IIT JEE , AIEEE Forum - Physics , Mechanics

anki016

two blocks of masses m1 and m2 are placed on a frictionless horizontal floor and are attached by a spring of spring const K .........on da block on right hand side a force of 2f and on da block on left a force of f N is applied ....both forces in da direction away from spring i.e against da spring.find da max extension in spring???????

vercitty

I forgot how to do this question. I havnt touched mechanics for some moths now lol. And also its night and my mind is half asleep. But i am trying. Tell whats wrong in this consept :

2f - kx = m1a1

kx - f = m2a2

When max extention a1 = a2

therefore by equating both equations : x = f/k ((2m2 + m1) / (m1 + m2))

Whats wrong in my answer. I remember this question was a lot harder than this.

paddy.dude

@vercitty
well why is that for max extension a1=a2

anki016

yes actlly da same is where i have struck.............my sir also just said it.........and left it for us ............in such other problem wen only one mass is given some velocity we assume max extension to be at v1 = v2 but why????????

little_genius

after max entension , both the blocks will move wid the same acceleration. so we ca consider both the blocks as a single system...
so acc of the blocks= a=(2f-f)/m1+m2= f/(m1+m2)
now , considr the left block..
we have ,
T-f= m1*a
so T =f+m1f/(m1+m2)...= f* (2m1+m2)/(m1+m2)
and so max extension= T/k....

we can also approach at the answer by considering the right block...
for that :
2f-T=m2*a
or T=2f-m2*a= 2f-m2*f/(m1+m2)= f* (2m1+m2)/(m1+m2)

so u see both ways we can get the answer.....

vercitty

Aiight, this question is easy.

And both mine and lil genius had something done wrong in the concepts.

I got the answer now and it is ( 2 f(2m1 + m2) / k (m1 + m2) )

The steps are hard to type lol. I'll tell the concept anyway. Obviously, the max extention is not when the accelerations are same. Infact it is max when the velocities are same. Just imagine, both the bodies will initially accelerate in the same direction and then reach the same velocity. Then any difference in velocity will bring the bodies nearer.

So we can imagine the question in two ways :

1) By finding the extention when bodies have the same velocity. Apply energy conservation. This is quite a mess.

Spring force + Kinetic Energy = Work done by the external force

2) By applying pseudo force so that we can consider a frame at the CM of the two bodies. In this frame, we will find both the bodies to move in opposite direction. Max extention is reached when both the bodies stop at reaching maximum elongation. This is a lot more easier.

a_cm= (2f-f)/(m1 + m2) = f/(m1 + m2)

Acceleration is towards right, so psedo force acts towards left. Consider the pseudo force in the left direction in both masses apart from the forces 2f and f. The force on one mass be F1 and F2 on other mass.

equation is 1/2 k(x1 + x2)² = F1x1 + F2x2

x1 + x2 is the max extention. Try solving this way and u will get my answer.

vercitty

wheres ur answer anki016 ?????? Tell it

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