anonymous
  • anonymous
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?
Physics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
|dw:1356870345732:dw|
anonymous
  • anonymous
Is it 0.1732m
anonymous
  • anonymous
I dont Know....any how Can u Show the Steps...:)...Some times It may Help .:)

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
Conservation of energy
anonymous
  • anonymous
Wat would be the K.E of Sys
anonymous
  • anonymous
We have to Consider Effective Mass Right ?
anonymous
  • anonymous
Initial K.E of the system is 3J
anonymous
  • anonymous
maximum extension is either 0.1732m or 0.15m
anonymous
  • anonymous
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
anonymous
  • anonymous
We have to Find the Compression After Collision
anonymous
  • anonymous
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
anonymous
  • anonymous
1/2 *6*(0.5)^2 ?????
anonymous
  • anonymous
I think |dw:1356871645704:dw|
anonymous
  • anonymous
It will be something like that after collision
naveenbabbar
  • naveenbabbar
Use conservation of momentum And the compression will be maximum when the relative velocity between the blocks is zero
anonymous
  • anonymous
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
anonymous
  • anonymous
U mean velocity of blocks are same
anonymous
  • anonymous
all*
anonymous
  • anonymous
Now, that gives 1/2 *kx^2 + 1/2 *9* (2/3)^2 =1/2 *6*(1)^2 x=0.1 m
anonymous
  • anonymous
@Vincent-Lyon.Fr
Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
I find 5 cm, but I am looking for a simpler answer than the one I have written so far.
anonymous
  • anonymous
can u Show..hw u Did ?
Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
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'?
anonymous
  • anonymous
Centre of mass
Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
Ok, so CoM is moving at velocity \(2v_o/3\), right?
Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
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?
Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
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.
anonymous
  • anonymous
v0 / 6 ???
Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
Velocities in CoM frame are: \(v^*_{M}=v_o-v(CoM)=v_o-2v_o/3=+v_o/3\) \(v^*_{2M}=v_o/2-v(CoM)=v_o/2-2v_o/3=-v_o/6\) which means that: M-block is moving to the right with speed \(v_o/3\) 2M-block is moving to the left with speed \(v_o/6\)
shubhamsrg
  • shubhamsrg
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.. !
Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
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.
anonymous
  • anonymous
@shubhamsrg The options are 5 / sqrt2 cm 10 sqrt [ 5/2]cm 10sqrt5 cm 5 cm
shubhamsrg
  • shubhamsrg
i got 10/sqrt2 i.e. 5 sqrt 2 .. hmm..
Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
If you follow the equations and the reasoning I wrote above, you will find 5 cm.
shubhamsrg
  • shubhamsrg
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.. ? :/
shubhamsrg
  • shubhamsrg
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..
shubhamsrg
  • shubhamsrg
been*
shubhamsrg
  • shubhamsrg
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! :(
anonymous
  • anonymous
I got 17.32 cm , 15 cm and 10 cm
Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
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
shubhamsrg
  • shubhamsrg
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/ !
shubhamsrg
  • shubhamsrg
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
shubhamsrg
  • shubhamsrg
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
shubhamsrg
  • shubhamsrg
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
Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
This is correct :) You are staying in lab-frame, so final KE is not zero. I was working in CoM-frame, so final KE* is zero. Both methods lead to 5 cm compression.

Looking for something else?

Not the answer you are looking for? Search for more explanations.