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 2 years ago
System B moving with Vo elastically Collides with system A at rest as Shown.At maximum Compression of Spring K.Velocity of M is?
 2 years ago
System B moving with Vo elastically Collides with system A at rest as Shown.At maximum Compression of Spring K.Velocity of M is?

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Shadowys
 2 years ago
Best ResponseYou've already chosen the best response.0Hmm, can't see the entire diagram though. Does it means that the first spring collided with the second while compressing to max?

Yahoo!
 2 years ago
Best ResponseYou've already chosen the best response.0@ghazi @VincentLyon.Fr @JFraser @phi

ghazi
 2 years ago
Best ResponseYou've already chosen the best response.1transfer whole kinetic energy of the system with mass 2m and 3m to the system with mass m and 2m , you will get it :)

Yahoo!
 2 years ago
Best ResponseYou've already chosen the best response.01/2 5M Vo^2 = 1/2 3M v^2

Yahoo!
 2 years ago
Best ResponseYou've already chosen the best response.0But i think....Spring Constant has Something to Play..)

ghazi
 2 years ago
Best ResponseYou've already chosen the best response.1yes spring is storing energy so you have to calculate that energy too and after that use the rule of collision, using conservation of energy and momentum. and don't forget to calculate stored potential energy of spring

sumanth4phy
 2 years ago
Best ResponseYou've already chosen the best response.3After collision assuming it is perfectly eastic the 2M will have a velocity of V. The Velocity of centre of mass is 2V/3. Now consider the motion of system A in centre of mass reference frame. When the spring is under maximum condition the velocities of both M, 2M will be zero in xcentr of mass reference frame. Hence V(absolute of M) = V(w.r.t to C.M) + V(velocity of Centr of mass) =>\[ V_{M}=V_{M,CM} + V_{CM}\] \[V_{M} = 0 + 2V/3 =2V/3\]

VincentLyon.Fr
 2 years ago
Best ResponseYou've already chosen the best response.0I have never solved that kind of problems. I would start with the initial 2M2M collision when the springs do not act yet. There would be exchange of KE and velocities of these two masses. Only then would springs start to compress.

sumanth4phy
 2 years ago
Best ResponseYou've already chosen the best response.3Exactly that is what I did..i have ignored the spring forces during the collision time of 2M2M.

Yahoo!
 2 years ago
Best ResponseYou've already chosen the best response.0@sumanth4phy ur answer is Correct..but i did nt understand..

Yahoo!
 2 years ago
Best ResponseYou've already chosen the best response.0Hw Velocity of centre of mass is 2V/3.??

sumanth4phy
 2 years ago
Best ResponseYou've already chosen the best response.3\[V_{cm,a} = (M*V_{M,a} + 2M*V_{2M,a})/3M\] = (0*M + V)*2M/3M = 2V/3

sumanth4phy
 2 years ago
Best ResponseYou've already chosen the best response.3small correction second equation should be = (M*0 +2M*V)/3M = 2V/3

Yahoo!
 2 years ago
Best ResponseYou've already chosen the best response.0Ok...then wat did u Do ?

sumanth4phy
 2 years ago
Best ResponseYou've already chosen the best response.3after collision the motion is a Simple Harmonic Motion...the velocities of masses M, 2M will change as function of cos() and are out of phase. One way to do is get the equations of motion for each mass individually..that would be cumbersome.... second method would be oscillation in C of Mass refrence frame where the motion of particles are in phase and velocities of both the particles are zero at maximum compression and maximum elongation....then add Velocity of centre of mass to get the velocity in ourrefrence frame. third and simple method is when under maximum compression both particles have sam evelocity so apply momentum conservation for system A \[(M+2M)*V_{\max compression} = (0*M + V*2M)\] You Get 2V/3
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