*Chapter Three *

**NEWTON’S LAWS OF MOTION **

**Newton’s First Law**

Newton’s first law, which states that a body remains in its state of rest or uniform motion in a straight line unless acted upon by an external force

**Inertia **

The property of bodies to resist change in state of motion is called inertia and it explains why cars have seat belts

Newton’s first law of motion suggests that matter has an in-built reluctance to change its state of motion or rest.

When a moving bus comes to an abrupt stop, the passengers lurch forward, i.e., tends to keep on moving. Likewise, when a bus surges forward, the passengers are jerked backwards, i.e., tend to resist motion. This property of bodies to resist change in state Of motion is called inertia and it explains why cars have seat belts

The mass of a body is a measure of its inertia. A larger mass requires a larger force to produce

a given acceleration or deceleration on it than a smaller mass. The larger mass therefore has a

greater inertia.

Newton’s first law of motion is also referred to as the law of inertia.

*Momentum’ *

A heavy-commercial vehicle requires a greater tractive force to start it moving when loaded than when empty. Likewise, a greater braking force is needed to bring to rest a heavy commercial vehicle than a small passenger car travelling at the same velocity. The vehicles each have a quantity called momentum which depends on the mass and the velocity of the vehicle. In the foregoing illustration, the heavy commercial vehicle has a greater momentum than the small car.

The momentum of a body is defined as the product of its mass and velocity. If m is the mass of a body in kg and v Its velocity in ms”, then;

momentum = mass (kg) x velocity (ms^{-1})

= mv

The SI unit of momentum is therefore kgms^{-1}. Momentum is a vector quantity, having both magnitude and direction. The direction of momentum is same as that of velocity of the body.

*Example 1 *

A van of mass 3 metric tonnes is travelling at a velocity of 72 kmh^{-1}. Calculate the momentum of the vehicle.

*Example 2 *

A car is moving at 36 kmh^{-1}. What velocities will double its momentum?

**Newton’s Second Law **

Newton’s second law of motion states that the rate of change of momentum of a body is’ directly proportional to the resultant external force producing the change, and takes place in the direction of the force. Thus; resultant force acting rate of change of momentum.

If the forces acting on the body are in equilibrium (balanced), then the resultant force acting on the body is zero, hence no change in momentum. This implies that the body under this condition will continue in its state of rest or uniform motion in a straight line (Newton’s first law).

*Relation between Force, Mass and Acceleration *

Consider a force F acting on a body of mass m for a time t. If its velocity changes from u to v, then;

change in momentum = final momentum – initial momentum

= mv-mu

:. Rate of change of momentum=

But acceleration a=

Hence, F mass x acceleration

So, F = krna, where k is a constant.

The Newton is that force which produces an acceleration of 1 ms^{-2} when it acts on a mass of 1 kg.

This definition gives; F = 1 N, a = 1 ms^{-2} and m = 1 kg.

Hence, substitution in F = kma leaves k = 1.

:. F=ma

Newton’s second law of motion can be verified by measuring the acceleration produced when various forces are applied to a frictionless trolley running on a friction-compensated runway. The trolley is taken to be of unit mass and the applied force is measured using identical elastic cords by taking the tension of the cord as a unit force when stretched by a certain

length.

*Example *3

What is the mass of an object which is accelerated at 3 ms^{-2} by a force of 125 N

*Example *4

A truck weighs 1.0 x 10^{5}N and is free to move. What force will give it an acceleration of 1.5 ms^{-2}? (Take g = 10 Nkg^{-l})

*Example5 *

A trolley of mass 1.5 kg is pulled along by an elastic cord and given an acceleration of2 ms^{-2}. Find the frictional force acting on the trolley if the tension in the cord is 5 N.

*Example 6 *

A car of mass 1 200 kg travelling at 45 ms^{-1} is brought to rest in 9 seconds. Calculate the average retardation of the car and the average force applied by the brakes.

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*Impulse *

When a force acts on a body for a very short time, the force is referred to as **an impulsive force**.

The result produced is known as the **impulse of the force**. Impulsive forces occur when two moving bodies collide, e.g., when two cars collide head-on or when a hammer strikes a stationary metal plate.

