At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions.

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions

|dw:1354024062491:dw|

conservation of momentum

But that Did nt work..)

|dw:1354026238842:dw|

1/2 5M Vo^2 = 1/2 3M v^2

But i think....Spring Constant has Something to Play..)

Exactly that is what I did..i have ignored the spring forces during the collision time of 2M-2M.

@sumanth4phy ur answer is Correct..but i did nt understand..

Hw Velocity of centre of mass is 2V/3.??

\[V_{cm,a} = (M*V_{M,a} + 2M*V_{2M,a})/3M\]
= (0*M + V)*2M/3M = 2V/3

small correction second equation should be = (M*0 +2M*V)/3M = 2V/3

Ok...then wat did u Do ?