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MOTION OF SYSTEM OF PARTICLES & COM - 04 | Elastic Collision | Physics | Class 11th/NEET/JEE

PW English Medium・62 minutes read

The lecture series delves into collisions, emphasizing the distinctions between elastic and inelastic interactions, alongside relevant formulas and principles such as momentum conservation and the coefficient of restitution. Students are encouraged to take personal notes to enhance retention while tackling various examples and problems related to calculating final velocities and energy lost during these collisions.

Insights

  • The lecture series is currently focused on collisions, particularly elastic and inelastic types, and includes five sessions before transitioning to topics like rotational motion and thermodynamics, emphasizing a structured learning approach.
  • Students are encouraged to create their own notes for each chapter instead of relying on online resources, as this method enhances their understanding and retention of key concepts and formulas necessary for exams.
  • Collisions are defined as interactions where the momentum of at least one object changes, with momentum being conserved in the absence of external forces; this principle applies to both elastic and inelastic collisions.
  • In elastic collisions, both momentum and kinetic energy are conserved, while inelastic collisions conserve momentum but result in kinetic energy loss, highlighting a fundamental difference in energy behavior during these interactions.
  • The lecture outlines specific equations for calculating final velocities in elastic collisions and emphasizes the importance of understanding the relationship between initial and final velocities, particularly in head-on and oblique collisions.
  • Practical examples and problems are provided to illustrate the application of collision principles, such as calculating final velocities and energy lost in various scenarios, reinforcing the theoretical concepts discussed in the lectures.

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Recent questions

  • What is a perfectly inelastic collision?

    A perfectly inelastic collision occurs when two colliding bodies stick together after the impact, moving with a common velocity. In this type of collision, momentum is conserved, but kinetic energy is not. The initial kinetic energy of the system is greater than the final kinetic energy, indicating that some energy is transformed into other forms, such as heat or sound, during the collision. An example of a perfectly inelastic collision is when a bullet embeds itself into a block of wood; both the bullet and the block move together after the collision, demonstrating the loss of kinetic energy and the conservation of momentum.

  • How do elastic collisions differ from inelastic collisions?

    Elastic collisions differ from inelastic collisions primarily in the conservation of kinetic energy. In elastic collisions, both momentum and kinetic energy are conserved, meaning that the total kinetic energy before and after the collision remains the same. This is exemplified by two billiard balls colliding, where they bounce off each other without any loss of energy. In contrast, inelastic collisions conserve momentum but not kinetic energy; some kinetic energy is transformed into other forms of energy, resulting in a loss of total kinetic energy. A common example of an inelastic collision is a car crash, where the vehicles crumple and some energy is dissipated as sound and heat.

  • What is the coefficient of restitution?

    The coefficient of restitution is a measure of the elasticity of a collision between two bodies, defined as the ratio of the relative velocity of separation to the relative velocity of approach. It quantifies how much kinetic energy remains after a collision compared to how much was present before the collision. A coefficient of restitution value of 1 indicates a perfectly elastic collision, where no kinetic energy is lost, while a value of 0 indicates a perfectly inelastic collision, where the bodies stick together and maximum kinetic energy is lost. Values between 0 and 1 represent partially elastic collisions, where some kinetic energy is conserved, and the coefficient provides insight into the nature of the collision.

  • What is momentum conservation in collisions?

    Momentum conservation in collisions refers to the principle that the total momentum of a closed system remains constant if no external forces act on it. In the context of collisions, this means that the total momentum before the collision is equal to the total momentum after the collision. This principle applies to both elastic and inelastic collisions, allowing for the calculation of final velocities of colliding bodies based on their initial velocities and masses. For example, in a head-on collision between two objects, the equation \( m_1 U_1 + m_2 U_2 = m_1 V_1 + m_2 V_2 \) can be used to determine the final velocities \( V_1 \) and \( V_2 \) after the collision, demonstrating the conservation of momentum.

  • How can I prepare for physics exams effectively?

