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Physics SS2 Second Term Lesson Note

LINEAR MOMEMTUM

CONTENT

1. Newton’s First Law of Motion, Impulse and Momentum
2. Newton’s First Law of Motion
3. Inertia
4. Applications of Newton’s First Law of Motion
5. Momentum
6. Impulse
7. Newton’s Third Law of Motion and Applications

 

Newton’s First Law of Motion, Impulse and Momentum

The influence of unbalanced forces and the laws governing the motion are discussed in this topic.

Newton’s First Law of Motion

The first law states that every object continues in its state of rest or uniform motion in a straight line unless acted upon by an external force.

Inertia

It is the tendency of a body to remain in its state of rest or uniform linear motion. Newton’s law of motion is called law of inertia.

Applications of Newton’s First Law of Motion

1. When a moving vehicle is suddenly brought to rest by the application of the brakes, the passengers suddenly jerk forward as they tend to continue in their straight line motion. That is why it is advisable to use a safety belt.
2. A car driver in a stationary car hit by another car from behind is likely to suffer neck injuries because when the car is hit, his body is pushed forward, but his head stays still and is jerked backward in relation to his body. It is advisable to have a headrest to protect the driver and passengers from injury.
3. A moving body comes to rest due to opposing forces such as air resistance, friction or pull of gravity.

Momentum (p)

The momentum of a body is defined as the product of its mass and its velocity. The S.I unit is kgms^-1 or Ns

P=mv

M = mass in kg,    V = velocity in ms^-1

Impulse

It is the product of the average force acting on a particle and the time during which it acts. It is numerically equal to change in momentum.

I=F×t,Ft=mv−mu.

mv is final momentum,     mu is initial momentum

The unit of impulse is Newton-second (Ns) or kgms ^{– 1}

Newton’s second law of motion: The rate of change of momentum of a body is directly proportional to the applied force and takes place in the direction in which the force acts.

Force α change in momemtumtimeF α mv−mutF α m(v−u)t

But v−ut=aF α ma,f=kma,ifk=1f=ma

a is acceleration in ms^-2

f is resultant force acting on the body and the unit is Newton.

Calculations:

1. A net force of magnitude 0.6N acts on a body of mass 40g, initially at rest.

Calculate the magnitude of the resulting  acceleration.

Solution:

f=0.6N, m=40g=0.04kg,u=0, a=?f=ma0.6=0.04×aa=0.60.04=15ms−2

2. A ball of mass 5.0kg hits a smooth vertical wall normally with a speed of 2ms^-1 and rebounds with the same speed. Determine the impulse experienced by the ball.

f=0.6N m=5.0kgu=2mls,

v=−2m/s (negative because of a change in direction)

Impulse=mv−muImpulse=5(−2−2)=5×(−4)=−20Ns

EVALUATION

1. A ball of mass 0.1kg approaching a tennis player with a velocity of 10ms^-1, is hit back in the opposite direction with a velocity of 15ms^-1. If the time of impact between the racket and the ball is 0.01s, calculate the magnitude of the force with which the ball is hit.
2. A body of mass 20kg is set in motion by two forces 3N and 4N, acting at right angles to each other. Determine the magnitude of its acceleration.

 

Newton’s Third Law of Motion and Applications

Newton’s Third Law of Motion

It states that Action and Reaction are equal and opposite. Or to every Action there is an equal and opposite Reaction.

Applications of Newton’s Third Law of Motion

1. Gun and bullet:

When a bullet is shot out of a gun the person firing it experiences the backwards recoil force of the gun. The recoil force of the gun ( reaction) is equal to the propulsive force(action) acting on the bullet.

Force is proportional to change in momentum,

The momentum of bullet is equal and opposite to the momentum of gun.

m_b – mass of bullet

v_b –  velocity of bullet

m_g –  mass of  gun

v_g – velocity of gun

mbvb=−mgvgvg=−mbvbmg

2. Rocket and jet propulsion:

The momentum of the stream of hot gases issuing out of the nozzle behind the jet or rocket impacts an equal and opposite momentum to the rocket or aeroplane which undergoes a forward thrust.

m_r – mass of rocket

v_r –  velocity of rocket

m_g –  mass of hot gas

v_g – velocity of hot gas

mrvr=−mgvgvr=−mbvbmg

Physics SS2 Second Term Lesson Note

Newton’s Third Law of Motion

This states that when a body acts on another body with a force F, the second body acts on the first body with an equal amount of force but in the opposite direction.

