Question

There are 100 identical sliders equally spaced on a frictionless track as shown in the figure. Initially, all the sliders are at rest. Slider 1 is pushed with velocity ‘v’ towards slider 2. In a collision the sliders stick together. The final velocity of the hundred stuck sliders will be:

$\begin{align}
  & (A)\dfrac{v}{99} \\
 & (B)\dfrac{v}{100} \\
 & (C)Zero \\
 & (D)v \\
\end{align}$

Answer 1

Hint: We will apply the conservation of momentum, on the initial system that comprises only one block and then the final system that comprises one hundred blocks stuck together. We can use this analogy because there is no loss of momentum in between any of the collisions up to the last one.

Complete step-by-step answer:
Let us first assign some terms that we are going to use in our calculations.
Since all the sliders are identical, let the mass of each slider be denoted by ‘m’. Also, the initial speed of the first slider is given to us as ‘v’.
Now, let the speed of the system, when all the hundred sliders have stuck together be given by ${{v}_{f}}$, then this speed can be calculated as follows:
Applying the conservation of momentum, we have:
Initial momentum of one block is equal to final momentum of one hundred blocks stuck together. Mathematically, this could be written as follows:
$\begin{align}
  & \Rightarrow mv=(m+m+m+.....................\text{ upto hundred times)}{{v}_{f}} \\
 & \Rightarrow mv=100m{{v}_{f}} \\
 & \therefore {{v}_{f}}=\dfrac{v}{100} \\
\end{align}$
Hence, the final velocity of the hundred stuck sliders comes out to be $\dfrac{v}{100}$.

So, the correct answer is “Option B”.

Additional Information: We can generalize the above velocity equation for ‘n’ number of identical blocks stuck together at the end as:
$\Rightarrow {{v}_{f}}=\dfrac{v}{n}$
This relation is valid only for this problem.

Note: We can also deduce some extra information from the given data in the problem. Since, the momentum is conserved it means that there is no external force like friction and air resistance. Also, since on every collision, the block sticks together, all these collisions are completely inelastic collisions.