If a force F acts on a body of mass m for a time t, then the impulse of the force or impulse is given by force x time.

That is, impulse = force x time

= Ft

From Newton’s second law;

F=

This can be rewritten as; Ft = mv – mu

Since mv – mu is the change in momentum produced in the body during the time t, the impulse of a force acting on a body during a given time interval is equal to the change in momentum produced in the body in that time.

The SI unit of impulse is Newton second (Ns).

Therefore, another unit of momentum is the Newton-second (Ns). Since the rate of change of momentum is equal to impulse, i.e., Ft = p,

F = /t. Thus, force can be defined as rate of change of momentum.

A plot of force F against time is as in figure 3.9. The area under the curve is Ft or change in momentum during collision.

Large impulsive force are produced when momentums of bodies change within a very short time.

For example, when the velocity of a car is suddenly brought down to zero in a collision, the impulsive force on the passengers is so great that it could be fatal.

Seat-belts and airbags help in safeguarding against severe injuries occasioned by inertial forward surge.

The same applies to collapsible bumpers and steering which also helps during head on collisions by cushioning impulsive forces during head-on collisions through collapsing.

*Example 7 *

The graph in figure 3.10 shows the force on a tennis ball when served during a game. Find the mass of the racket with a velocity of 40 ms^{-1} (Assume the ball is stationary before it is struck)

*Example 8*

(a) Determine the change in momentum produced when a force of3.5 x 10^{3} N acts on a body which is at rest for 0.02 seconds.

(b) What velocity will be given to the body if it has a mass of 20 kg?

*Example 9 *

The valve of a gas cylinder containing 15 kg of compressed gas is opened and the cylinder empties in 1 hour and 20 minutes. If the gas issues from the exit nozzle with an average velocity of 30 ms ^{-1}, find the force exerted on the cylinder.

*Example 10 *

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.

*Example 11 *

A ball of mass 35 g travelling horizontally at 20 ms^{-1} strikes a wall at right angles and rebounds with a speed of 16 ms^{-1}. Find the impulse exerted on the ball.

Newton’s Third Law

Newton’s third law of motion states action and reaction are equal and opposite.

The law tells us that forces do not occur singly but due to action and reaction, they occur in pairs.

As the floor is part of a large mass (earth), the acceleration produced on it is not noticeable.

The force due to gravity W is the action force while that acting normally upwards is the reaction

force R. Since there is no resultant motion;

R=W=mg

*Note: *

The action force is produced by the block of wood when its weight is exerted on the table. The reaction is the equal force exerted by the table top on the block of wood. Hence. ‘action always begs for a reaction!’

Weight of a Body in a Lift

A passenger in a lift (elevator) experience forces against the feet, depending on the direction of motion and the acceleration of the lift. Consider a body of mass m on a weighing machine in a lift.

LAW OF CONSERVATION OF LINEAR MOMENTUM

The law of conservation of linear- momentum, which states that for a system of colliding bodies, the total linear momentum remains constant, provided no external forces act

*Example 14 *

A body A of mass 5 kg moving with a velocity of 3 ms^{-1} collides head-on with another body B of mass 4 kg moving in the opposite direction at 6ms^{-1}. If after the collision the bodies move together (coalesce), calculate the-common velocity v.

*Example 15 *

A bullet of mass 0.005 kg is fired from a gun of mass 0.5 kg. If the muzzle velocity bullet is 350 ms^{-1}, determine the recoil velocity of the gun.

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**Collisions **

One common property of any system of colliding bodies that the total momentum is conserved.

After collision, bodies may:

(i) fuse and move together in one direction, .

(ii) separate and move in different directions, or,

(iii) separate and move in the same direction.

*Elastic Collisions *

An elastic collision is one in which both kinetic energy and momentums are conserved.

*Inelastic Collisions *

An inelastic collision is one in which momentum is conserved but kinetic energy is not. The collision of lumps of plasticine or a bullet fired from a gun getting embedded into a block is cases of perfectly inelastic collisions. The characteristic of this type of collision is that after the collision:

(i) the total mass is the sum of the masses of the individual bodies.

