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

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1353427046355:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@experimentX @siddhantsharan

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.01/2 mv^2 = 1/2 kx^2

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Why did you close this question?

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.1for maximum compression..whole kinetic energy is transferred to spring energy...that helps ?

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.1and in such case,,effective mass = m1 m2 / (m1 + m2) am i helping ?

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.1i mean are you getting ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0wat is effective mass

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.12 masses are connected to a spring on 2 ends..say m1 and m2.. then the effective mass is m1 m2/ (m1 + m2)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Wait....I mean....I did nt get...wat is effective mass shuld nt it be m1 + m2

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.1effective mass or can be also called reduced mass of the system.. never heard of it ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So.... 1/2 k x^2 = 1/2 effective mass * v^2

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.1well reduced mass is no big deal..has a simple derivation.. you wanna know ?

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.1there isnt much need for the derivation for you to know since you're JEE aspirant..but still i can tell if you want..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yes..i am interested in it

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.1kk..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..

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.1all 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 ?

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.1tell me if not getting anything ? we are not done yet with the thing..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yup....i can follow u..)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Tell me a Scenario where we will use Effective Mass

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.1now..m1m2 / (m1+ m2) = M (reduced or effective mass) so we have F2/M = a or F2 = Ma so what do you conclude ?

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.1see,,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..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Nw..i get it...)...thxx....)

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.1hmm,,whats the conclusion you can draw then?
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