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anonymous
 4 years ago
The Spring mass System moves with a Speed of 1m/sand has a Complete inelastic collision woth a mass resting..Find the maximum compression of spring after collison?
anonymous
 4 years ago
The Spring mass System moves with a Speed of 1m/sand has a Complete inelastic collision woth a mass resting..Find the maximum compression of spring after collison?

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1356870345732:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I dont Know....any how Can u Show the Steps...:)...Some times It may Help .:)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Conservation of energy

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Wat would be the K.E of Sys

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0We have to Consider Effective Mass Right ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Initial K.E of the system is 3J

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0maximum extension is either 0.1732m or 0.15m

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0But I am not sure which equation is correct 1/2 kx^2 + 1/2 *6*(0.5)^2 = 1/2*6*1^2 1/2 kx^2 = 1/2 6*1^2

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0We have to Find the Compression After Collision

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Now I think it is 1/2 kx^2 + 1/2 *6*(0.5)^2 = 1/2*6*1^2 which gives x=0.15 m

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I think dw:1356871645704:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0It will be something like that after collision

NaveenBabbar
 4 years ago
Best ResponseYou've already chosen the best response.0Use conservation of momentum And the compression will be maximum when the relative velocity between the blocks is zero

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Also keep in mind that just after the collision the spring exerts a force on the two 3kg masses and speeds them up so their final velocity is > than 0.5 m/sec

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0U mean velocity of blocks are same

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Now, that gives 1/2 *kx^2 + 1/2 *9* (2/3)^2 =1/2 *6*(1)^2 x=0.1 m

VincentLyon.Fr
 4 years ago
Best ResponseYou've already chosen the best response.2I find 5 cm, but I am looking for a simpler answer than the one I have written so far.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0can u Show..hw u Did ?

VincentLyon.Fr
 4 years ago
Best ResponseYou've already chosen the best response.2How do you usually solve these problems? Do you stay in the lab's frame of reference, or to you go to the 'CoM frame of reference'?

VincentLyon.Fr
 4 years ago
Best ResponseYou've already chosen the best response.2Ok, so CoM is moving at velocity \(2v_o/3\), right?

VincentLyon.Fr
 4 years ago
Best ResponseYou've already chosen the best response.2Wait! It is easier to consider the first collision in lab frame. After inelastic collision, new mass 2M will move at \(v_o/2\), right?

VincentLyon.Fr
 4 years ago
Best ResponseYou've already chosen the best response.2Now, new problem in CoM frame is:dw:1356887265577:dw Do you agree? Now, working out KE of system is easy. When spring is fully compressed, both masses will be at rest and KE will be zero. PE will lead you to compression of spring.

VincentLyon.Fr
 4 years ago
Best ResponseYou've already chosen the best response.2Velocities in CoM frame are: \(v^*_{M}=v_ov(CoM)=v_o2v_o/3=+v_o/3\) \(v^*_{2M}=v_o/2v(CoM)=v_o/22v_o/3=v_o/6\) which means that: Mblock is moving to the right with speed \(v_o/3\) 2Mblock is moving to the left with speed \(v_o/6\)

shubhamsrg
 4 years ago
Best ResponseYou've already chosen the best response.1energy is never conserved in case of in elastic collision. momentum is conserved, also, reduced mass = (3*3)/(3+3) = 3/2 so question simplifies to a mass 3/2 kg attached to a spring of spring constant 200N/m travels with a speed 1m/s colliding with 3 kg mass from consvn of momentum (3/2) (1) = (3/2 +3)v v = 1/3 now, loss in KE must have been converted to spring energy 1/2 (200)(x^2) = 1/2 (3/2) 1^2  1/2 (3/2 +3) (1/3^2) am getting 7cm approx.. and there is high chance i have made mistakes.. !

VincentLyon.Fr
 4 years ago
Best ResponseYou've already chosen the best response.2I do not understand how you determine the reduced mass. After the collision, there are 2 masses. One is M, the other is 2M. The reduced mass is 2M/3 = 2 kg.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@shubhamsrg The options are 5 / sqrt2 cm 10 sqrt [ 5/2]cm 10sqrt5 cm 5 cm

shubhamsrg
 4 years ago
Best ResponseYou've already chosen the best response.1i got 10/sqrt2 i.e. 5 sqrt 2 .. hmm..

