Center of Mass and Collisions: Comprehensive Physics Notes
Meta Description: Master advanced collision mechanics, elastic and inelastic impacts, Newton’s law of restitution, rigorous step-by-step mathematical derivations, detailed vector diagrams, and numerical problem-solving.
Table of Contents
- 1. Definition of Collision
- 2. Law of Conservation of Kinetic Energy
- 3. Types of Collisions and Definitions for Each One
- 4. One-Dimensional (Head-On) Elastic Collision & Detailed Derivations
- 5. Head-on Inelastic Collision with Mathematical Expression
- 6. Perfect Inelastic Collision & Velocity Derivation
- 7. Loss in Kinetic Energy During Inelastic Collision
- 8. Coefficient of Restitution & Newton’s Law of Restitution
- 9. Conceptual Understanding Questions
- 10. Solved Numerical Problems
1. Definition of Collision
Definition: A collision is an isolated, dynamic event in which two or more bodies exert strong mutual impulsive forces on each other for an extremely short interval of time, resulting in a substantial redistribution of momentum and energy.
Physical contact is not strictly required for a collision to occur; interactions can be mediated by repulsive or attractive fields, such as electrostatic or magnetic forces between subatomic particles. During the impact phase, contact forces rise rapidly to a high peak value before decreasing to zero as the bodies separate. Because the duration of impact is infinitesimally small, external forces like gravity and friction are neglected during the collision interval.
2. Law of Conservation of Kinetic Energy
Definition: The law of conservation of kinetic energy states that the total kinetic energy of a closed, isolated system remains invariant before and after an interaction, provided no mechanical energy is dissipated into non-conservative forms such as heat, sound, or permanent deformation.
Mathematical Expression
$$\sum K_{\text{initial}} = \sum K_{\text{final}}$$
$$\frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2$$
During physical impacts, kinetic energy is momentarily converted into internal elastic potential energy during the maximum deformation stage. In perfectly elastic systems, this potential energy is fully restored into kinetic energy upon separation.
3. Types of Collisions and Definitions for Each One
Definition: Collisions are broadly classified into elastic, inelastic, and perfectly inelastic categories based on whether mechanical kinetic energy is conserved and how the bodies interact post-impact.
1. Elastic Collision
An interaction in which both total linear momentum and total kinetic energy are rigorously conserved, with zero net thermal or structural energy loss ($e = 1$).
2. Inelastic Collision
An interaction where total linear momentum is conserved, but total kinetic energy decreases due to energy dissipation into heat, sound, or deformation ($0 < e < 1$).
3. Perfectly Inelastic Collision
An extreme inelastic interaction where colliding bodies stick together upon impact and move onward with a single unified common velocity ($e = 0$).
4. One-Dimensional (Head-On) Elastic Collision & Detailed Derivations
Definition: A head-on elastic collision is a direct, one-dimensional impact along a single line of action between two bodies where both linear momentum and kinetic energy remain perfectly conserved.
Complete Step-by-Step Derivation
Step 1: Conservation of Linear Momentum
Total momentum before collision equals total momentum after collision:
$$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$$
Rearranging terms to group masses together:
$$m_1 (u_1 – v_1) = m_2 (v_2 – u_2) \quad \text{— (Equation 1)}$$
Step 2: Conservation of Kinetic Energy
Total kinetic energy before collision equals total kinetic energy after collision:
$$\frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2$$
Cancelling out the factor of $\frac{1}{2}$ and grouping terms:
$$m_1 (u_1^2 – v_1^2) = m_2 (v_2^2 – u_2^2) \quad \text{— (Equation 2)}$$
Step 3: Algebraic Expansion of Kinetic Energy Equation
Using the algebraic identity $(a^2 – b^2) = (a – b)(a + b)$, Equation 2 can be rewritten as:
$$m_1 (u_1 – v_1)(u_1 + v_1) = m_2 (v_2 – u_2)(v_2 + u_2) \quad \text{— (Equation 3)}$$
Step 4: Establishing Relative Velocity Relationship
Dividing Equation 3 by Equation 1 ($\frac{\text{Eq. 3}}{\text{Eq. 1}}$):
$$\frac{m_1 (u_1 – v_1)(u_1 + v_1)}{m_1 (u_1 – v_1)} = \frac{m_2 (v_2 – u_2)(v_2 + u_2)}{m_2 (v_2 – u_2)}$$
Simplifying the terms yields Newton’s law of restitution for elastic impacts:
$$u_1 + v_1 = v_2 + u_2 \implies u_1 – u_2 = v_2 – v_1$$
This proves that the relative velocity of approach ($u_1 – u_2$) equals the relative velocity of separation ($v_2 – v_1$).
Step 5: Solving for Final Velocity $v_1$
From Step 4, express $v_2$ as: $v_2 = u_1 – u_2 + v_1$. Substituting this into Equation 1:
$$m_1(u_1 – v_1) = m_2((u_1 – u_2 + v_1) – u_2)$$
Solving algebraically for $v_1$ yields:
$$v_1 = \left(\frac{m_1 – m_2}{m_1 + m_2}\right) u_1 + \left(\frac{2m_2}{m_1 + m_2}\right) u_2$$
Step 6: Solving for Final Velocity $v_2$
Similarly, substituting $v_1 = v_2 – u_1 + u_2$ into the momentum equation yields:
$$v_2 = \left(\frac{2m_1}{m_1 + m_2}\right) u_1 + \left(\frac{m_2 – m_1}{m_1 + m_2}\right) u_2$$
5. Head-on Inelastic Collision with Mathematical Expression
Definition: A head-on inelastic collision is a direct impact along a single line of motion where linear momentum is conserved, but total kinetic energy decreases due to internal thermal and structural dissipation.
