/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 38 \(=\) A set on \(n\) identical... [FREE SOLUTION] | 91Ó°ÊÓ

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\(=\) A set on \(n\) identical cubical blocks lies at rest parallel to each other along a line on a smooth horizontal surface. The separation between the near surface of any two adjacent block 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 \(t=\frac{(n-1)}{v} L\) (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 \(\frac{v}{ }\)

Short Answer

Expert verified
Options (a) and (c) are correct.

Step by step solution

01

Identify Initial Conditions

Let's consider the initial condition of the cubical blocks. We have a total of \( n \) blocks, each at rest, with only the first block moving with speed \( v \). The distance between the near surfaces of any two adjacent blocks is \( L \). All collisions are completely inelastic, meaning after each collision, the colliding blocks move together.
02

Calculate Time for Last Block to Move

Since all collisions are inelastic, each block will take an equal amount of time \( \frac{L}{v} \) to reach and collide with the next block, as each collision results in stillness until the preceding block impacts it. For \( n-1 \) blocks, the time it takes for the last block to move is calculated as the sum of these times: \[ t = (n-1) \times \frac{L}{v} = \frac{(n-1)L}{v}. \] This matches option (a).
03

Understand Centre of Mass Speed

Since all blocks eventually stick together and the original velocity was \( v \), the final speed of the center of mass of the system involving all \( n \) blocks will be the initial momentum divided by the total mass, which remains the velocity \( v \). Hence, the center of mass will have the final speed \( v \), as stated in option (c).
04

Identify Correct Options

Based on our analysis, we verify that the times and speeds for the scenarios match options. (a) matches the derived time and (c) matches the speed of the center of mass as derived.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Kinematics
Kinematics is the branch of physics focused on the motion of objects without considering the causes of the motion. In problems involving inelastic collisions, such as the one with cubical blocks, the kinematic equations help us understand how long it takes for each block to start moving.

For the blocks, kinematics boils down to calculating the time intervals between collisions.
  • Each collision is inelastic, meaning the blocks stick together after impact.
  • Initially, only the first block travels with speed \( v \).
  • Successive collisions happen at intervals of \( \frac{L}{v} \) since each block takes this time to reach the next.
This time interval is crucial in determining when the last block begins moving. By adding up each of these equal time intervals, we can use the formula \( \frac{(n-1)L}{v} \) to determine the time after which the last block will start moving.
This method uses a straightforward application of kinematic principles, breaking down the motion step-by-step. It simplifies the analysis of systems undergoing sequential inelastic collisions.
Momentum Conservation
Momentum conservation is a fundamental principle of physics stating that the total linear momentum of a closed system remains constant, provided no external forces act on it. In the case of cubical blocks, it is key to understanding the dynamics after each collision.

In fully inelastic collisions where blocks stick together:
  • The momentum before collision equals the momentum after the collision.
  • If the first block has momentum \( mv \) (where \( m \) is mass and \( v \) is speed), momentum conservation ensures the combined system's momentum doesn't change.
As each collision occurs, the combined mass of the blocks increases, altering the speed of the system. However, the initial momentum still governs, ensuring that the system adheres to the conservation rules.
For example, once all blocks have collided and are moving together, the system's net velocity remains \( v \) due to the uniform distribution of momentum across a larger mass. Understanding this proves that momentum is conserved, validating the final speed predictions.
Center of Mass Velocity
The center of mass (COM) is a conceptual point in a system where the entire mass is thought to be concentrated. The velocity of the center of mass can give insights into the movement of the whole system, particularly after multiple inelastic collisions.

In the block scenario:
  • Initially, only the first block moves with speed \( v \).
  • As each block joins the movement due to collisions, they collectively have a redistributed mass moving with uniform velocity.
  • The velocity of the center of mass for the entire system then becomes \( v \) after completion of all collisions.
This happens because the speed of the COM is the total momentum divided by total mass, which remains equal to the initial speed \( v \) even as mass increases.
So, after all blocks collide and move together, the COM moves uniformly with the original speed. This illustrates how the center of mass velocity stays constant and validates the conclusions about system motion post-collisions.

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Most popular questions from this chapter

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