    To prepare for physics exams effectively, it is crucial to develop a strong understanding of the concepts and principles covered in the course. One effective strategy is to create short notes for each chapter, summarizing key formulas, definitions, and examples. This practice not only aids in retention but also helps in organizing information for quick review. Additionally, solving practice problems and past exam questions can reinforce understanding and application of concepts. Engaging in group study sessions can also be beneficial, as discussing topics with peers can clarify doubts and enhance learning. Lastly, ensuring a consistent study schedule and taking breaks to avoid burnout will contribute to a more effective and productive exam preparation process.

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Summary

00:00

Understanding Elastic and Inelastic Collisions

  • The lecture series on the system of particles and center of mass has reached its fourth session, focusing on collisions, specifically elastic and inelastic collisions, and the coefficient of restitution, with a total of five lectures planned for this chapter before moving on to rotational motion and other topics like heat, thermodynamics, and gravitation.
  • Students are advised to prepare their own short notes for each chapter rather than relying on internet resources, as this practice aids in better retention of formulas and concepts, which is crucial for exam preparation.
  • A collision is defined as an event where the momentum of at least one body changes due to the interaction with another body, indicating that momentum is conserved in the absence of external forces.
  • There are two main types of collisions: elastic and inelastic. In both cases, momentum is conserved, meaning the initial momentum equals the final momentum, regardless of the type of collision.
  • In elastic collisions, both momentum and kinetic energy are conserved, while in inelastic collisions, momentum is conserved but kinetic energy is not; the initial kinetic energy is greater than the final kinetic energy, indicating energy loss.
  • A perfectly inelastic collision occurs when two colliding bodies stick together and move with a common velocity after the collision, exemplified by a bullet embedding into a block.
  • Elastic collisions can be further classified into head-on and oblique collisions. In head-on collisions, the velocity vectors of the colliding bodies are aligned along the line of impact, while in oblique collisions, they are not.
  • The momentum conservation equation for a head-on elastic collision is expressed as \( m_1 U_1 + m_2 U_2 = m_1 V_1 + m_2 V_2 \), where \( m_1 \) and \( m_2 \) are the masses, and \( U_1, U_2, V_1, V_2 \) are the initial and final velocities of the bodies.
  • The kinetic energy conservation equation for the same collision is given by \( \frac{1}{2} m_1 U_1^2 + \frac{1}{2} m_2 U_2^2 = \frac{1}{2} m_1 V_1^2 + \frac{1}{2} m_2 V_2^2 \), which can be simplified to show the relationship between the squares of the velocities before and after the collision.
  • The lecture emphasizes the importance of understanding the differences between elastic and inelastic collisions, particularly in terms of energy conservation, and provides mathematical equations to illustrate these principles.

18:19

Understanding Elastic Collisions and Velocity Changes

  • The equation \( u_1 - u_2 = v_2 - v_1 \) represents the relationship between the relative velocities of two bodies before and after a head-on elastic collision, where \( u_1 \) and \( u_2 \) are the initial velocities, and \( v_1 \) and \( v_2 \) are the final velocities.
  • The relative velocity of approach, defined as \( u_1 - u_2 \), is equal to the relative velocity of separation, defined as \( v_2 - v_1 \), indicating that momentum and kinetic energy are conserved in elastic collisions.
  • For two colliding bodies with masses \( m_1 \) and \( m_2 \), the formulas for their final velocities after an elastic collision are given by \( v_1 = \frac{m_1 - m_2}{m_1 + m_2} u_1 + \frac{2m_2}{m_1 + m_2} u_2 \) and \( v_2 = \frac{m_2 - m_1}{m_1 + m_2} u_2 + \frac{2m_1}{m_1 + m_2} u_1 \).
  • To simplify the calculation of final velocities, if \( m_1 \) is the mass of the first body and \( m_2 \) is the mass of the second body, the first term in \( v_1 \) and \( v_2 \) formulas represents the effect of the mass difference on the initial velocities.
  • An example problem involves two particles of mass \( m \) and \( 2m \) moving in opposite directions with velocities \( v \) and \( 2v \); the final velocities can be calculated using the provided formulas.
  • For the specific case of a 2 kg ball colliding with another object, if the initial velocities are \( 5 \, \text{m/s} \) and \( 2 \, \text{m/s} \), the final velocity \( v_1 \) can be calculated as \( v_1 = \frac{2 - 3}{5} \cdot 5 + \frac{6}{5} \cdot 2 = \frac{7}{5} \, \text{m/s} \).
  • In a scenario where a smooth sphere of mass \( m \) collides with another sphere of mass \( m \) at rest, the final velocity of the second sphere can be derived from the formula \( v_2 = \frac{2m}{m + m} u = u \), indicating it moves with the same velocity as the first sphere.
  • The calculations for final velocities involve substituting the masses and initial velocities into the formulas, ensuring to account for the direction of motion by assigning positive and negative values appropriately.
  • Understanding these principles and formulas is crucial for solving problems related to elastic collisions in one dimension, as they provide a systematic approach to determining the outcomes of such interactions.

