In this illustration we will discuss about
Impulse by a colliding particle on a rod. In this situation we are given that a small
particle of mass m moving horizontally on a smooth surface collides elastically with
a uniform rod of mass 2m, and length l we can see in the figure, at a point D in the
direction normal to the length of rod. Given that CD is equal to small d and we are
required to find the impulse imparted by particle on rod, and also we are required to find the
speeds of ball and rod. So in this situation we can see, as the particle
is colliding in the rod is free to move here about the center of mass of this rod, or a
point which is on table or on a smooth surface, on which the whole system is placed, we can
conserve angular momentum and as no external force is present linear momentum can also
be conserved, and given that collision is elastic, so we can also make use of coefficient
of restitution for ease in calculations. So here in solution we can start with, using,
conservation of, angular momentum, about C, here we can write we have, about point C which
is there on the ground, and in this situation this is the given state which is before collision,
and if you draw the state after collision we can see if this is the rod, and after collision
say, the ball will experience an impulse J backward and it will impart an impulse J on
the rod due to which say the rod starts moving with a speed v one and due to the angular
impulse of this J rod will attain an angular speed and say the balls speed which was initially
u with which it was coming after collision its speed changes to v say this is the state
which is after collision. So after collision for the whole system we can conserve angular
momentum so we can write the initial angular momentum about C was m u d that is only that
of the ball because the rod was at rest and after collision ball and rod both were having
angular momentum so here we can write for ball it is m v d as final speed of ball I
have taken as v for rod as it is rotating about center so we write it I omega so its
moment of inertia will be two m L square by twelve multiplied by omega say this is our
first equation and we can write by linear momentum conservation here we can directly
write the initial momentum was m u and final momentum will be m v plus two m v one this
is your second equation you are getting and we can write as e is equal to one because
collision is elastic in this situation so here we are using e is equal to one so we
can write the coefficient of restitution equation as the approach velocity in this case is u
and after collision the separation velocity can be given as for the point on rod it is
moving with velocity v one plus d omega because this separation is given to us as d so here
this is v one plus d omega minus v so this is v one plus d omega minus v that’s the
our third equation. And from these three equations we can solve for the values of v v one and
omega so here we can write from equations if we just have a look here the mass can be
cancelled so here we are getting the value of u is v plus v one and here we are getting
u as v one plus d omega minus v so if we re arrange these two equations to eliminate the
value of u this will be v plus v one can be equated to v one plus d omega minus v so we
are able to relate these equations to get the value v one and u one by one we can solve
like here in two and three from two and three we get in this situation from these values
we are getting we just add up these equations and eliminate the value of v this gives us
two u is equal to three v one plus d omega say this is our equation four. And we can
also get a value relation between u and v if we eliminate the value of v one so if we
eliminate the value of v one it can be given as using two into three minus two see what
this will result this will be two u minus u is u equal to this two v one minus two v
one gets cancel out this is two d omega minus this will be minus two v minus v is three
v so this is giving us a value here v is equal to two d omega minus u by three say this is
our equation five. Now we can write from equation five and one we just substitute the value
of v from here to here see what we are getting here m gets cancelled out and we are having
u d is equal to v d where v is already given to us is two by three d square omega minus
one by three u d plus this is plus one by six l square omega, if we simplify this relation
we are getting the value of omega from here the value of omega we are getting is eight
u d by l square plus four d square so we are required to find the speeds of ball and rod
so this is the angular speed of rod which we have obtained can be taken as one answer
and if we substitute the value of omega over here we can write now in continuation from
five the value of v we are getting is two by three d omega we substitute as eight u
d by l square plus four d square minus one by three u then on simplifying this value
we are getting the value of v as one by three u multiplied by twelve d square minus l square
divided by four d square plus l square so this is another result of the problem. For
v we are getting the final speed of ball and if we calculate the final speed of rod that
can be obtained either from equation two or we can obtain it from equation four so we
can write from equation four we get the value of v one as one by three times two u minus
d omega so here you can substitute the value of omega which we have obtained from here
so on substituting the value of omega just for lack of space I am substituting and you
can verify the result I am directly writing it is given as two u l square divided by thrce
of four d square plus l square that’s the result of this problem for the final velocity
of rod after collision. At last if you wish to calculate the impulse imparted by rod then
here we can calculate the impulse J as it is the change in momentum of ball or that
of rod so for ball we can directly use J can be given as m times u minus v and here the
value of v you can substitute and verify the result finally you will be getting is four
m u l square divided by thrice of four d square plus l square that is the result of this problem
that is the way how we have analyzed and mainly we have used linear and angular momentum conservation
and we have made use of the coefficient of restitution in this case.

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For temperature: kelvin scale
for force: newton scale
for workk: joule scale
for physics: Physicsgalaxy scale!

My concepts have been made stronger than ever thanks to these videos!

2. Shery Singh says:

Dear Sir I think you have considered the value of "v" in the angular momentum and linear momentum positive instead according to me according to ball later direction it should be
mud=Iw- mvd and linear momentum be as- mu= 2mv1- mv. If sir you could pls correct me it would be a great help can pls explain me where i am going wrong so as that I can match to ur solution.

3. M.A. choudhury says:

sir if we conserve A.M about the point where the ball starts moving,will we get the answer?

Sir how u justify velocity of approach is equal to velocity of separation of road as it cm velocity has not considered since in collision velocity of cm has taken

5. Hrs says:

sir if you conserve momentum outside along the line of impact the question can be solved much easier and in shorter time

6. Jyotisman Rath says:

Can we conserve the momentum along any other point!?

7. SYED AQUIB OFFICIAL says:

Sir, how to calculate impulse using rod.

sir in question it is not mentioned that C is the COM of rod. therefore MOI in equation 1 would be different

9. lions roar says:

there is mistake in application of coefficient of restitution

10. Prasanna Bartakke says:

These are a very good set of problems for revising all the major concepts for JEE ADVANCED. Thank you Sir.

11. Gourav Shaw says:

amazing playlist… all illustration were amazing
my favorite problems 43 39 37 34…
thank you sir