Question

A set of n-identical cubical blocks lies 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 :

• A. The last block starts moving at t = n(n-1)$\frac{L}{2v}$
• B. The last block starts moving at t = (n-1)$\frac{L}{v}$
• C. After all collisions the centre of mass of the system will have a speed $\frac{v}{n}$
• D. After all collisions the centre of mass of the system will have a speed v.

Answer 1

Answer: (A), (C)

A, C

Answer 2

The problem describes a series of completely inelastic collisions between identical cubical blocks on a smooth horizontal surface. We need to determine the time when the last block starts moving and the final speed of the center of mass of the system.

Let 'm' be the mass of each identical cubical block.

Part 1: Time when the last block starts moving

1. Collision 1 (Block 1 hits Block 2):

   • Block 1 moves with speed 'v' towards Block 2, which is at rest. The distance is L.
   • Time taken for Block 1 to reach Block 2: $t_1 = \frac{L}{v}$.
   • After the collision, Block 1 and Block 2 stick together (completely inelastic collision).
   • By conservation of momentum: $mv = (m+m)v_1 \Rightarrow v_1 = \frac{v}{2}$.
   • So, a combined block of mass 2m moves with speed $\frac{v}{2}$.

2. Collision 2 (Combined (1+2) hits Block 3):

   • The combined block (mass 2m, speed $\frac{v}{2}$) moves towards Block 3, which is at rest. The distance is L.
   • Time taken for this combined block to reach Block 3: $t_2 = \frac{L}{(v/2)} = \frac{2L}{v}$.
   • Total time from $t=0$ until Block 3 is hit: $T_2 = t_1 + t_2 = \frac{L}{v} + \frac{2L}{v} = \frac{3L}{v}$.
   • After this collision, the combined block (1+2) and Block 3 stick together.
   • By conservation of momentum: $(2m)(\frac{v}{2}) = (2m+m)v_2 \Rightarrow mv = 3mv_2 \Rightarrow v_2 = \frac{v}{3}$.
   • So, a combined block of mass 3m moves with speed $\frac{v}{3}$.

3. Generalizing the pattern:

   • After the k-th collision (when k blocks have combined), the combined mass is km, and its velocity is $\frac{v}{k}$.

   • The time taken for this combined block to travel distance L and hit the (k+1)-th block is: $t_{k+1} = \frac{L}{(v/k)} = \frac{kL}{v}$.

   • The total time from $t=0$ until the k-th collision (i.e., when k blocks have combined, and the (k+1)-th block is hit) is the sum of times for all preceding travels: $T_k = \sum_{i=1}^{k} t_i = \frac{L}{v} + \frac{2L}{v} + \frac{3L}{v} + \dots + \frac{kL}{v}$

     $T_k = \frac{L}{v} (1 + 2 + 3 + \dots + k) = \frac{L}{v} \frac{k(k+1)}{2}$.

4. Time for the last block (n-th block) to start moving:

   • The last block (n-th block) starts moving when it is hit by the combined (n-1) blocks. This corresponds to the $(n-1)$-th collision in our sequence (where k=n-1 blocks have combined and hit the n-th block).

   • So, we need to find $T_{n-1}$:

     $T_{n-1} = \frac{L}{v} \frac{(n-1)((n-1)+1)}{2} = \frac{L}{v} \frac{(n-1)n}{2} = \frac{n(n-1)L}{2v}$.

   • Therefore, option (A) is correct.

Part 2: Speed of the centre of mass after all collisions

1. System: All 'n' identical blocks.

2. Initial State (at t=0):

   • Mass of the system = $nM$.
   • Initial velocity of Block 1 = $v$.
   • Initial velocity of Blocks 2 to n = $0$.
   • Initial momentum of the system ($P_{initial}$) = $mv + (n-1)m(0) = mv$.

3. Conservation of Momentum:

   • The surface is smooth, meaning there are no external horizontal forces acting on the system of 'n' blocks.
   • Therefore, the total momentum of the system is conserved.

4. Final State (after all collisions):

   • All 'n' blocks have collided and stuck together (completely inelastic collisions).
   • The final mass of the combined system is $nM$.
   • Let the final velocity of this combined system be $V_f$.
   • Final momentum of the system ($P_{final}$) = $(nM)V_f$.

5. Applying Conservation of Momentum:

   $P_{initial} = P_{final}$

   $mv = (nM)V_f$

   $V_f = \frac{mv}{nM} = \frac{v}{n}$.

   • Since all blocks are moving together as a single entity, the speed of the center of mass of the system will be equal to the final velocity of this combined block.
   • Therefore, the centre of mass of the system will have a speed $\frac{v}{n}$.
   • Thus, option (C) is correct.