Question

A set of $n$ identical cubical blocks lie at rest parallel to each other along a line on a smooth horizontal surface. The separation between the near surfaces of 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
(i) The last block starts moving at $t=\frac{(n-1)L}{v}$.
(ii) The last block starts moving at $t=\frac{n(n-1)L}{2v}$.
(iii) The center of mass of the system will have a final speed $v$.
(iv) The center of mass of the system will have a final speed $\frac{v}{n}$.

• A. (i),(iv)
  
• B. (ii),(iii)
  
• C. (ii),(iv)
  
• D. (i),(iii)
  

Answer 1

The correct option is C (ii),(iv)

Hint: “Conservation of linear momentum”

Formula used:

$P_i=P_f$

$t=\frac{L}{v}$

Solution:

The time after which the 1st collision takes place $=\frac{L}{v}$

Velocity after the $1$st collision,
$mv=(m+m)v_1\Rightarrow v_1=\frac{v}{2}$

Time between the $1$st and the $2$nd collision,

$=\frac{L}{\frac{v}{2}}=\frac{2L}{v}$

Velocity after the $2$nd collision,

$2m\times \frac{v}{2}=(2m+m)v_2\Rightarrow v_2=\frac{v}{3}$

Time between the 2nd and the 3rd collision,

$=\frac{L}{\frac{v}{3}}=\frac{3L}{v}$

Total time up to the $(n-1{)}^{th}$ collision is

$\frac{L}{v}+\frac{2L}{v}+\frac{3L}{v}+\dots +\frac{(n-1)L}{v}=\frac{n(n-1)L}{2v}$

Velocity of the center of mass after the $(n-1{)}^{th}$ collision $=\frac{v}{n}$

Final Answer: (d)