That is, for every action, there is an equal but opposite reaction.

 

Applications of the Newton Third Law of Motion

1. Garden sprinkler:

This has a Z-shaped tube mounted on a pivot and through which water flows into the tube.As water is forced out of the open end of the tube, the tube is pushed backward with a equal but opposite reaction. This way, the tube is able to spin round and round and water the field all around it.

2. Weightlessness or weight loss in a lift:

A person standing on a weighing machine in a lift descending with a certain acceleration will experience weight loss and if the downward acceleration of the lift is equal to the prevailing acceleration due to gravity at that location, the person becomes weightless and float around in the lift.

 

 

 

(Note: the value of R represents the reading on the weighing machine on which the person stands). In this case, the reading on the machine is greater than the usual weight W. The person will feel heavier as the lift ascends.

1. When the lift is ascending but at constant velocity.

 

R−mg=NetforceR−mg=0R=mg

(Note that in this case, the reading recorded on the weighing machine is the exact weight of the body at that location)

iii. When the lift is descending with acceleration a

Mg−R=maR=mg−maR=m(g−a)

In this case, the person will feel lighter as the lift descend.

However, if the lift is descending with acceleration a  =  g

R=m(g−g)R=m(0)R=0

In this case, the person will experience weightlessness (free fall). In fact, he will float around in the lift

 

Worked Problems:

1. A tight rope walker of mass 60 kg stands in the middle of a rope and such that at his feet, the rope makes angle 5⁰ to the horizontal. Calculate the tension in the rope.

mg=Tsin5+Tsin560×10=T(2×0.087)6000.174=TT=3448N

Worked Example 2:

A brave but inadequate rugby player is being pushed backward by an opposing player who is exerting a force of 800 N on him. The mass of the losing player is 90.0 kg and he is accelerating at 1.2 m/s². What is the force of friction between the losing player’s feet and the grass?

 

ma=800−fr90×1.2=800−frfr=800−108fr=692N

Worked Example 3:

A woman of mass 57 kg stands on a weighing machine inside a lift ascending at 0.2 m/s². What is the reading on the machine?

 

R−mg=maR=ma+mgR=m(a+g)R=57(0.2+10)R=57×10.2R=581.4N

 

GENERAL EVALUATION

1. A bullet of mass 0.045kg is fired from a gun of mass 9kg, the bullet moving with an initial velocity of 200m/s. Find the initial backward velocity of the gun.

 

MECHANICAL ENERGY

CONTENT

1. The Principle of Conservation of Linear Momentum
2. Types of Collisions

 

The Principle of Conservation of Linear Momentum

In a closed system of colliding bodies, the total momentum after the collision is equal to the total momentum before the collision provided there is no net external force acting on the system.

Case 1: Consider two bodies A and B of masses m₁ and m₂ moving in the same direction with velocities u₁ and u₂ respectively. After collision their velocities were V₁ and V₂ as shown below.

 

Applying the principle of conservation of linear momentum,

M1U1+M2U2=M1V1+M2V2

Case 2: Consider two bodies A and B of masses m₁ and m₂ moving towards each other with velocities u₁ and u₂ respectively. After collision there velocities were V₁ and V_{2 .}

M1U1+M2U2=M1V1+M2V2

Assuming M1U1>M2U2,

Applying the principle of conservation of linear momentum,

M1U1+M2U2=(M1+M2)V

Physics SS2 Second Term Lesson Note

CALCULATIONS:

Example 1:

A trolley of mass 4kg moving on a smooth horizontal platform with a speed of 1.0ms^-1 collides perfectly with a stationary trolley of the same mass on the same platform. Calculate the total momentum of the two trolleys immediately after the collision.

Solution:

M_{1 =} 4kg,  U₁ = 10m/s,   U₂ = 0,   M_{2 =} 4kg

Applying the principle of conservation of linear momentum,

Total momentum before the collision  = total momentum after the collision.

M1U1+M2U2=M1V1+M2V2SinceU2=0,M2U2=0

Momentum after collision (M1V1+M2V2)=M1U1=4×1=4.0kgms−1

Example 2:

A ball of mass 0.5kg moving at 10ms^-1 collides with another ball of equal mass at rest. If the two balls move off together after the impact, calculate their common velocity.

