Very elastic collision

Classical Mechanics Level 2

Suppose there are $21$ objects placed one by one such that mass of every object (except the first one) is $two$ times the object placed before it.

Now, if the first object collides with the second object with a velocity of $1ms^{-1}$ and then the second one collides to the third and this process continues and the last object gains a velocity of $p$ $ms^{-1}$ .

Now what is the value of $⌊p \times 10^4⌋$ ?

Note:

Assume all the collisions to be elastic.
$⌊⋅⌋$ denotes the floor function .

The answer is 3.

1 solution

A Former Brilliant Member
Jun 4, 2020

Applying the Energy-Momentum conservation equations successively, we get the following expression for the velocity of the $i^{th}$ object :

$v_i=(\frac{2}{3})^{i-1}$ .

So the velocity of the $21^{st}$ object is $(\frac{2}{3})^{20}\approx 0.0003007$ , and $\lfloor 0.0003007\times 10^4\rfloor =\boxed 3$ .

Very elastic collision

The answer is 3.

1 solution

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