# mod07lec60 In the previous lecture we discussed collision
and in this lecture that follows that one I am going to solve 2 problems on collision
in two dimensions and will give you a feeling for the collision problems and how the momentum
is transferred from one colliding particle to the other colliding particle So in the 1st problem we see that a ball of
mass M collided with another identical ball at rest and the collision is elastic What
that means is that the kinetic energy is not lost Show if the collision is not head on
That means the the point at which the collision takes place is not along the line of motion
and I will explain this in the picture in a minute Then the two balls
move at 90 degrees to each other after the collision So what is happening in this problem is there
is a ball at rest and this is in the plane of this screen and another ball comes in with
the centre moving like this and collides with this ball Now when it collides let me make
it here this is not head on That means the line along which it was moving the point of
contact where the collision takes place is not on that line You can see from from the
collision point of view that as it collides the forces are going to be perpendicular The impact that the 2 balls have this force
is generated or the impulse generated is going to be perpendicular to their common surface
at the point and therefore it is going to be in the direction like this perpendicular
Since ball 2 was initially at rest is going to move exactly in the direction of the force
Ball 1 which was coming in experiences a an impulse like this and therefore its momentum
which was in the direction to the right is going to change a bit and it is going to move
like this So finally the picture that emerges is that
one ball is moving like this and the other ball is moving like this And this angle we
want to show is 90 degrees This problem can be solved using the principle of conservation
of momentum and principle of conservation of energy So what we are given is that this this ball
which was coming in with some initial velocity let us say U collides with another ball which
was off line and after that let us say this ball the 2nd ball let me call this 2 call
this 1 it starts moving at this angle we have already decided is going to be along the line
of impulse and let us call this angle theta 2 And ball 1 starts moving at some other angle
this way which I will call theta 1 So this is ball 1 moving this way And we wish
to show that theta 1 plus theta 2 is equal to pi by 2 What are the equations that we
have at our disposal That is momentum conservation Of and for good bookkeeping I will call the
initial direction in which the ball was moving X direction and the perpendicular direction
to be Y direction Since there is no external force momentum is conserved and therefore
PX total is same as it was earlier Earlier the momentum was MU and after the
collision the 1st ball is moving with velocity let us say V1 and this is moving with V2 MV1
cosine of theta 1 plus the momentum of the 2nd ball in the X direction is going to be
MV2 cosine of theta 2 And you can see that M cancels and therefore I have U equals V1
cosine of theta 1 plus V2 cosine of theta 2 I have got one equation from momentum conservation I have got another equation from momentum
conservation that is PY total should also remain the same And what was PY total to start
with There was no momentum component in the Y direction and therefore I am going to have
0 equals MV1 sine theta1 minus MV2 sine theta2 I have put a minus sign here already anticipating
that the Y components are in the opposite direction If you put plus finally you will
get V1 equal to minus some V2 So that does not really matter And this equation gives me V1 sine theta1
minus V2 sine theta2 equals 0 That is equation number 2 The 3rd equation comes from energy
conservation and that says that initial energy MU square should remain the same even after
the collision because it is an elastic collision plus one half MV2 square And again half cancels
M cancels and I have U square equals V1 square plus V2 square That is my 3rd equation Let me write them
again The 3 equations I have is U equals V1 cosine of theta 1 plus V2 cosine of theta
2 I have 0 equals V1 sine theta1 minus V2 sine theta2 And U square equals V1 square
plus V2 square Equation 1 equation 2 equation 3 How many unknowns are there I have unknowns
U is supposed to be known I have V1 V2 theta1 and theta2 4 unknowns So I cannot get all 4 quantities by solving
these 3 I will eat some other information But I can find relationships and that is what
we are trying to find here We are trying to find and we want to show that theta1 plus
theta2 they are going to be such that they always add up to pi by 2 Beyond that I cannot
really find all those 4 quantities or 4 unknowns individually So let us now solve these equation I square U equals V1 cosine theta1 plus V2
cosine of theta 2 I square the left hand side I square the right hand side and I get U square
equals V1 square cosine square theta 1 plus V2 square cosine square theta 2 plus 2 V1
V2 cosine of theta 1 cosine of theta 2 But U square from equation 3 I already know is
V1 square plus V2 square because it is an elastic collision This comes out to be V1 square cosine square
theta1 plus V2 square cosine square theta2 plus 2 V1 V2 This is V2 cosine theta 1 cosine
theta 2 I will take V1 square and V2 square to the right hand side and therefore I will
get 0 equals minus V1 square sine square theta1 minus V2 square sine square theta2 plus 2
V1 V2 cosine theta1 cosine theta2 which I write as minus V1 sine square theta1 plus
V2 square sine square theta2 plus 2 V1 V2 cosine theta1 cosine theta2 This I can further why because I already know
from 2nd equation that V1 sine theta minus V2 sine theta is 0 so I am I will bring it
use that equation and bring it in that form I will write this as minus V1 sine theta1
minus V2 sine theta2 whole square plus 2 V1 V2 cosine theta1 cosine theta2 And to make
the this term the term here in red box and the new equation I am writing I have to subtract
here minus 2 V1 V2 sine theta1 sine theta2 Everything sets well Now I already know that
V1 sine theta1 minus V2 sine theta2 is 0 So this term is going to be 0 So I am left with
0 equals 2 V1 V2 which will take common cosine of theta 1 cosine of theta 2 minus sine of
theta1 sine of theta2 So I get 0 equals 2 V1 V2 cosine of theta
1 cosine of theta 2 minus sine of theta1 sine of theta2 which is nothing but 2 V1 V2 cosine
