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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?
 one year ago
 one year ago
Yahoo! Group Title
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?
 one year ago
 one year ago

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Yahoo! Group TitleBest ResponseYou've already chosen the best response.0
dw:1356870345732:dw
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.0
Is it 0.1732m
 one year ago

Yahoo! Group TitleBest ResponseYou've already chosen the best response.0
I dont Know....any how Can u Show the Steps...:)...Some times It may Help .:)
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.0
Conservation of energy
 one year ago

Yahoo! Group TitleBest ResponseYou've already chosen the best response.0
Wat would be the K.E of Sys
 one year ago

Yahoo! Group TitleBest ResponseYou've already chosen the best response.0
We have to Consider Effective Mass Right ?
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.0
Initial K.E of the system is 3J
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.0
maximum extension is either 0.1732m or 0.15m
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.0
But 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
 one year ago

Yahoo! Group TitleBest ResponseYou've already chosen the best response.0
We have to Find the Compression After Collision
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.0
Now 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
 one year ago

Yahoo! Group TitleBest ResponseYou've already chosen the best response.0
1/2 *6*(0.5)^2 ?????
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.0
I think dw:1356871645704:dw
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.0
It will be something like that after collision
 one year ago

naveenbabbar Group TitleBest ResponseYou've already chosen the best response.0
Use conservation of momentum And the compression will be maximum when the relative velocity between the blocks is zero
 one year ago

gleem Group TitleBest ResponseYou've already chosen the best response.0
Also 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
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.0
U mean velocity of blocks are same
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.0
Now, that gives 1/2 *kx^2 + 1/2 *9* (2/3)^2 =1/2 *6*(1)^2 x=0.1 m
 one year ago

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@VincentLyon.Fr
 one year ago

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

Yahoo! Group TitleBest ResponseYou've already chosen the best response.0
can u Show..hw u Did ?
 one year ago

VincentLyon.Fr Group TitleBest ResponseYou've already chosen the best response.2
How 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'?
 one year ago

Yahoo! Group TitleBest ResponseYou've already chosen the best response.0
Centre of mass
 one year ago

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

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

VincentLyon.Fr Group TitleBest ResponseYou've already chosen the best response.2
Now, 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.
 one year ago

VincentLyon.Fr Group TitleBest ResponseYou've already chosen the best response.2
Velocities 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 \(v_o/6\)
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.1
energy 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.. !
 one year ago

VincentLyon.Fr Group TitleBest ResponseYou've already chosen the best response.2
I 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.
 one year ago

Yahoo! Group TitleBest ResponseYou've already chosen the best response.0
@shubhamsrg The options are 5 / sqrt2 cm 10 sqrt [ 5/2]cm 10sqrt5 cm 5 cm
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.1
i got 10/sqrt2 i.e. 5 sqrt 2 .. hmm..
 one year ago

VincentLyon.Fr Group TitleBest ResponseYou've already chosen the best response.2
If you follow the equations and the reasoning I wrote above, you will find 5 cm.
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.1
ohh..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.. ? :/
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.1
maybe 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..
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.1
how 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! :(
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.0
I got 17.32 cm , 15 cm and 10 cm
 one year ago

VincentLyon.Fr Group TitleBest ResponseYou've already chosen the best response.2
I 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
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.1
maybe,,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/ !
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.1
am 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
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.1
i'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
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.1
at 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
 one year ago

VincentLyon.Fr Group TitleBest ResponseYou've already chosen the best response.2
This 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.
 one year ago
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