conservation

Two bodies A and B of masses m and 2m respectively are placed on smooth floor.They are connected by a light spring of stiffness k.A third body C of mass m moves with velocity v along line joining A and B collides elastically with A.If l be the natural length of spring,then find minimum separation betweeen blocks

6 Answers

1
Akshat Mishra ·

As far as the collision is concerned

its given elastic and A,C both have same mass.

You can easily check from the elastic collision formula that after collision body C will stop & body A will have velcity v

after that I cant imagine the scenario exactly [1]

·

@ Akshat ... you are right abt the collision.

Try to imagine what is happening after collision .. I can post the solution, but I think you guys should think a bit more [1], thats gonna help u in JEE

1
Ashish Sharma ·

We have to apply the conservation of momentum & conservation of energy to sovle this problem .. thats sure [3]

@Lokesh Please post the solution ...

·

Look this problem can be solved theoretically as well as conceptually :)

Just a little bit of thinking will solve lots of calculations ....

First the theoretically solution :-

Momentum Conservation :-
mVa + 2m Vb = mV
=>Va + 2 Vb = V ---------------(1)

Energy Conservation :-
1/2 * mVa2 + 1/2 * 2m Vb2 + 1/2 KX2= 1/2 mV2
=>Va2 + 2 Vb2 + K/m X2 = V2 ---------------(2)

At the point of minimum seperation between blocks .. the compression in the spring will be maximum.
so the sum of kinetic energy stored in the two blocks have to be minimum (see equation 2, RHS is constant)

Say Function F = Va2 + 2 Vb2
=> F = (V-2 Vb)2 + + 2 Vb2 ( from eqn 1)
Diffrentiating F w.r.t Vb and equating to zero gives Vb = V/3
which gives Va = V/3

From eqn 2 we can get X = √2m/3K V

·

Conceptual Solution :-
Analysing the whole thing from Frame of reference of COM of system
Please Note that mass of A is half of that of B

At time t=0, the spring is uncompressed and A will start moving with some velocity (here its V) towards B

after some time as the spring gets compressed, it will appy some force on B and B will also start moving forward, so A will deccelerate and B will accelerate ( think why [1] .. its very simple )
but as A has greater velocity the distance between A & B will keep on decreasing ...
untill ......
velocity of A wrt COM becomes 0

because B is continuously accelerating in forward direction ...... think why
and upto this time A was moving in forward direction and after this point it will change its direction of motion (wrt COM)

so at this point of time Va = VCOM = mV/(m+2m) = V/3
after this 1-2 line calculations and you get the result [1]

1
Ashish Sharma ·

WOW !!!!
Thanks Lokesh .... that was such a brilliant explanation
....especially the 2nd solution

I just loved it [1]

Your Answer

Close [X]