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

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Homework Help Overview

The problem involves a chain of 10 blocks connected by a rope, which is struck by two additional blocks in an elastic collision. The scenario is set on a frictionless table, and the main inquiry is about the movement and speed of the blocks after the collision.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the implications of elastic collisions, momentum transfer, and the conditions affecting the movement of the blocks. Questions arise regarding the assumptions about the mass of the blocks and the nature of the ropes connecting them.

Discussion Status

Some participants have offered insights into the momentum transfer during the collision, while others are exploring the effects of the ropes' properties and the initial conditions of the blocks. There is an ongoing examination of how the collision dynamics play out, particularly regarding the last block in the chain.

Contextual Notes

Participants note the assumption that all blocks have the same mass and that the ropes are of uniform length, but there is uncertainty about whether the ropes are elastic. This may influence the outcome of the collision and the subsequent movement 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??
 

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