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.
| 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.
*** Conservation of momentum definition by centre of mass
We have just written the equation
Vcom = (m₁v₁+m₂v₂+m₃v₃+…………..+mₙvₙ)/(m₁+m₂+m₃+……….+mₙ)
Now if total external force on the system is zero then Fnet = 0
What does it mean individual force on the particles are zero ? answer is Big no it indicate that total force on the system is zero.
It also indicate that centre of mass acceleration is zero that is if Fnet = 0 then acom = acceleration of centre of mass = 0 .
It also indicates that Vcom = velocity of centre of mass is constant
Now from the equation
Vcom = (m₁v₁+m₂v₂+m₃v₃+…………..+mₙvₙ)/(m₁+m₂+m₃+……….+mₙ) = constant so arrange
m₁v₁+m₂v₂+m₃v₃+…………..+mₙvₙ = (m₁+m₂+m₃+……….+mₙ)*constant
here mass total mass is also constant so finally we can write
m₁v₁+m₂v₂+m₃v₃+…………..+mₙvₙ = constant
p₁+p₂+p₃+……………………+pₙ = constant or we can write
Psystem = constant
Hence this prove that if total external force on a system Fnet =0 is zero then momentum of the system remains constant.
So momentum will remain conserved and this is called conservation of momentum. I hope you have got the concept behind conservation of momentum.
Conservation of momentum for internal particle
Don’t be confuse internal particle momentum can change one ball will hit other ball and there momentum can change, But the whole system momentum will not change, It will remain constant forever if external force on the system is zero.
Conservation of momentum by Newton’s second law
Newton’s second law state that rate of change of momentum is equal to net force
Fnet = ma = mdv/dt = dp/dt ( we know p = mv or dp = mdv partial differentiation )
Fnet = dp/dt ( This is applicable for a single particle as well as for a system )
Now if external force Fnet = 0 then we can also write dp/dt = Fnet = 0 so
dp = 0 or p = constant or Psystem = constant.
Hence Newton’s second law conservation of momentum.
Conservation of momentum application
When we fire a bullet conservation of momentum is apply here before firing bullet gun and bullet momentum is zero, as we fire bullet then bullet got a momentum with high speed to cancel this momentum gun get backward same momentum so that momentum is conserved.
How momentum is produce ? force is cause to produce the momentum without force momentum can’t be produce.
Rocket lunch also work on conservation of momentum it thrash the smoke back to get momentum forward so that the momentum can be conserved.
There are so many application of conservation of momentum.
Numerical problems based upon conservation of momentum.
See the picture below and find the velocity of mass M.
As from the question nuclei was initially at rest no any external force acting on nuclei system hence its initial momentum will be zero . Now as the alpha particle will come out so it will get momentum in forward direction,now to remain momentum conserve remaining nuclei will get jerk in backward direction.
So we can apply the conservation of momentum as net external force Fnet =0 on the nuclei.
Hence initial momentum = final momentum or Pi = Pf so we can write here Pi =0
0 = mv -Mu (here direction of velocity of nuclei is in opposite so it is minus )so u = mv/M this will be velocity of remaining nuclei in the backward direction.
Similarly you can use conservation of momentum for gun case initially gun was at rest so Pi = Pf here Pi =0 hence use the equation .
0 = mv -Mu or u = mv/M
I hope you have got how we use here conservation of momentum.
How to solve question by Conservation of momentum
From first question car is already moving with velocity v, It is not necessary always system or body will be in rest position, here car and cat to be consider as system.
Cat drop on the roof of moving car with very slowly u =0 or zero velocity. So here momentum of the system will remain constant.
Hence apply conservation of momentum Pi = Pf now Pi is initial momentum of the system Pf is final momentum of the system.
Pi = initial momentum of cat + initial momentum of car = m*0 +M*v = Mv
Pf = final momentum of cat and car together = (m+M)u here u is velocity of car after landing of cat on roof of car. Some momentum of car will be taken by cat so some momentum of car will decrease.
now equate the initial and final momentum Pi = Pf
Mv = (m+M)u or u = Mv/(m+M) this will be new speed of car after landing cat.
Now from second question trolley and man are initially at rest. As shown from the picture man is standing on trolley.
there is no any external force on the system hence momentum of the system will remain constant, so we can apply conservation of momentum.
Pi = Pf here Pi =0 because initially system is at rest
Pf = final momentum of man + final momentum of trolley .
Here i will ask what do you think about final momentum of man ? is it will be mass of man and product of its velocity that is 20*2 = 40 ? Big no
If you think like that then you are wrong but why ?
Here speed of man given with respect to trolley and if suppose trolley will move with velocity u in backward direction it will be with respect to ground.
So two different frame of reference you are doing calculation which is wrong . So we will have to take same frame of reference for trolley and man ground as frame of reference.
Hence we need to calculate velocity of man with respect to ground which will be (2-u) because man is moving with 2m/s forward on trolley but trolley is moving backward with u velocity with respect to ground.
Hence a person standing on the ground and observing velocity of man observer will observe slow speed of man and that will be (2-u).
Hence we can find the final momentum Pf = 20*(2-u) -10*u = 40 -20u-10u = 40-30u
Pi = Pf = 0 = 40-30u or u = 40/30 = 4/3 = 1.33m/s
So this will be speed of trolley in backward direction.
Hence this is the concept you will solve question by conservation of momentum.
From the above picture question is given as
Q m slide on 10m without any friction and 10m is kept on frictionless surface . Find the velocity of m just reaches the bottom surface.
Solution Think and tell me we can apply here conservation of momentum .Is net force on the system is zero ? Answer is no net force is not zero, normal contact force between mass m and 10m is zero but mg external force is acting on the mass m which is external force.
What is the law of the conservation of momentum?
So it is clear that external net force is not zero. Then what do you think now we can apply conservation of momentum yes or no.
Answer is yes we can apply conservation of momentum how ? see mg external force is acting along y axis but net force along x axis is zero so we can apply conservation of momentum along x axis, this is concept remember, momentum is a vector quantity so Conservation of momentum can apply along any particular direction whenever net force along that direction is zero.
Now suppose when m block reaches at bottom its velocity is v with respect to ground and this block pushes 10m in back direction hence 10m velocity is u in backward direction with respect to ground.
Conservation of momentum along x direction.
Pix = Pfx initially system is at rest hence Pix = 0 now final momentum
Pfx = mv – 10mu = 0 or v = 10u now here two variable v and u but only one equation so we will find one more equation.
Now think when block reaches at bottom point then mass m has kinetic energy and 10m has also kinetic energy, from where this kinetic energy has came. So we see that block m lost its potential energy and that potential energy converted into kinetic energy.
So we can apply conservation of mechanical energy. Yes we can apply conservation of mechanical energy here because mg force is conservative force but you should remember that conservation of mechanical energy is apply only where all forces must be conservative and no any external non conservative force acting on the system.
So we can apply here conservation of mechanical energy Mi = Mf
Ui +KEi = Uf +KEf
mgR +0 = 0+ 1/2mv² +1/210m*u²
gR = v²/2 +10u²/2 now replace u = v/10 and put g = 10m/s², R = 22m and solve the equation
v² = 20*20 v = 20m/s this is velocity of mass m at bottom.
So you have got the concept for conservation of momentum use in problem.
Now we will continue in next post. I hope you have enjoyed learning Conservation of momentum. If you like comment and share, thanks for reading and sharing learn and grow.
Dated 17th Nov 2018
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