Elastic collission one dimension - Work, Energy & Power Video Lecture - Class 11

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elastic collision in one dimension so
when we say elastic collision we mean to
say that there is one object say of mass
m1 there is another body of mass M 2 so
initially M 2 2 is at rest M 1 moves it
hits M 2 then both of them starts moving
with some velocity say M 1 and M 2 so in
this case before collision and after
collision so before collision one ball M
1 is moving with velocity u 1 the other
goal M 2 is at rest that means its
velocity is zero after collision M 1 is
moving with some velocity V 1 and M 2 is
also started moving with some velocity B
now what we will see here is we will try
to find out the values of V 1 and V 2 in
terms of the known quantities that is M
1 is 2 and u 1 we will see that with
what velocity does the two objects start
moving after an elastic collision since
it is elastic collision therefore
momentum and sign and kinetic energy
both will be conserved now as per
momentum conservation the total linear
momentum before collision that is M 1 u
1 plus 0 because u 2 is 0 should be
equal to the total linear momentum after
collision that is m1 v1 plus m2 v2 so
this is our first equation now we apply
kinetic energy conservation now
according to in kinetic energy
conservation before collision the total
kinetic energy will be half M 1 u 1
square this should be equal to the total
kinetic
energy after pollution that is half em 1
V 1 square plus half M 2 V 2 square so
from here we can cancel all the house so
we get M 1 u 1 square is equal to M 1 V
1 square plus M 2 V 2 square so let us
say this is my equation 2 now what we do
we multiply equation 1 with V 2 so
multiplying equation 1 with we do we
will get M 1 u 1 V 2 is equal to M 1 V 1
V 2 plus M 2 V 2 square so say let us
say this is my equation C now what do I
do this is my equation 3 and this is my
equation 2 we see in both this equation
we got a common term M 2 V 2 square so
let us do one thing let us subtract
equation 3 from equation 2 so equation 2
minus equation 3 this will give M 1 u 1
square minus M 1 u 1 V 2 is equal to M 1
V 1 square minus M 1 V 1 V 2 because
these two terms will get cancelled like
so now here we can write M 1 G 1 into u
1 minus V 2 is equal to M 1 V 1 V 1
minus V T so M 1 M 1 will get cancelled
on both sides so we can write u 1 square
minus V 1 square is equal to u 1 V 2
minus V 1 V 2 or we can write u 1 square
minus V 1 square is equal to V 2 u 1
minus P 1
now we can write it as a square minus B
Square is a plus B into a minus B so u 1
plus T 1 into u 1 minus V 1 this is
equal to V 2 u 1 minus V 1
now you can cancel you 1 minus P 1 which
u 1 minus V 1 so this becomes V 2 is
equal to u 1 plus V 1 that is equation
number 4 now let us put equation 4 in
equation 1 so what do we get we get m1
u1 is equal to m1 v1 plus m2 u 1 plus t1
right it's nothing but simple
mathematics so we get m1 u1 is equal to
m1 v1 plus m2 u 1 plus M 2 3 1 again v1
let us take V 1 comment we take we will
take the terms with v1 on one side since
we take V 1 common if we are left with M
1 plus M 2 this becomes equal to M 1 u 1
minus M 2 u 1 so we have taken the u1
terms on the same side and b1 terms on
the same side now from this we get V 1
is equal to u 1 into m1 minus m2 divided
by m1 plus m2 so this is one thing which
we were intending to find out that is
this is the velocity with which ball 1
will move after collision so this is
this V 1 is the velocity with which ball
will move right so before
now let us find the velocity that is the
true with which ball the ball number two
that is called B will move after
collision so from Equation 4 we get v2
is equal to u1 plus v1 now it is very
simple
v1 you already know so we will put this
value of b1 in this equation so we get
this is equal to you find will take you
one common so this will be one plus m1
minus m2 divided by m1 plus m2 so this
will become m1 plus m2 plus m1 minus m2
divided by m1 plus m2 so this M 2 and M
2 will get cancelled so V 2 will become
2 M 1 u 1 divided by M 1 plus M 2 so
these are the values of velocities of
the two balls after an elastic collision
right so if you see very carefully we
did nothing we just applied conservation
of momentum and conservation of kinetic
energy and simple mathematics to solve
the equation we will now look at certain
special cases of what we did just now we
did just now collision in one dimension
for an elastic collision so we look at
the special cases the first case is when
m1 is equal to m2 then m1 is equal to m2
then what will be the value of V 1 and B
2 that is with what velocity then the
objects will move from the previous
expression we found that V 1 is equal to
u 1 m1 minus m2 divided by m1 plus m2
now if m1 is equal to
this will become 0 so B 1 B 0 and what
about V 2 B 2 was equal to 2m 1 u 1
divided by M 1 plus M 2 so this will
become 2 M 1 u 1 divided by 2 m 1 right
because M 1 and M 2 are equal so this
will become u 1 that means in this case
what will happen if both the masses are
equal in that case it would be something
like this this both are having the same
mass now initially M 2 is at rest M 1 is
moving after collision M 1 will come to
rest and M 2 will start moving with the
velocity of U 1 that means after
collision V 1 is equal to 0 means after
collision this km to rest and after
collision V 2 became equal to u 1 that
means it started moving with the
velocity with which this came before
collision fine now let us look at the
case two where M 2 is very very greater
than M 1 that means the second mass is
very very greater in that case what
would be the value of V 1 in that case V
1 would be we can write it as u 1 let us
say I write it as minus u1 m2 minus m1
divided by m1 plus m2
now if M 2 is very very greater than m1
then m2 minus m1 will be equal to m2
similarly M 2 plus M 1 will also be
equal to M 2 why if suppose you have a
very large number say one zero zero zero
zero zero zero zero with this if you
multiply a very small number the answer
will almost be equal to the large number
correct similarly if you subtract a very
small number with a very large number
the answer will almost be equal to the
large number so we know the m2 minus m1
will almost be equal to m2 and m2 plus
m1 will also be almost equal
M 2 so M 2 M 2 will cancel and we will
say minus u 1 that means the first
object that means the velocity of the
first ball will remain the same but it
will be in the opposite direction and
what about V 2 for V 2 it will be 2 m1
m2 divided by m1 plus m2 for this case
again M 2 plus M 1 will almost be equal
to M 2 so this we can write as 2 M 1 u 1
divided by M 2
now if M 2 is very very large that means
if the denominator is very large say it
is almost tending to infinity that is
this value will tend to 0 so this entire
value will tend to 0 if M 2 is very very
large than M 1 that means the velocity
of the second object will almost come to
0 right so in this case when the second
object is very very massive when
compared to the first object after
collision the first object will move
with the same velocity but in opposite
direction and the second object will not
move at all it will remain at rest
that means the first object will come it
will collide with the second object and
it will revert back whereas the second
object will remain as it was thank you
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