VincentLyon.Fr
 4 years ago
Best ResponseYou've already chosen the best response.2If you follow the equations and the reasoning I wrote above, you will find 5 cm.

shubhamsrg
 4 years ago
Best ResponseYou've already chosen the best response.1ohh..there is reduced mass also after the collision ! :O didnt notice that.. allow me to try again then.. reduced mass before = 3/2 reduced mass after= 2 so now, 3/2 = 2v v= 3/4 then 1/2 (200)x^2 = 1/2 (3/2) 1^2  1/2 (2)(3/4)^2 now getting 4.3 cms approx.. where am i going wrong.. ? :/

shubhamsrg
 4 years ago
Best ResponseYou've already chosen the best response.1maybe conservation of energy isnt applicable right ? as, not necessary all change in Ke would have been converted to spring energy.. heat must have also be generated! hmm..

shubhamsrg
 4 years ago
Best ResponseYou've already chosen the best response.1how about this approach : if we apply COLM only on right block in the spring mass system : 3(1) = (3+3)v v = 1/2 (to the right i.e. original direction) dw:1356965408871:dw now, applying consn of energy , 1/2 (3)(1^2) + 1/2 (6) (1/2^2) = 0 + 0 + 1/2 (200) x^2 now 15 cms! :(

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I got 17.32 cm , 15 cm and 10 cm

VincentLyon.Fr
 4 years ago
Best ResponseYou've already chosen the best response.2I do not know where you are going wrong, because I do not understand what laws you are following or in what frame of reference you are working. If you understand my drawing above and what I wrote: "Now, working out KE of system is easy. When spring is fully compressed, both masses will be at rest and KE will be zero. PE will lead you to compression of spring." You will get: \(KE\,^*=\Large\frac{1}{2}\normalsize M \Large(\frac{v_o}{3})^2 \normalsize + \Large\frac{1}{2}\normalsize (2M) \Large(\frac{v_o}{6})^2 \normalsize = \Large\frac{1}{12}\normalsize M v_o^2\) Then \(PE=\Large\frac{1}{12}\normalsize M v_o^2=\Large\frac{1}{2}\normalsize k \;\Delta l ^2\) will lead to \(\Delta l\) = 5 cm

shubhamsrg
 4 years ago
Best ResponseYou've already chosen the best response.1maybe,,here is where i went wrong, both blocks during max compression will have some common velocity, 3(1) + 6(1/2) = 3v + 6v 2/3 =v so consn of energy will modify like this : 1/2(3)(1^2) + 1/2(6) (1/4) = 1/2(3+6)(4/9)+ 1/2(200) x^2 ohh! i got 5 cms! \m/ !

shubhamsrg
 4 years ago
Best ResponseYou've already chosen the best response.1am sorry i didnt look at your method yet,, i wanted to do this by some other way,,and voila ! to others and mostly to my surprise,,i was able to! :D

shubhamsrg
 4 years ago
Best ResponseYou've already chosen the best response.1i'll try to explain my approach now : first let us consider the inelastic collision between the 2 blocks, applying COLM , 3(1) = (3+3)v so v=1/2 now we are reduced to the question dw:1356966700681:dw and we need to find max compression in this case

shubhamsrg
 4 years ago
Best ResponseYou've already chosen the best response.1at max compression, both blocks will acquire a common velocity, since no external force is acting, momentum will be conserved. 3(1) + 6(1/2) = 3v + 6v v = 2/3 so, initial energy was 1/2 (3) (1^2) + 1/2(6) (1/2^2) final energy = 1/2(3)(2/3)^2 + 1/2(6) (3/2)^2 + 1/2 (200)x^2 equate both and find x

VincentLyon.Fr
 4 years ago
Best ResponseYou've already chosen the best response.2This is correct :) You are staying in labframe, so final KE is not zero. I was working in CoMframe, so final KE* is zero. Both methods lead to 5 cm compression.
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