NEWTON’S LAW OF MOTION

NEWTON’S LAWS OF MOTION

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

 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 of the force.

F α mv –mu

t

F α m (v –u)

t

F α ma

F = kma

Where k =1

F =ma

MOMENTUM

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

Impulse is the product of a force and time. It is also defined as the change in momentum. Thus both momentum and impulse have ‘Ns’ as unit

F = m (v-u)/t

Ft = mv – mu (where ‘mv-mu’ is the change of momentum)

F x t = I (Ns)

Newton’s third law of motion states that to every action, there is an equal but opposite reaction. A 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, if: m= mass of bullet, v= velocity of bullet, M=mass of gun, V= velocity of the recoil of the gun.

Then, the velocity, V, of the recoil of the gun is given by:

MV = mv

V = mv/M

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 force, the total momentum after collision remains constant. The principle is true for both elastic and inelastic collision.

COLLISION

There are two types of collision- elastic and inelastic.

In elastic collision the two bodies collide and then move with different velocities. Both momentum and kinetic energy are conserved e.g. 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

In an inelastic collision, the two bodies join together after the collision and with the same velocity. Here, momentum is conserved but kinetic energy is not conserved 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 = v (m₁ +m₂)

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

V_A              V_B

 

M_A                        M_B                        M_A M_B

BEFORE COLLISION             AFTER COLLISION

M_AV_A + M_BV_B = V (M_A + M_B)

V= common velocity

V= M_AV_A + M_BV_B

(M_A + M_B)

 

TWO BODIES TRAVELLING IN OPPOSITE DIRECTION

=

M_A    M_B                                           M_A M_B

M_AV_A – M_BV_B = V (M_A + M_B)

V= M_AV_A + M_BV_B

M_A + M_B

 

COLLISION BETWEEN A STATIONARY AND MOVING BODY

 

V_A                                         =                 V

M_A            M_B                             M_AM_B

 

The momentum of a stationary body is zero because velocity is zero

M_AV_A + 0 = V (M_A + M_B)

V =   M_AV_A

M_A + M_B

EXAMPLE

1. Two moving toys of masses 50kg and 30kg 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.

M_AV_A + M_BV_B = V (M_A + M_B)

V= M_AV_A + M_BV_B

(M_A + M_B)

V = (50 x 5) + (30 x 3)

50 + 30

V = 250 + 90

80

V = 340

80

V = 4.05 m/s

2. Two balls of masses 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 ball has a speed of 1m/s in the opposite direction. What is the speed of the second ball after collision

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

 

 

0.5              0.3                                 0.5              0.3

Before                                                         After

3×0.5 + (0.3 x-4) = 0.5 (-1) + 0.3v

• – 1.2 = -0.5 + 0.3v

0.3v = 2.0 – 1.2

V = 0.8 / 0.3

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?

Solution

Momentum   gun = momentum of bullet

MV = m v

10 x V = 0.002 x 400

V = 0.002 x 400

10

V= 0.8 m/s

ONLINE WORK

1. Derive from Newton’s law the relationship between Force, mass and acceleration
2. State Newton laws of motion and explain the consequences of each law
3. State the principle of conservation of linear momentum.
4. 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).

ASSIGNMENT

SECTION A

1. A force acts on a body for 0.5s changing its momentum from 16kgms^-1 to 21kgms^-1. Calculate the magnitude of the force (a) 42N (b) 37N (c) 32N (d) 10N
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 machine gun with mass of kg fires a 50g bullet at a speed of 100m/s. The recoil speed of the machine gun is (a) 0.5m/s (b) 1.0m/s (c) 1.5m/s (d) 2.0m/s
4. When taking a penalty kick, a footballer applies a force of 30N for a period 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 jet engine develops a thrust of 270N when the velocity of the exhaust gases relative to the engine is 300m/s, what is the mass of the gas ejected per second? (a) 81.00kg (b) 9.00kg (c) 0.90kg (d) 0.09kg

SECTION B

1. An object of mass 5kg slides down a smooth plane at an angle of 25⁰. If the object starts from rest, find; (i) its velocity after 3m (ii) its momentum 3m from the starting point (iii) the force causing it moves (g=m/s²)
2. State the law of conservation of linear momentum. A 3kg rifle lies 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.
3. A bullet of mass 120g is fired horizontally into a fixed wooden block with a speed of 20m/s. The bullet is brought to rest in the block in 0.1s by a constant resistance. Calculate the: (i) magnitude of the resistance (ii) the distance moved by the bullet in the wood.

 

See also

PROJECTILES AND ITS APPLICATION

DERIVATION OF EQUATONS OF LINEAR MOTION

SCALAR AND VECTOR QUANTITIES

POSITION

ELECTROMAGNETIC FIELD