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
• B. The last block starts moving at
• C. The centre of mass of the system will have a final speed v
• D. The centre of mass of the system will have a final speed

Answer 1

Answer: (B), (D)

The correct options are:

B The last block starts moving at

D The centre of mass of the system will have a final speed

Time required to cover a distance ‘L’ by first block $=\frac{L}{v}$

Now first and second block will stick together and move with $\frac{v}{2}$ velocity (by applying conservation of momentum) and combined system will take time $\frac{L}{\frac{v}{2}}=\frac{2L}{v}$ to reach up to block third.

Now these three blocks will move with velocity $\frac{v}{3}$ and combined system will take time $\frac{L}{\frac{v}{3}}=\frac{3L}{v}$ to reach upto the block fourth.

So, total time $=\frac{L}{v}+\frac{2L}{v}+\frac{3L}{v}+...\frac{(n-1)L}{v}=\frac{n(n-1)L}{2v}$ and velocity of combined system having n blocks is $\frac{v}{n}$