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

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Yahoo!Best ResponseYou've already chosen the best response.0
@experimentX @siddhantsharan
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
1/2 mv^2 = 1/2 kx^2
 one year ago

VincentLyon.FrBest ResponseYou've already chosen the best response.0
Why did you close this question?
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.1
for maximum compression..whole kinetic energy is transferred to spring energy...that helps ?
 one year ago

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

shubhamsrgBest ResponseYou've already chosen the best response.1
i mean are you getting ?
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.1
2 masses are connected to a spring on 2 ends..say m1 and m2.. then the effective mass is m1 m2/ (m1 + m2)
 one year ago

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

shubhamsrgBest ResponseYou've already chosen the best response.1
effective mass or can be also called reduced mass of the system.. never heard of it ?
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.0
So.... 1/2 k x^2 = 1/2 effective mass * v^2
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.1
well reduced mass is no big deal..has a simple derivation.. you wanna know ?
 one year ago

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

Yahoo!Best ResponseYou've already chosen the best response.0
Yes..i am interested in it
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.1
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..
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.1
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 ?
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.1
tell me if not getting anything ? we are not done yet with the thing..
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.0
Yup....i can follow u..)
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.0
Tell me a Scenario where we will use Effective Mass
 one year ago

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

shubhamsrgBest ResponseYou've already chosen the best response.1
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..
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.0
Nw..i get it...)...thxx....)
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.1
hmm,,whats the conclusion you can draw then?
 one year ago
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