of theta1 plus theta2 which immediately tells me that cosine of theta1 plus theta2 is equal
to 0 or theta1 plus theta2 equals pi by 2 Notice that in this problem I right from the
beginning used the momentum conservation equation and energy conservation equation Just a comment
on that So far I have used these momentum conservation
equations derived this condition just to give you a little more feel for for how the momentum
is important Let me look at it more carefully from the point of view of the impulse which
is given by ball 1 to ball 2 and as I commented earlier the impulse given to ball 2 is going
to be in the direction in which it moves which is at an angle theta2 from the initial direction
of ball 1 and ball 1 moves like this with velocity V1 had an angle theta1 And the ball 2 is moving with theta 2 V2 So
if the impulse is J the impulse of ball 1 is going to be minus K J vector is going to
be MV2 and therefore JX is going to be MV2 cosine of theta 2 and JY is going to be MV2
sine of theta 2 The change change in momentum of ball 1 is delta PX is going to be MV1 cosine
of theta 1 minus MU This is the X component of the momentum after
the collision MU is the initial X component of the momentum and this is minus JX and this
is minus MV2 cosine of theta 2 And if I cancel M I it immediately gives me the equation I
used earlier V1 cosine of theta 1 plus V2 cosine of theta 2 is equal to U Similarly delta PY is equal to V1 sine of
theta1 with M here minus initial momentum was 0 and this should be minus JY which is
equal to minus MV2 sine theta2 which gives you V1 sine theta1 plus V2 sine theta2 is
equal to 0 This this sign problem is coming because again I am taking V1 and V2 to be
in the same direction but that is not a problem So what you see here is that that this equation
of momentum conservation actually comes from Newton’s 3rd law where J are acting in opposite
equal and opposite way on the 2 balls The next problem I am going to solve is again
a two dimensional problem and it is based on the familiar game that you all play is
carom So I want to say a carom striker has diameter 4 point 1 centimetres
and and weighs 15 gram It is moving with speed 2 metres per second along x axis
It strikes a stationery carom coin that
diameter 3 point 1 centimetre and weighs 5 grams The distance
of the centre of the coin from the x axis is 2 centimetres Find the velocity
of the coin and striker after
the collision assuming it to be perfectly elastic That means there is no energy loss
So what you are given is here is a carom coin 3 point 1 centimetre in diameter and here
is a striker bigger in size its diameter is 4 point 1 centimetre It is moving along the x axis And this fellow
is at a distance 2 centimetres from the x axis This comes in and hits the coin and the
coin will start moving Now notice the difference from the previous problem Here the distances
are given So we do have knowledge about one of the angles So what happens is when this striker strikes
the coin the striker was moving like this The distance here is 2 centimetres There is
going to be an impulse on the coin in the direction shown by green And since the coin
was stationery to start with it is going to move exactly in that direction So it is going
to move at an angle theta from the x axis in the direction of this impulse and that
we can easily find out Because you are given the radius here is one
half of 3 point 1 1 point 55 centimetres and the radius of the striker here is 4 point
1divided by 2 and that is 2 point 05 So I know this distance in the Triangle the thick
green line to be 2 point 5 plus 1 point 55 3 point 6 centimetres and the perpendicular
to be 2 centimetres So you immediately have sine of theta is equal to 2 over 3 point 6
which is 20 over 36 or five ninths So theta at which the coin is going to move is sine
inverse of five ninths which is roughly 33 point 75 degrees So we already know that the coin is going
to move at an angle 33 point 75 degrees from the x axis Now let us write the equations
Momentum conservation This tells me that the momentum in both X
and Y direction is going to be conserved Let me take this X and perpendicular direction
to be Y So initial momentum is 15 times 10 raised to minus 3 since the units of mass
are going to be grams all throughout I do not have to write minus 3 here So 15 times 2 is the initial momentum in the
X direction is going to be equal to 5 grams is the mass of the coin and suppose it moves
with velocity V1 V1 cosine theta plus 15V2 cosine of let us say striker now moves at
an angle file cosine of file That is my equation 1 So this is PX For PY initial momentum in
the Y direction is 0 And this is going to be 5V1 sine of theta
again anticipating that they will move in opposite directions and Y direction is going
to be 15V2 sine of file That is my equation number 2 Equation number 3 is going to be
energy conservation and that will give me one half times 15 times 2 square is 4 that
is the initial kinetic energy is going to be one half times 5 V1 square plus one half
times 15 V2 square How many unknowns do we have I already know
theta So I do not know file I do not know V2 and I do not know V1 3 unknowns and 3 equations
Because we already have information about theta using conservation principles I can
get these 3 unknowns And now what remains is just solving these equations Let me simplify
them right here So I have 30 15 times 2 is 30 is equal to
5 V1 cosine of 33 point 75 degrees plus 15 V2 cosine of file That is my equation 1 I
have 0 equals 5 V1 sine of 33 point 75 degrees sine we have already calculated is five ninths
minus 15 V2 sine of file That is my equation number 2 and I have half cancels this half
cancels out here So I can write 60 equals 5 V1 square plus 15 V2 square That is my equation
number 3 I solve these which I will leave for you and
get answers that V1 comes out to be 2 point 5 metres per second V2 comes out to be 1 point
39 metres per second and file comes out to be 19 point 5 degrees I leave the solution
getting the solution for you for these equations So you saw again that with the information
available I could 1st find out in which direction coin is going to move Because it was stationery so it moved in the
direction along the normal of the 2 common surfaces at the point where where the coin
and the striker were meeting And once I found that direction I had one angle already And
now what we are left with were 3 unknowns and those we could solve with the help of
conservation equations