35:41

Collision Analysis and Energy Loss Calculation

  • The problem involves a 1 kg block with an initial velocity \( U \) colliding with another block moving at \( U/3 \), and the goal is to find the mass \( m \) of the second block after the collision.
  • The formula for the final velocity \( V_1 \) after the collision is given by \( V_1 = \frac{m_1 - m_2}{m_1 + m_2} U_1 + \frac{2m_2}{m_1 + m_2} U_2 \), where \( U_2 = 0 \) and \( U_1 = U \).
  • Substituting the values into the formula, we find \( V_1 = \frac{1 - m}{1 + m} U \) and set \( V_1 = -U/3 \) to account for the direction of motion.
  • Cross-multiplying leads to the equation \( -1 - m = 3 - 3m \), which simplifies to \( 2m = 4 \), resulting in \( m = 2 \) kg for the second block.
  • In a separate problem, a body of mass \( 4M \) collides with a stationary body of mass \( 2M \) in a head-on elastic collision, and the fraction of energy lost by the colliding body is to be determined.
  • The initial kinetic energy of the moving body \( A \) is calculated as \( KE_i = \frac{1}{2} \times 4M \times U^2 = 2MU^2 \).
  • After the collision, the final velocity \( V_1 \) of body \( A \) is calculated using the same formula, yielding \( V_1 = \frac{4M - 2M}{6M} U = \frac{U}{3} \).
  • The final kinetic energy of body \( A \) after the collision is \( KE_f = \frac{1}{2} \times 4M \times \left(\frac{U}{3}\right)^2 = \frac{4M U^2}{18} = \frac{2MU^2}{9} \).
  • The energy lost is calculated as \( KE_i - KE_f = 2MU^2 - \frac{2MU^2}{9} = \frac{18MU^2 - 2MU^2}{9} = \frac{16MU^2}{9} \).
  • The fraction of energy lost is then \( \frac{\text{Energy lost}}{\text{Initial energy}} = \frac{\frac{16MU^2}{9}}{2MU^2} = \frac{16}{18} = \frac{8}{9} \), confirming that the correct option is \( \frac{8}{9} \).

