A set of n identical cubical blocks lies at r | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Since the collisions are perfectly inelastic,all the blocks will stick together one by one and move as a combined mass.Let the mass of each block

Vedclass · Accepted Answer

The last block starts moving at $t = \frac{n(n - 1)L}{2v}$

Vedclass · Answer

(B) Since the collisions are perfectly inelastic,all the blocks will stick together one by one and move as a combined mass.Let the mass of each block be $m$.Time required for the first block to cover distance $L$ is $t_1 = \frac{L}{v}$.After the first collision,the first and second blocks stick together. By conservation of momentum,$mv = (2m)v_1$,so $v_1 = \frac{v}{2}$.The time taken by this combined system to cover the next distance $L$ is $t_2 = \frac{L}{v/2} = \frac{2L}{v}$.After the second collision,the three blocks move with velocity $v_2 = \frac{mv}{3m} = \frac{v}{3}$.The time taken to cover the next distance $L$ is $t_3 = \frac{L}{v/3} = \frac{3L}{v}$.Continuing this process,the time taken to cover the distance between the $(n-1)$-th and $n$-th block is $t_{n-1} = \frac{(n-1)L}{v}$.The total time taken for the last block to start moving is the sum of these intervals:$T = \frac{L}{v} + \frac{2L}{v} + \frac{3L}{v} + ... + \frac{(n-1)L}{v} = \frac{L}{v} (1 + 2 + 3 + ... + (n-1)) = \frac{L}{v} \cdot \frac{(n-1)n}{2} = \frac{n(n-1)L}{2v}$.Thus,option $(b)$ is correct.

Similar Questions

$A$ particle of mass $M$ moving along the $X$-axis with speed $V_0$ collides with and sticks to another particle of mass $m$ moving along the $Y$-axis with speed $V_0$. What will be the velocity of the combined mass after the collision?

$A$ $2 \ kg$ block slides on a horizontal surface at a speed of $4 \ m/s$ and strikes an uncompressed spring. The kinetic friction force is $15 \ N$ and the spring constant is $10,000 \ N/m$. By how many $cm$ will the spring be compressed?

The friction coefficient between the horizontal surface and each of the blocks shown in the figure is $\mu = 0.2$. The collision between the blocks is perfectly elastic. What is the separation between the blocks when they come to rest? (in $cm$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module