A set of n-identical cubical blocks lie at rest along a line on a smooth horizontal surface. The separation between any two adjacent blocks is L. The block at one end is given a speed V towards the next one at time t = 0. All collisions are completely inelastic, then
A
The last block starts moving at t = n (n – 1) 
B
The last block starts moving at t = (n – 1) 
C
The centre of mass of the system will have a final speed v/n
D
The centre of mass of the system will have a final speed v
Text Solution
Verified by ExpertsThe correct answer is:
CHECK THE SOLUTION.
(a, c)
mv = nv ′ m ⇒ v ′ = 
time for first collisen is
t 1 = 
( 2nd block)
2nd collisions , t 2 = 
= 2t 1 (3rd block)
So t = t 1 + 2t 1 + 3t 1 + at 1 ...........(n–1) t 1 .
t = t 1 [1 + 2 + 3] .......................(n–1)]
= 
= 
So t = 
n (n – 1)
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