(ii) the bodies end up with a common velocity.

In inelastic collisions, kinetic energy is lost because the bodies undergo some deformation.

Also, some of the energy is transformed to heat, sound or light.

*Example 16 *

A bullet of mass 10 g travelling horizontally at a speed of 100 ms ^{-1 }embeds itself in a block of wood of mass 990 g suspended from a light inextensible string so that it can swing freely.

Find:

(i) the velocity of the bullet and block immediately after collision.

(ii) the height through which the block rises.

*Example 17 *

A minibus of mass 1 500 kg travelling at a constant velocity of 72 kmh^{-1 }collides 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.

(b) the distance moved after the impact.

(c) the impulsive force.

(d) the change in kinetic energy.

Some Applications of the Law of Conservation of Momentum

*Rocket and Jet Propulsion *

A rocket propels itself forward by forcing out its exhaust gases. The hot exhaust gases are pushed out of the exhaust nozzle at high velocity and gain momentum in one direction. The rocket thus gains an equal momentum in the opposite direction. The rate at which the momentum changes provides the forward thrust on the rocket.

*Note: *

The rocket engine uses liquid hydrogen as its fuel and ‘liquid oxygen for combustion. It moves faster in the outer space, where there is no air resistance, than in the earth’s atmosphere.

A jet engine works on the same principle as the rocket engine but requires air which provides oxygen for combustion. The jet engine also requires a large mass of air to push out of its exhaust nozzles, so as to provide greater thrust.

*The Garden Sprinkler *

The garden sprinkler operates on the same principle as the engine discussed above. The pressure ;’ of the water in the pipe causes the water to be ejected through the nozzles at high velocity. The ejected water gains momentum and causes the sprinkler to rotate as in figure 3.24.

**FRICTION **

Friction is the resistance a body experience when it tend to move over another body

Static/ limiting friction is the maximum force between the surfaces before a body start moving

Sliding/ dynamic friction is the maximum force between the surfaces when the body is in motion

**Molecular Explanation of Friction **

smooth surfaces would look very rough when viewed under a powerful microscope.

This is due to some molecules on the surface lying on top of one another forming ‘tiny hills’.

The pressure at these points is quite enormous and the molecules making the ‘tiny hills’ which are in contact tend to stick together

For the bodies to move over each other work has to be done to overcome the interlocking between the ‘bumps’ and the ‘troughs’. The force opposing the work being done constitutes friction.

** **

**Laws of Friction **

Experimental results on friction between solids are summed up under the following laws:

(i) Frictional force between two surfaces oppose their relative motion.

(ii) Frictional force is independent of the area of contact of the given surfaces when the normal reaction is constant.

(iii) . Frictional friction is directly proportional to the normal reaction R.

(iv) Kinetic friction is independent of relative velocity.

(v) Frictional force is dependent on the nature of the surfaces in contact.

From the third law above,

F

F=

Where is the coefficient of friction

NB RESULTANT FORCE = FORCE APPLIED – FRICTION

Resultant force causes the body to accelerate

*Example 18 *

A wooden box of mass 30 kg rests on a rough floor. The co-efficient of friction between the floor and the box is 0.6.

(a) Calculate the force required to just move the box.

(b) If a force of200 N is applied to the box, with what acceleration will it move? (Take g = 10 ms^{-2})

*Example 19 *

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 co-efficient of friction between the surface and the block. (Take g = 10 ms^{-2})

**Methods of Minimising Friction **

It may not be possible to achieve a completely frictionless surface, but friction can be greatly minimized using the following:

*Rollers *

Rollers are placed between two rough surfaces so that when one body is to slide, friction is reduced. Rollers may be placed between the floor and heavy crates to enable the crates to slide,

Rollers may also be used when a marine vessel is being launched. They work on the principle that rolling friction is less than sliding friction.

*Ball Bearings *

Ball bearings reduce the friction for rotating axles. They are used extensively in machinery and are made of hard steel to prevent wear

Grease must be used together with ball bearings to lubricate the rolling action.

