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

[StarPhysics]

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 [itex]v[/itex] 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: [tex] P_{i}=P_{f} \rightarrow m_{1}v_{1i}+...+m_{n}v_{ni}=m_{1}v_{1f}+...+m_{n}v_{nf} [/tex]
Moreover, the colission is completely elastic, which means: [tex]E_{k(i)}=E_{k(f)}[/tex]

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: [tex]P_{i}=mv+mv=2mv [/tex].
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 [itex]2v[/itex] (please, correct me if I am wrong)
However, then, I have to calculate [itex]P_{f}=m(v_{1}+...+v_{12})[/itex], but I don't see how to calculate the relation between all the velocities... In other words, I have the following system:
[itex]P_{i}=P_{f} \rightarrow 2v=v_{1} + ... + v_{12}[/itex]
[itex]E_{k(i)}=E_{k(b)} \rightarrow 2v=v_{1} + ... +v_{12}[/itex]
I post an image so that you can get an idea about how the collision is.

Thanks.

 

[Kishlay]

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...!

 

[haruspex]

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?

 

[Kishlay]

to whom are you telling this thing??

 

[Nickie]

The effect of an elastic collision between two blocks and a chain can be determined by using the principles of conservation of momentum and conservation of kinetic energy. In this scenario, we have a chain of 10 blocks connected by a rope and two additional blocks colliding with the chain at one end. The collision is assumed to be elastic, meaning that there is no loss of kinetic energy during the collision.

First, we need to determine the initial and final linear momentum of the system. Initially, the velocity of the chain of 10 blocks is zero, so the initial momentum (P_i) is equal to the momentum of the two colliding blocks (m_1v). After the collision, the final momentum (P_f) will be equal to the sum of the individual momentums of all 12 blocks (m_1v_1 + m_2v_2 + ... + m_{12}v_{12}).

Using the principle of conservation of momentum (P_i = P_f), we can set up the following equation:
m_1v = m_1v_1 + m_2v_2 + ... + m_{12}v_{12}

Next, we need to consider the conservation of kinetic energy. Since the collision is elastic, the initial kinetic energy (E_k(i)) is equal to the final kinetic energy (E_k(f)). We can set up the following equation:
m_1v^2 = m_1v_1^2 + m_2v_2^2 + ... + m_{12}v_{12}^2

Now, we have two equations and 12 unknowns (the final velocities of each block). However, since the blocks are connected by a rope, they will all have the same final velocity (v_f). This is because the rope will prevent any relative motion between the blocks and will ensure that they all move together as one system.

Using this information, we can simplify the equations to:
m_1v = 12m_1v_f
m_1v^2 = 12m_1v_f^2

Solving for v_f, we get:
v_f = v/12

This means that after the collision, all 12 blocks will move together with a final velocity of v/12. This is because the mass of the two colliding blocks is much smaller compared to the mass of the chain, so their contribution to the final velocity