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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