Lubrication

This is the application of oil or grease between moving parts.

*Air Cushion *

Air cushioning is done by blowing air into the space between surfaces. This prevents the surface coming into contact. The hovercraft uses air cushion to move with greatly reduced frictional force. Also air cushion is used in air tracks to produce a frictionless runway

**Applications of Friction**

*Walking *

Walking is made easier by friction. Pavements are made rough and tyres treaded to increase friction.

*Motor Vehicles *

Rotating tyres push backwards against the road surface. Friction opposes this force and the resultant force enables the vehicle to move.

*Brakes *

Friction between the brake drum and the brake lining halts the vehicle.

*Matchstick *

Friction between the matchstick head and the rough surface develops heat, igniting the stick head.

Friction can also be a nuisance. It causes wear, tear and noise between moving parts of a system, hence the need for lubrication in machines. Friction also, causes energy loss since work has to be done against it.

VISCOSITY

It is more difficult to wade through water than to move the same distance in open air space.

A steel ball dropped in a cylinder full of glycerine takes longer to reach the bottom than when dropped into the cylinder full of water.

This frictional resistance to motion in fluids is called **viscosity**. It is defined as the force which opposes the relative motion between the layers of the fluid. Glycerine has higher viscosity than water.

Terminal Velocity

*EXPERIMENT *3.5: *To investigate the relationship between the viscous drag F and velocity *v

*Apparatus *

Tall measuring cylinder (1 000 ml), ball bearing, glycerine, stop watch, metre rule, rubber bands.

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*Procedure *

- Fill the measuring cylinder with glycerine.
- Fix narrow horizontal rubber bands labelled 1, 2, 3,4, 5 and 6 at equal intervals along the cylinder, as shown in figure 3.33.
- Introduce a small ball bearing gently into the liquid (first dip the ball into glycerine).
- Measure the time of fall from the surface to the band labelled 1.
- Repeat for bands 2, 3, 4, 5 and 6.
- Determine the time of fall between each pair of rubber bands, i.e., between land 2, 2 and 3,3 and 4, 4 and 5, and, 5 and 6.
- Determine the velocity for each pair of bands and record the results in table 3.5.

*Table *3.5

TIme of | Distance between | TIme between | Velocity between |

.fall to | bands | bands | bands |

level | | | |

1 = | |||

2 = | 1 and 2 | ||

3 = | 2 and 3 | ||

4 = | 3 and 4 | ||

5 = | 4and5 | ||

6 = | 5 and 6 |

- Plot a graph of velocity against time of fall.

*Observation *

The ball bearing moves with increasing velocity when released into the liquid. The velocity of the ball between bands 4, 5 and 6 appears not to change.

A plot of velocity against time is as shown in figure 3.34.

*Explanation *

The forces acting on the ball when it is moving in a liquid are:

(i) its weight mg, acting vertically downwards.

(ii) the viscous drag F due to the liquid, acting vertically upwards.

(iii) the upthrust U due to the liquid, acting vertically upwards.

These forces are shown in figure 3.35.

When the ball enters into the liquid, mg > F + U and the resultant downward force therefore accelerates the ball towards the bottom of the cylinder. The viscous drag F however increases with the velocity and soon mg becomes equal to upward force (F + U). The resultant force is now zero and the ball attains a steady velocity called terminal velocity vto The terminal velocity is the constant velocity attained when the sum of the upward forces equals the weight of the object falling in the fluid.

A plot of velocity against time for a body falling through different liquids is shown in figure 3.36.

Stokes’Law

Stokes established that when a small object such as a steel sphere of radius r is dropped through a column of liquid and moves with a velocity v, it experiences a force which is directly proportional to:

(i) the radius r of the sphere.

(Ii) the velocity v of the sphere.

So, F rv. Hence, F = krv, where k is a constant.

Stokes found that k=6 where 11 is called the co-efficient of viscosity.

:. F = 6

This is the expression for Stokes’ law. The law holds when:

(i) the radius of the ball is small compared to the extent of liquid surface.

(ii) the ball does not create turbulence in the liquid as it falls.