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# Law of Conservation of Momentum

Before we talk about conservation of momentum, let us first of  all  define momentum. Momentum is the products of mass and velocity. For example if a body of mass, M, is moving with velocity v, then the momentum of the body is = m x v.

From the Newton’s second and third law of motion, the law of conservation of momentum is formulated.

Law of conservation of momentum: it state that  in close system of colliding objects, the total momentum is conserved , provided there is no net external force acting on them or it can also be stated as follow:

The total momentum of an isolated or closed system of colliding bodies remains constant, thus if two or more collide in a closed system, the total momentum before collision is equal to total momentum after collision.

By closed or isolated system we mean that in the system there is no external force acting on the system.

# Equation of law of conservation of momentum

Let u1 u2, v1, v2 be the initial and final velocities of the two colliding bodies with mass m1 and m2, therefore the law of conservation of momentum can be written mathematically as:

M1u1 +m2u2 = m1v1 + m2v2 (law of conservation of momentum)

The above equations of law of conservation of momentum show that, action of body1 and on body2 is equal to the reaction body2 on body1, and that both forces acted for an exactly equal time.

The equation of law of conservation of momentum above is peculiar to the collision which is elastic one. But for inelastic one, where the two colliding bodies moves with common velocity after collision, that equation of law of conservation of momentum can be written as:

m1u1 + m2u2 = v(m1 +m2) only in inelastic collision where v1= v2 =v

# Elastic and Inelastic collision

In law of conservation of momentum we consider, the two type of collision, elastic and inelastic collision

In an elastic collision, both momentum and kinetic energy are conserved , this means that for two colliding bodies with mass m1, and m2, initial velocities u1 and u2 and final velocities v1 and v2, then

m1u1 +m2u2 =m1v1 +m2v2 (law of conservation of momentum in an elastic collision)

½ m1u12 +½m2u22 =½m1v12 + ½m2v22 (Energy conservation in an elastic collision)

The example of perfect elastic collision is a ball which bounce off the ground back to it original height.

# Inelastic collision

In case of inelastic collision momentum is conserved but kinetic energy is not. This is because in inelastic collision kinetic energy decreases as some are converted to heat, sound or elastic potential energy

In complete inelastic collision the two objects or bodies join together after collision and move together with common velocity v, so the equation of law of conservation of  can be written as:

M1u1 +m2u2 =v(m1 +m2)

Kinetic energy before collision = ½m1u12 +½m2u22

Then after collision kinetic energy =½ (m1 + m2)v2

Consider a special case when the body m2 is at rest, then

K.E1 = ½m1u12

K.E2 = ½(m1 + M2)v2 $but\quad v\quad =\frac { { m }_{ 1 }{ u }_{ 1 } }{ { m }_{ 1 }{ +\quad m }_{ 2 } }$ $therefore\quad K.{ E }_{ 2 }=\frac { 1 }{ 2 } (\frac { { m }^{ 2 }{ u }^{ 2 } }{ { m }_{ 1\quad }+\quad { m }_{ 2 } } )$ $then\quad \frac { { K.E }_{ 1 } }{ { K.E }_{ 2 } } =\frac { { { m }_{ 1 } } }{ { m }_{ 1 }{ \quad +\quad m }_{ 2 } }$

This shows that the final kinetic K.E2 of the body is less than the initial energy, K.E1. This statement in general is also true even if u2 is not zero.

Examples on the law of conservation of energy

1. A body of mass 4kg moving with a velocity of 10m/s, collides with a stationary body of mass 6kg. If the two bodies move together after the collision, calculate their common velocity.

Solution, m1 =4kg, u1 =10m/s, u2 = 0, m2 =6kg, then using

M1u1 +m2u2 = v(m1 +m2) =4 x 10 + 6 x 0 = v (4 + 6)

40 +0  = 10v

10v = 40

V = 4m/s (common velocity)

1. Example 2.

Object A of mass 20kg moving with a velocity of 3m/s makes a head-on collision with object B , mass 10kg moving with a velocity of 2m/s in opposite direction. If A and B stick together after collision, calculate their common velocity in direction of A

Solution

mA = 20kg, mB =10kg, uA = 3m/s, uB =2m/s, then using

mAuA +mBuB = v (mA + mB) = 20 x 3 + 10 x(- 2) = v ( 20 + 10 )

60 – 20 = v( 30) = 40 = 30v

V = 40/30 = 1.33m/s

V = 1.33m/s

Example 3

What is the change in momentum of body if a force of 6.0N acts on it for 5s

Force = the rate of change of momentum, which is given as $force\quad =\quad \frac { change\quad in\quad momentum }{ time } \quad that\quad is\quad \\ \\ F\quad =\frac { MV-Mu }{ t } ,\quad and\quad f\quad =6.0N,\quad t=\quad 5s\\ \\$ $then\quad \\ 6\quad =\frac { mv-mu }{ 5 } \\ mv-mu\quad =\quad 6\times 5\quad =30Ns\\ \\ \\$ 