Lesson Notes By Weeks and Term - Senior Secondary School 2

NEWTON'S LAWS OF MOTION

SUBJECT: PHYSICS

CLASS:  SS 2

DATE:

TERM: 1st TERM

REFERENCE TEXT

• New School Physics by M.W Anyakoha
• SSCE WAEC Past Questions
• UTME Past Questions

WEEK FIVE

TOPIC: NEWTON'S LAWS OF MOTION

CONTENT

• Newton’s laws motion
• Conservation of linear momentum

NEWTON’S FIRST LAW OF MOTION states “that everybody continues in its state of rest or of uniform motion in straight line unless it is acted upon by a force.”

This simply means that a body at rest will remain permanently there or a body moving with uniform velocity on a straight line will continue moving forever if it were possible for all the external opposing forces to be eliminated.

The tendency of a body to remain at rest or, if moving, to continue its motion in a straight line is called the inertia of the body. That is why Newton’s first law is otherwise referred to as the law of inertia

There are consequences of this law. For example, when a car had a head on collision with another car or the driver suddenly applies the brake, the passengers are likely to be injured when they hit the windscreen.

The reason is that an external force will only stop the car but not the passengers who tend to continue their linear motion. This necessitated, the use of safety precautions e.g seat belt

Also, if a stationary car is knocked forward from behind, the passengers may sustain neck injuries as their bodies tend to move forward in relation to the car while their neck move backward. Modern cars have rests to prevent this incident

NEWTON’S SECOND LAW OF MOTION states “that 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.” This implies that when a heavy body and a light one are acted upon by the same force for the same time, the light body will build up a grater velocity than the heavy one. But they gain the same momentum

F α mv -mu

t

F α m (v –u)

t

F = K m (v –u)

t

where k is a constant, if v = 1, u = 0, t= 1, m =1 and F=1 ,then k=1

but a = v – u

t

F= kma    but k= 1

F =ma   …………………………………….   1

Momentum of a body is the product of the mass and velocity of the body.

The S.I unit of momentum is  kgm/s.

IMPULSE

The impulse of a force and time. This impulse is also equal to the change in momentum and they therefore share the same unit (Ns)

F = m (v-u )

t

Ft = mv – mu    …………………………..  2

F x t = I  (Ns)   ………………………..  3

mv- mu = change in momentum  …………………… 4

NEWTON’S THIRD LAW OF MOTION states that to every action, there is an equal but opposite reaction.  when a book is placed on a table, the downwards weight (force) of the book on the table is balanced by the upwards reaction of the table on the book.

Another practical demonstration of this law can be observed when a bullet is fired from a gun, the person holding it experiences the backward recoil force of the gun (reaction) which is equal to the propulsive force (action ) acting  on the bullet.

According to Newton second law of motion, force is proportional to change in momentum,therefore the momentum of the bullet is equal and opposite to the momentum of the gun i.e.

mass of bullet x muzzle velocity = mass of gun x recoil velocity,

hence, for a bullet of mass in and muzzle velocity , v the velocity v of the recoil of the gun is given by

MgVg  =mbvb  ……………………………………….. 5

Vg = mbvb /Mg

EVALUATION

1. State Newtons laws of  motion
2. Mention and explain the consequencies of each law stated above

CONSERVATION OF LINEAR MOMENTUM

The principle of conservation of linear momentum  states that when two or more bodies  collide, their  momentum remain constant  provided  there is  no  external  force  acting on the system.

This  implies  that in a closed  or  isolated system where there is no external  forces , the total  momentum  after collision  remains constant.

The principle is true for both elastic and inelastic collision.

TYPES OF COLLISION

There are two types of (a) collision- elastic and   (b) inelastic.

IN ELASTIC COLLISION  where  the two bodies and then  move with different velocities , both momentum and kinetic  energy are conserved eg collision between  gaseous particles , a ball which rebounds to its original height etc .If the two colliding bodies have  masses m1and m2 initial  velocities u1 and u2 and final  velocities v1 and v2 , , the  conservation  principle can be mathematically expressed as

m1u1 + m2u2  =  m1v1 + m2v2 ……………………………………. 6

IN  AN INELASTIC COLLISION , the  two bodies  join  together  after the collision and with the same  velocity momentum is conserved but  kinetic  energy   is not  conversed because  part of it has been converted  to heat or sound  energy, leading to deformation

Thus, the  conversation  principle can be re-written  as

m1U1 + m2U2 = V12( m1 +m2)   ……………………………….7

V12 = common velocity

Since momentum is a vector quantity, all the velocities must be measured  in the same direction , assigning positive signs to the forward velocities and negative signs to the backward or opposite  velocities .

