SATHEE: Work Energy And Power Question 78

Work Energy And Power Question 78

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 [IIT 1995]

Options:

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

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

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 $\frac{v}{n}$

Show Answer

Answer:

Correct Answer: B

Solution:

Since collision is perfectly inelastic so all the blocks will stick together one by one and move in a form of combined mass. Time required to cover a distance L by first block $= \frac{L}{v}$

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

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

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

$= \frac{n (n - 1) L}{2 v}$ and velocity of combined system having n blocks as $\frac{v}{n}$ .

Work Energy And Power Question 78

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Answer:

Solution:

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Work Energy And Power Question 78

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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