52:50

Collision Dynamics and Velocity Calculations

  • To calculate the velocities \( V_1 \) and \( V_2 \) after a collision, start with the formula for \( V_1 \): \( V_1 = \frac{m_1 - m_2}{m_1 + m_2} U_1 + \frac{2m_2}{m_1 + m_2} U_2 \). If \( m_2 \) is negligible compared to \( m_1 \), then \( V_1 \) simplifies to \( U_1 \), meaning the velocity of the heavier mass remains unchanged after the collision.
  • For \( V_2 \), use the formula: \( V_2 = \frac{m_2 - m_1}{m_1 + m_2} U_2 + \frac{2m_1}{m_1 + m_2} U_1 \). If \( m_2 \) is negligible, this simplifies to \( V_2 = 2U_1 \), indicating that the smaller mass moves at twice the velocity of the larger mass after the collision.
  • An example scenario involves a truck moving at 50 m/s colliding with a stationary bike. After the collision, the truck continues at 50 m/s, while the bike accelerates to 100 m/s, which is twice the truck's speed.
  • In the case of two identical bodies colliding, their velocities are exchanged post-collision. If one body has velocity \( U_1 \) and the other \( U_2 \), after the collision, the first body moves with \( U_2 \) and the second with \( U_1 \).
  • If \( m_2 \) is significantly larger than \( m_1 \) and \( U_2 \) is 0, the smaller mass \( m_1 \) will move backward with velocity \( U_1 \) after the collision, while \( m_2 \) remains at rest.
  • When \( m_1 \) is much larger than \( m_2 \) and \( U_2 \) is at rest, \( m_1 \) retains its velocity \( U_1 \) after the collision, while \( m_2 \) moves forward at \( 2U_1 \).
  • A question example involves two identical balls colliding elastically. If ball A moves at 0.5 m/s and ball B at 0.3 m/s, after the collision, ball A will move at 0.3 m/s and ball B at 0.5 m/s, effectively swapping their velocities.
  • In another scenario, a ball moving horizontally with speed \( V \) strikes a pendulum bob of equal mass at rest. The bob will rise to a height \( H \) calculated using the formula \( H = \frac{V^2}{2g} \), where \( g \) is the acceleration due to gravity.
  • For a pendulum bob released from a horizontal position, when it strikes another bob at rest, the speed of the second bob after the collision can be determined using conservation of energy principles, leading to a speed of \( V = \sqrt{20} \) m/s, approximately 4.47 m/s.
  • Lastly, when a spring is involved in an elastic collision, the maximum compression of the spring can be analyzed by considering the initial velocities and masses of the colliding objects, applying conservation of momentum and energy principles to find the compression during the collision.

01:10:15

Momentum and Energy in Collision Dynamics

  • Before the collision, a mass \( m \) gains an initial velocity \( V_0 \). After the collision, it moves with a new velocity \( U \), while a mass \( 2m \) also moves with the same velocity \( U \). The conservation of linear momentum states that the initial momentum \( m \cdot V_0 \) equals the final momentum \( (m + 2m) \cdot U \), leading to the equation \( V_0 = 3U \), which simplifies to \( U = \frac{V_0}{3} \).
  • The conservation of mechanical energy is applied to analyze the system. The initial kinetic energy is given by \( \frac{1}{2} m V_0^2 \), and the potential energy of the spring is zero at its natural length. After the collision, both masses \( m \) and \( 2m \) move with velocity \( \frac{V_0}{3} \), resulting in their kinetic energies being \( \frac{1}{2} m \left(\frac{V_0}{3}\right)^2 \) and \( \frac{1}{2} (2m) \left(\frac{V_0}{3}\right)^2 \) respectively, while the potential energy stored in the spring is \( \frac{1}{2} k x^2 \).
  • By equating the initial kinetic energy to the sum of the final kinetic energies and the potential energy in the spring, the equation \( m V_0^2 = \frac{1}{9} m V_0^2 + \frac{2}{9} m V_0^2 + \frac{1}{2} k x^2 \) is derived. Simplifying this leads to \( k x^2 = \frac{2m V_0^2}{3} \), resulting in the displacement \( x = \sqrt{\frac{2m}{3k}} V_0 \).
  • Two homework problems are assigned: one involves a ball moving at 2 m/s towards a wall moving at 1 m/s, requiring the calculation of the ball's velocity after an elastic collision. The second problem involves a proton colliding with a heavy particle at rest, where the proton bounces back with \( \frac{4}{9} \) of its kinetic energy, prompting the calculation of the mass of the heavy particle. Students are encouraged to write short notes after each lecture for better retention.
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