M_{1 =} 0.5kg,  U₁ = 10m/s,   U₂ = 0,   M_{2 =} kg

Solution:

Applying the principle of conservation of linear momentum,

Total momentum before the collision  = total momentum after the collision.

M1U1+M2U2=(M1+M2)V0.5×10+0.5×0=(0.5+0.5)V5=1.0VV=5m/s

 

EVALUATION

1. State the following:

(a) Newton’s third law of motion

(b) Principle of conservation of linear momentum.

2. A ball P of mass 0.25kg loses one-third of its velocity when it makes a head-on collision with an identical ball Q at rest. After the collision, Q moves off with a speed of 2m/s in the original direction of P. Calculate the initial velocity of P.

 

Types of Collisions

There are two major types of collisions, elastic and inelastic collisions.

Elastic Collision:

In an elastic collision both momentum and kinetic energy are conserved. This means that for two colliding bodies with masses m₁ and m₂  and initial velocities u₁ and u₂ and final velocities after collision V₁ and V₂ ,

M1U1+M2U2=M1V1+M2V212M1U21+12M2U22=12M1V21+12M2V22

An example of perfectly elastic collision is a ball which bounces off the ground back to its original height.

Inelastic Collision:

In this case momentum is conserved but kinetic energy is not conserved. The energy lost is usually converted to heat, sound or elastic potential energy.

M1U1+M2U2=M1V1+M2V212M1U21+12M2U22≠12M1V21+12M2V22

Example 3:

A body of mass 5kg moving with a velocity of 20m/s due south hits a stationary body of mass 3kg. If they move together after collision with a velocity v due south, find the value of v.

Solution:

Applying the principle of conservation of linear momentum,

Total momentum before the collision = total momentum after the collision.

M1U1+M2U2=(M1+M2)V5×20+3×0=(5+3)V100=8VV=12.5m/s

 

EVALUATION

1. Distinguish between perfectly elastic collision and perfectly in elastic collision.
2. A tractor of mass kg is used to tow a car of mass kg. The tractor moves with a speed of 3.0m/s just before the towing rope becomes taut. Calculate the

(i) speed of the tractor immediately the rope becomes taut.

(ii) loss in kinetic energy of the system just after the car starts moving.

MECHANICAL ENERGY: SIMPLE MACHINES

CONTENT

1. Definition of Machine and Simple Machine
2. Mechanical Advantage/Force Ratio
3. Velocity Ratio
4. Efficiency of a Machine
5. Lever
6. Pulley
7. Inclined Plane
8. Hydraulic Press
9. Screw
10. Wheel and Axle
11. Gear

Physics SS2 Second Term Lesson Note

Definition of Machine and Simple Machine

An arrangement by which work can be done conveniently on a load or against a resistance is known as machine 

A simple machine is a machine in its simplest form. They are devices that use energy to do work. The work is being done by the machine when a small effort is used to overcome a large resistance.

A machine can be defined as a tool or devices that allows a force (or effort) applied to one point to overcome a resisting force (or load) at another point.

Simple machine can be classified into different categories namely the lever, the pulley, the inclined plane, the wedge, the wheel and axle, the screw, the hydraulic press.

Some common examples of simple machines are, scissors, drill brace, the shovel (a form of lever), the pulley at the top of a flagpole, the steering wheel of an automobile (a form of wheel and axle), and the wheelchair ramp (a form of inclined plane). An everyday example of a complex machine is the can opener, which combines a lever (the hinged handle), a wheel and axle (the turning knob), and a wedge (the sharpened cutting disk).

 

Mechanical Advantage/Force Ratio

Mechanical advantage/Force ratio of a machine is defined as the ratio of the load to the effort.

Mechanical Advantage =LoadEffort=Le

If the laod is bigger than the effort, the mechanical Advantage is greater than one.

In pratice,all machines have some friction in them and this reduces the efficiency. Part of the work put into a machine is thus always wasted in overcoming friction and in moving some parts of the machine. Thus, no machine is hundred percent (100%) efficient.

 

Velocity Ratio

The velocity ratio of a machine is defined as the distance moved by the effort to the distance moved by the load.

Velocity Ratio =Distance moved by EffortDistance moved by Load=dedL

Physics SS2 Second Term Lesson Note

Efficiency of a Machine

The efficiency of a machine is defined as the ratio of work obtained from the machine to work put into the machine expressed in percentage.

