust gases. The hot gases are pushed through the exhaust nozzle at high velocity, therefore gaining momentum to move forward.
2. 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 the weight of bodies in contact and the force between them. This relationship is called the 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

1. 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 = µ F_n

   F_n = weight = 50 × 10 = 500 N

   F_f = 0.30 × 500 = 150 N

2. 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/s²)

   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; thus, the following statements should be taken as approximate descriptions:

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

Applications of friction

1. Match stick
2. Chewing food
3. Brakes
4. Motion of motor vehicles
5. Walking

Methods of reducing friction

1. Rollers
2. Ball bearings in vehicles and machines
3. Lubrication / oiling
4. 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:

1. The force required to just move the box
2. If a force of 200 N is applied, with what acceleration will the box move?

Solution

1. Frictional force F_f = µ F_n = µ (mg)

   = 0.6 × 30 × 10 = 180 N

2. The resultant force = 200 – 180 = 20 N

   From F = ma, then 20 = 30 a

   a = 20 / 30 = 0.67 m/s²

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 F + U = mg, where F is viscous force, U is upthrust, and mg is weight.