TWO BODIES MOVING IN THE SAME DIRECTION BEFORE COLLISION

VA              VB

MA                          MB                              MA MB

BEFORE COLLISION             AFTER COLLISION

MAVA + MBVB = VAB (MA + MB )…………………………. 8

VAB= COMMON VELOCITY

V= MAVA + MBVB

MA+  MB

TWO BODIES TRAVELLING IN OPPOSITE DIRECTION

=

MA   MB                             MA MB

MAVA – MBVB = VAB( MA+ MB )  ………………………………………. 9

VAB= MAVA – MBVB

MA + MB

COLLISION BETWEEN A  STATIONARY  AND MOVING BODY

VA                                          =    V

MA                        MB                             MAMB

The momentum  of a stationary  body is zero

MAVA + 0 = MAVA / MA + MB.

Worked example.

1. Two  moving toys of masses 50kg  and 30 kg  are traveling on the same plane  with speeds of 5 m/s  and 3 m/s respectively in  the same  direction  .If they  collide and  stick  together , calculate  their  common  velocity.

MAVA + MBVB = V( MA +MB

V = MAVA + MBVB

MA + MB

= 50  X 5  + 30 X 3

50 + 30

= 250 + 90

80

= 340

80

= 4.05 m/s

2    Two  balls of masses  0.5 kg and 0.3kg move towards  each other  in the same  line at  speeds of  3 m/s  and 4 m/s  respectively .After  the collision , the  first  balls  has a  speed of 1m/s in the opposite direction

What is the  speed of the second  ball after collision?

3m/              4m/s                    1m/s              V

.5             .3                                  .5                   .3

Before                                                            After

3x0.5 + (0.3 x-4 ) = 0.5 (-1) + 0.3v

1.5 -1.2 = -0.5 + 0.3 v

0.8 /0.3 =V,                           V = 2.7m/s

3    A gun of mass 100kg  fires a  bullet of mass 20g at a speed of 400m/s .What is the  recoil  velocity of the gun ?

Momentum   gun = momentum of bullet

M V = m v

10 x V = 0.002 x 400

V = 0.002  x 400

10

V= 0.8 m/s.

EVALUATION

1. State the principle of conservation of linear momentum.
2. Explain elastic and inelastic collision, and give two example of each

GENERAL EVALUATION

1.  State Archimedes principle.
2.  A 15kg monkey hangs from a cord suspended from the ceiling of an elevator. The cord can withstand a tension of 200N and breaks as the elevator accelerates. What was the elevators minimum acceleration (g=10m/s2).

WEEKEND ASSIGNMENT

1        A ball of mass 0.5kg moving at 10m/s collides with another ball of equal mass at rest. If the two balls move off together after  the impact, calculate their common velocity    (A) 0,2m/s  (B)0.5m/s  (C) 5m/s  (D) 10m/s.

2        A ball of mass 6kg moving with a velocity of 10m/s collides with a 2kg ball moving in the opposite direction with a velocity of 5m/s. After the collision the two balls coalesce

and move in the same direction. Calculate the velocity of the composite body.

(A) 5m/s  (B) 6.25m/s  (C) 8.75m/s  (D) 12m/s

3        A constant force of 5N acts for 5 seconds on  a mass of 5kg initially at rest. Calculate the final momentum. (A) 125kgm/s  (B) 25kgm/s  (C) 15kgm/s  (D) 5kgm/s.

4        When taking a penalty kick, a footballer applies a force of 30N  for a periods of 0.05S. If the mass of the ball is 0.075kg, calculate the speed with which the ball moves off. (A) 4.5m/s  (B) 11.25m/s  (C) 20m/s  (D) 45m/s.

5     A body of mass 40kg changes its velocity from 10m/s to 80m/s in 10 seconds. Calculate the force acting on the body.    (A) 480N  (B) 380N  (C) 280N  (D) 180N.

THEORY

1. State the law of conservation of linear momentum. A 3kg rifle lays on a smooth table when it suddenly discharges, firing a bullet of 0.02kg with a speed of 500m/s. Calculate the recoil speed of the gun.
2. Distinguish between:

(a) elastic and inelastic collisions        (b) Inertial mass and weight

(c)  Derive from Newtons law the relationship between Force, mass and acceleration.