What is the effect of an elastic collision between two blocks and a chain?

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In an elastic collision involving a chain of 10 blocks and two incoming blocks, the momentum and kinetic energy must be conserved. Initially, the chain is at rest, and the incoming blocks have a combined momentum of 2mv. After the collision, it is suggested that the momentum will be transferred to the last block, causing it to move while the others remain stationary. The discussion raises questions about the behavior of the system, particularly regarding the elasticity of the ropes connecting the blocks and the implications when the last block is struck. The interaction dynamics become complex once the ropes are taut, affecting the subsequent motion of the blocks.
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Homework Statement



We have a chain of 10 blocks, all of them joined by a thin rope and placed in a straight line.
Suddenly, other two blocks collide with v speed at one end with the chain of the 10 blocks.
It is assumed that the table is frictionless and the collision is elastic.
The main question is: after the colission, which blocks are going to move and at what speed?.

Homework Equations



Since we have an elastic collision, we have to take into account the following equations: P_{i}=P_{f} \rightarrow m_{1}v_{1i}+...+m_{n}v_{ni}=m_{1}v_{1f}+...+m_{n}v_{nf}
Moreover, the colission is completely elastic, which means: E_{k(i)}=E_{k(f)}

The Attempt at a Solution



Initially, the velocity of the blocks of the chain is zero, which means that the linear momentum is the following one: P_{i}=mv+mv=2mv.
We also know that the collision is elastic, and, therefore, (all the blocks have the same weight), the final velocity of the whole system, should be exactly 2v (please, correct me if I am wrong)
However, then, I have to calculate P_{f}=m(v_{1}+...+v_{12}), but I don't see how to calculate the relation between all the velocities... In other words, I have the following system:
P_{i}=P_{f} \rightarrow 2v=v_{1} + ... + v_{12}
E_{k(i)}=E_{k(b)} \rightarrow 2v=v_{1} + ... +v_{12}
I post an image so that you can get an idea about how the collision is.
CHAIN.jpg

Thanks.
 
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first of all, since the collision is elastic, the momentum of the block will be transferred to the last block and the rest block will be at rest, so you know the momentum of last i.e MV and the rest 9 blocks will be at rest, i think tis will be enough for you...!
 
If I understand the problem correctly, you have to assume the blocks are all the same mass and the ropes are are all the same length (or, rather, that none is longer than the one preceding). Just consider the first collision to begin with: how will the two blocks move after? As Kishlay says, you will be able quickly to move on to considering the point at which the last block has been hit. But now it gets interesting. We are not told whether the ropes are elastic. If not, what happens when the rightmost rope becomes taut?
 
to whom are you telling this thing??
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
I was thinking using 2 purple mattress samples, and taping them together, I do want other ideas though, the main guidelines are; Must have a volume LESS than 1600 cubic centimeters, and CAN'T exceed 25 cm in ANY direction. Must be LESS than 1 kg. NO parachutes. NO glue or Tape can touch the egg. MUST be able to take egg out in less than 1 minute. Grade A large eggs will be used.
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