conservation of momentum
Conservation of momentum is very important topic of Physics because conservation of momentum concept state second law of Newton.We will see how Newton's law was derived from conservation of momentum.In our previous post we have already studies about conservation of momentum formula and its basic concept, You can refer the previous post for basic concept and definition of conservation of momentum concept for conservation of momentum. In this topic we will cover the application of conservation of momentum as well as its numerical problems.So lets start.
Conservation of momentum for a system
Linear conservation of momentum, we know that any rigid bodies system is made by collection of many small particles, Take an example of rotating wheel of a car, Another example rotating ceiling fan.
When we observe this rotating bodies, Then we found that there are many small particle, Which are rotating with different speed in different radius.
If anybody ask find the speed of the rotating wheel,Then what will be your answer, Which small particle speed you will answer this is really difficult, So to make this simple, we answer the speed of center of mass of the rigid body.
Because whole mass of the rigid body is concentrated at the centre of mass. we have already learn in previous post . So whole body is consider as a system and its internal small particle conservation of momentum gives the whole body conservation of momentum.
We will see how these two conservation of momentum is same, just wait see the below picture.
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| Conservation of momentum |
In Centre of mass post we have studies that how to find the centre of mass for small point mass, So i am just using that concept.
We know that centre of mass general formula is given as rcom (centre of mass) .
rcom = (m₁r₁+m₂r₂+m₃r₃+......................+mₙrₙ)/(m₁+m₂+m₃+................+mₙ)
Now differentiate this equation with respect to time on both side then we will get
d(rcom)/dt = (m₁dr₁/dt+m₂dr₂/dt+m₃dr₃/dt+..............+mₙdrₙ/dt)/(m₁+m₂+m₃+..........+mₙ)
we know dr/dt = velocity = v so we can write the above equation as
Vcom = (m₁v₁+m₂v₂+m₃v₃+..............+mₙvₙ)/(m₁+m₂+m₃+..........+mₙ)
Now arranging the equation we can write like this.
(m₁+m₂+m₃+..........+mₙ)Vcom = m₁v₁+m₂v₂+m₃v₃+..............+mₙvₙ
(m₁+m₂+m₃+..........+mₙ) = total mass of rigid body =M as shown in above picture so we can write as
MVcom = m₁v₁+m₂v₂+m₃v₃+..............+mₙvₙ now as shown in above motion picture momentum of different small particles m₁v₁, m₂v₂, m₃v₃ hence If we want to write momentum of the system can write Psystem = MVcom . p = mv
Whenever we want to write momentum of a rigid body then we take whole mass of the body and velocity of centre of mass product as state above.Hence now we can write
→ → → → →
Psystem = p₁+p₂+p₃+........................+pₙ
Hence total momentum of a body is equal to vector sum of momentum of individual particles.