It also defined as the ratio of work output  of the machine to the total work input expressed in percentage.

Efficiency =Work outputWork input×100

Derivation of the Formula

Efficiency E=MAVR×100

The efficiency of a machine can be determined by taking ratio of the work output to work input of the machine or its velocity ratio and mechanical advantage.

Therefore:

Efficiency (E)=Work outputWork input×100

Efficiency (E)=Work done in loadWork done in effort×100

Since Work=force×distance

Efficiency

(E)=Load(L)×distance moved by load(dL)Effort(E)×distance moved by effort(de)×100

Therefore, E=L×dLe×(de)×100

But MA=Leand1VR=dLde

Therefore, E=M.AV.R×100

Mechanical Advantage (M.A)=Le

Efficiency (E)=MechanicalAdvantageVelocityRatio×100

Example 1:

A system of lever with velocity ratio 30 overcomes resistance of 2500 Newton when an effort of 125 Newton is applied to it, calculate

• The mechanical advantage of the system.
• It’s efficiency.

Solution:

Mechanical Advantage =LoadEffort=2500125=20N

Efficiency =M.AV.R×100

Efficiency =2030×100=2003=66.75

GENERAL EVALUATION

1. Explain what is meant by a machine.
2. Define the terms: mechanical advantage, velocity ratio and efficiency as applied to a machine.
3. Show that the efficiency E, the force ratio M.A and the velocity ratio V.R of a machine are related by the equation E=M.AV.A×100
4. Explain why the efficiency of a machine is usually less than 100%.

 

Lever

In a lever the relative positions of Force(F), Effort(e) and Load(L) may vary and this leads to different types of lever. The lever operates on the principle of moment.

First Order Lever

In first order lever, the fulcrum is between the load and the effort e.g crowbar, claw hammer,  pliers, scissors, see-saw e.t.c

 

 

Physics SS2 Second Term Lesson Note

Taking moment about F gives;

Clockwise moment = anticlockwise moment

Y×L=X×ELE=XY=M.A=V.R

Second Order Lever

In second order lever, the load is between the effort and the fulcrum. E.g wheelbarrow, bottle opener, nut cracker[/vc_wp_text][/vc_column][/vc_row][vc_row][vc_column width=”1/3″][vc_hoverbox image=”918″ primary_title=”” hover_title=”Lesson Notes” hover_btn_title=”click” hover_btn_color=”warning” hover_add_button=”true” hover_btn_link=”url:https%3A%2F%2Fultimosedusupport.com%2Fprimary-one-2%2F|||”]Primary School lesson notes, all subjects.[/vc_hoverbox][/vc_column][vc_column width=”1/3″][vc_hoverbox image=”912″ primary_title=”” hover_title=”Lesson notes” hover_btn_title=”click” hover_btn_color=”warning” hover_add_button=”true” hover_btn_link=”url:https%3A%2F%2Fultimosedusupport.com%2Fjss1-2%2F|||”]Lesson  notes and plan for secondary schools[/vc_hoverbox][/vc_column][vc_column width=”1/3″][vc_hoverbox image=”914″ primary_title=”” hover_title=”Question Bank” hover_btn_title=”click” hover_btn_color=”warning” hover_add_button=”true” hover_btn_link=”url:https%3A%2F%2Fultimosedusupport.com%2Fprimary-one%2F|||”]Question bank for primary[/vc_hoverbox][/vc_column][/vc_row][vc_row][vc_column width=”1/2″][vc_hoverbox image=”920″ primary_title=”” hover_title=”Question Bank” hover_btn_title=”click” hover_btn_color=”success” hover_add_button=”true” hover_btn_link=”url:https%3A%2F%2Fultimosedusupport.com%2Fjss1-question-bank%2F|||”]Question Bank for Secondary School[/vc_hoverbox][/vc_column][vc_column width=”1/2″][vc_hoverbox image=”923″ primary_title=”” hover_title=”work sheet” hover_btn_color=”warning” hover_add_button=”true” hover_btn_link=”url:https%3A%2F%2Fultimosedusupport.com%2Fwork-sheet%2F|||”]Worksheet for all levels[/vc_hoverbox][/vc_column][/vc_row][vc_row][vc_column][vc_wp_text]

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