Lesson Notes By Weeks and Term - Senior Secondary School 2
NEWTON'S LAWS OF MOTION
NEWTON'S LAWS OF MOTION
SUBJECT: PHYSICS
CLASS: SS 2
DATE:
TERM: 1st TERM
REFERENCE TEXT
WEEK FIVE
TOPIC: NEWTON'S LAWS OF MOTION
CONTENT
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
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.
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
GENERAL EVALUATION
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
(a) elastic and inelastic collisions (b) Inertial mass and weight
(c) Derive from Newtons law the relationship between Force, mass and acceleration.
READING ASSIGNMENT
New Sch Physics FOR SSS-ANYAKOHA PAGES-161—165.
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