Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

The system Shown ( the spring is without tension) travels towards the wall and has an elastic collision . The maximum Compression of the spring is ?

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

Join Brainly to access

this expert answer

SIGN UP FOR FREE
|dw:1353427046355:dw|
1/2 mv^2 = 1/2 kx^2

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

Why did you close this question?
@Vincent-Lyon.Fr
for maximum compression..whole kinetic energy is transferred to spring energy...that helps ?
and in such case,,effective mass = m1 m2 / (m1 + m2) am i helping ?
i mean are you getting ?
wat is effective mass
2 masses are connected to a spring on 2 ends..say m1 and m2.. then the effective mass is m1 m2/ (m1 + m2)
ok..)
Wait....I mean....I did nt get...wat is effective mass shuld nt it be m1 + m2
effective mass or can be also called reduced mass of the system.. never heard of it ?
No..)
So.... 1/2 k x^2 = 1/2 effective mass * v^2
right?
seems right..
well reduced mass is no big deal..has a simple derivation.. you wanna know ?
there isnt much need for the derivation for you to know since you're JEE aspirant..but still i can tell if you want..
Yes..i am interested in it
kk..newton's 3rd law states that F1 = -F2 right ? both taken in vector form.. F1 will be one acting on 1st body,,F2 on second..
ok...
all expressions are in vector form below : m1 a1 = - m2 a2 =>a1 = (-m2/m1) a2 now relative acceleration between the bodies a = a2 - a1 = a2 (1 + m2/m1) = m2 a2 ( m1 + m2)/(m1 m2) following ?
??
tell me if not getting anything ? we are not done yet with the thing..
Yup....i can follow u..)
Tell me a Scenario where we will use Effective Mass
right here!!
Another..Eg
now..m1m2 / (m1+ m2) = M (reduced or effective mass) so we have F2/M = a or F2 = Ma so what do you conclude ?
see,,any 2 body system can be converted to 1 body system with this effective mass... force acting between them will also act between this one body system..
Nw..i get it...)...thxx....)
you sure?
Yes.
hmm,,whats the conclusion you can draw then?

Not the answer you are looking for?

Search for more explanations.

Ask your own question