91Ó°ÊÓ

Two spheres \(A\) and \(B\) move on a smooth horizontal surface with the same velocity \(V\) and have some separation between them. A third sphere \(C\) is moving in opposite direction on same surface with same speed. All the spheres are of equal? mass. The collisions are elastic. \(V_{\mathrm{CM}}\) represents the centre mass velocity of all the three spheres. Column II represnt the values after all the possible impacts have occurred. $$ \begin{array}{|l|l|} \hline {\text { Column I }} & {\text { Column II }} \\ \hline \begin{array}{l} \text { i. } \text { If } A \text { and } B \text { are not } \\ \text { connected to each other } \end{array} & \begin{array}{l} \text { a. } \\ \text { is } V / 3 \end{array} \\ \hline \begin{array}{l} \text { ii. If } A \text { and } B \text { are connected } \\ \text { to each other by a } \\ \text { massless rigid rod } \end{array} & \begin{array}{l} \text { b. } V_{\mathrm{CM}} \text { after all the } \\ \text { collision is } \frac{V}{3} \end{array} \\ \hline \begin{array}{l} \text { iii. If } A \text { and } B \text { are connected } \\ \text { by an ideal string } \end{array} & \begin{array}{r} \text { c. } \\ \text { bystem }(B+C) \text { is not } \\ \text { conserved } \end{array} \\ \hline \begin{array}{r} \text { iv. } \text { If } A \text { and } B \text { are connected } \\ \text { by an ideal spring which } \\ \text { is initially unstretched } \end{array} & \text { d. } \begin{array}{l} \text { Momentum of system } \\ (A+B) \text { is conserved } \end{array} \\ \hline \end{array} $$

Step by step solution

01

Understanding initial conditions

We have three spheres \( A, B, \) and \( C \). \( A \) and \( B \) are moving with velocity \( V \) on a smooth horizontal surface. \( C \) is moving in the opposite direction with the same velocity \( V \). All spheres have the same mass.

02

Analyze case i (Spheres A and B are not connected)

As the collisions are elastic, after the collision with sphere \( C \), the velocities will be exchanged due to symmetry and equal masses. Since the velocities are reversed, the initial velocity of the Center of Mass \( V_{\text{CM}} \) is \( \frac{V+V-(-V)}{3} = \frac{3V}{3} = V \). After all collisions, since the velocities exchange symmetrically, \( V_{\text{CM}} \) becomes zero. Therefore, column II corresponds to option a: \( V/3 \).

03

Analyze case ii (Spheres A and B connected by a rigid rod)

When connected by a rigid rod, \( A \) and \( B \) together act as a single body. After the encounters with \( C \), while impulses will exchange between spheres \( A, B, \) and \( C \), \( V_{\text{CM}} \) for the overall system will be unchanged due to symmetry in an enclosed system. So, it remains \( \frac{V}{3} \) (option b).

04

Analyze case iii (A and B connected by an ideal string)

In this scenario, linear momentum may initially be exchanged between \( C \) and \( A+B \) as a system, facilitated by the string. However, because string tension can affect motion, this setup allows for momentum conservation constraints to hold, leading to no steady momentum exchange. This makes option c appropriate, where bystem \((B+C)\) is not conserved.

05

Analyze case iv (A and B connected by an ideal spring)

The spring only allows for oscillatory exchange of momentum within \( A \) and \( B \) without external influences. Elasticity causes an internal balancing effect, so no net momentum is gained or lost within \( A + B \). Momentum is thus internally conserved as described in option d (momentum of system \( (A+B) \) is conserved).

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Key Concepts

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

Center of Mass Velocity

When dealing with multiple objects, like our spheres, it's important to understand the concept of center of mass velocity, denoted by \( V_{\text{CM}} \). Essentially, this is the average velocity at which the entire system's mass is moving.

To calculate \( V_{\text{CM}} \), you need to consider the mass and velocity of each object in the system. Like in our exercise, if all objects have the same mass \( m \), the formula simplifies to the average of their velocities. With three spheres, two moving with velocity \( V \) and one moving with \(-V\), the calculation would be \( V_{\text{CM}} = \frac{V + V - V}{3} = 0 \), meaning initially, their average velocity is balanced.

It's crucial in understanding how systems' components interact since it provides a macroscopic view of the system's movement. After certain interactions, like exchanges of velocity due to collisions, \( V_{\text{CM}} \) may change depending on the scenario's setup.

Momentum Conservation

Momentum conservation is a fundamental principle in physics, especially in the context of collisions. It states that the total momentum of a closed system (no external forces) remains constant over time.

In our scenario for spheres \( A \) and \( B \), whether they are connected or independent, momentum conservation plays a key role. This principle allows us to predict the outcome of interactions. When \( A \) and \( B \) are connected by a rigid rod or a spring, internal forces exist. However, due to the system being closed (no external forces acting on the objects), these internal exchanges do not alter the overall momentum of \( A + B \) or \( C \).

This is particularly important in understanding option (d), where the momentum of \( (A+B) \) remains conserved, indicating that their combined momentum before and after collisions is the same.

Collisions in Mechanics

Collisions are events where two or more bodies exert forces on one another for a relatively short time. Mechanically, these events are fascinating because they involve exchanges of momentum and energy.

In elastic collisions, like our exercise with spheres, both momentum and kinetic energy are conserved. This type of collision means that the objects essentially 'bounce off' each other without any permanent deformation or heat generation. For our spheres \(A, B,\) and \( C\), they exchange velocities due to their equal masses and elastic interaction, resulting in a symmetric reversal of their initial velocities.

This symmetry is why \( V_{\text{CM}} \) changes in specific scenarios, as, after each collision and velocity exchange, the spheres end up moving in different directions, visibly altering the system's overall momentum distribution.

Rigid Rod and Ideal Spring Mechanics

Understanding the effects of connections like rigid rods and ideal springs is key in mechanics, especially in collision scenarios.

- **Rigid Rod**: When objects like spheres \( A \) and \( B \) are connected by a rigid rod, they act as a single entity. This rod restricts any relative movement between the spheres, ensuring that they maintain a constant distance. Hence, for collision analysis, they must be treated as one object. This influences the momentum conservation seen in options where they act collectively, as with \( V_{\text{CM}} = \frac{V}{3} \).

- **Ideal Spring**: Springs introduce elasticity between objects. When \( A \) and \( B \) are connected by an ideal spring, they can undergo oscillatory motions, meaning they can move back and forth around a mean position. The spring stores energy when stretched or compressed, but this exchange is only internal. The overall momentum remains consistent in the absence of external forces, as seen in option (d).

Connections like these dictate how impulse and momentum transfer or conserve, forming the basis for analyzing complex motion scenarios in mechanical processes.