Mathematical Expression
$$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$$
$$v_2 – v_1 = e(u_1 – u_2) \quad \text{where } 0 < e < 1$$
Because mechanical energy is lost during deformation, the relative speed of separation is reduced by factor $e$ compared to the initial approach speed.
6. Perfect Inelastic Collision & Velocity Derivation
Definition: A perfectly inelastic collision is a direct impact in which the colliding bodies lock or stick together permanently upon contact, continuing to move onward as a single combined mass with a unified common velocity ($e = 0$).
Step-by-Step Derivation
Step 1: Application of Conservation of Linear Momentum
Since no external impulsive forces act on the system during impact, total initial momentum equals total final momentum:
$$m_1 u_1 + m_2 u_2 = (m_1 + m_2) v$$
Step 2: Solving for Common Final Velocity ($v$)
Isolating velocity $v$ on one side of the equation:
$$v = \frac{m_1 u_1 + m_2 u_2}{m_1 + m_2}$$
7. Loss in Kinetic Energy During Inelastic Collision
Definition: The loss in kinetic energy ($\Delta K$) represents the net mechanical energy deficit converted irreversibly into thermal energy, sound, and structural deformation work during an inelastic impact.
Step-by-Step Derivation
Step 1: Energy Difference Expression
$$\Delta K = K_{\text{initial}} – K_{\text{final}}$$
$$\Delta K = \left( \frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 \right) – \frac{1}{2} (m_1 + m_2) v^2$$
Step 2: Substitution and Simplification
Substituting the common velocity formula for a perfectly inelastic collision and simplifying algebraically yields:
$$\Delta K = \frac{1}{2} \left(\frac{m_1 m_2}{m_1 + m_2}\right) (u_1 – u_2)^2$$
8. Coefficient of Restitution & Newton’s Law of Restitution
Definition: The coefficient of restitution ($e$) is a dimensionless material property defined as the ratio of the relative velocity of separation after impact to the relative velocity of approach before impact.
Mathematical Expression
$$e = \frac{\text{Relative Velocity of Separation}}{\text{Relative Velocity of Approach}} = \frac{v_2 – v_1}{u_1 – u_2}$$
For perfectly elastic collisions, $e = 1$; for perfectly inelastic collisions, $e = 0$; and for general real-world inelastic collisions, $0 < e < 1$.
9. Conceptual Understanding Questions
Definition: Conceptual questions are theoretical prompts designed to test fundamental qualitative comprehension and analytical reasoning regarding conservation laws and collision dynamics.
- Why is linear momentum conserved during a collision even when external forces like gravity act on the system?
Answer: Internal impulsive contact forces are orders of magnitude larger than external forces, making external forces negligible over the infinitesimally short time interval ($\Delta t \to 0$). - Can a collision occur without physical contact between the bodies? Provide an example.
Answer: Yes, via electrostatic repulsion between alpha particles and gold atomic nuclei or gravitational slingshot maneuvers in celestial mechanics. - What happens to the total kinetic energy in a perfectly inelastic collision?
Answer: The lost kinetic energy dissipates into thermal energy, acoustic energy, and permanent structural deformation work. - Is it possible for the coefficient of restitution $e$ to be greater than 1? Explain.
Answer: No, in classical macroscopic physics $e \le 1$. Values exceeding 1 would imply spontaneous release of internal chemical or nuclear energy. - Two identical masses undergo a 1D elastic head-on collision where one is stationary. What happens?
Answer: They completely exchange velocities; the striking mass stops entirely and the target mass moves off with the initial velocity of the striker. - Why does a cricket fielder pull their hands backward while catching a fast-moving ball?
Answer: Pulling hands back increases impact duration ($\Delta t$), which drastically reduces the average impact force ($F_{\text{avg}} = \frac{\Delta p}{\Delta t}$). - Does the center of mass of an isolated system accelerate if internal collisions take place?
Answer: No, internal action-reaction forces cancel out internally, leaving the acceleration of the center of mass strictly zero. - Why is the relative velocity of separation zero in a perfectly inelastic collision?
Answer: Because both bodies lock together and travel onward with a unified common velocity. - How does the mass ratio affect energy transfer in a 1D elastic collision?
Answer: Maximum kinetic energy transfer occurs when colliding bodies have equal masses ($m_1 = m_2$). - Is kinetic energy conserved during the maximum deformation stage of an inelastic collision?
Answer: No, relative velocity drops to zero and kinetic energy is temporarily stored as internal potential energy.
10. Solved Numerical Problems
Definition: Solved numerical problems are quantitative practice exercises applying physics formulas to calculate concrete values for velocities, energy losses, and impact parameters.
- Problem 1: A ball of mass $2\ \text{kg}$ moving at $10\ \text{m/s}$ collides head-on with a stationary ball of mass $3\ \text{kg}$. If they stick together, find their common velocity.
Solution: $v = \frac{(2)(10) + (3)(0)}{2 + 3} = \frac{20}{5} = 4\ \text{m/s}$. - Problem 2: Calculate the loss in kinetic energy for Problem 1.
Solution: $\Delta K = \frac{1}{2} \left(\frac{(2)(3)}{2 + 3}\right) (10 – 0)^2 = (0.6)(100) = 60\ \text{J}$. - Problem 3: A particle of mass $1\ \text{kg}$ at $5\ \text{m/s}$ collides elastically and head-on with a stationary $4\ \text{kg}$ particle. Find $v_1$.
Solution: $v_1 = \left(\frac{1 – 4}{1 + 4}\right)(5) = -3\ \text{m/s}$ (rebounds at $3\ \text{m/s}$). - Problem 4: Find final velocity $v_2$ of the $4\ \text{kg}$ particle from Problem 3.
Solution: $v_2 = \left(\frac{2(1)}{1 + 4}\right)(5) = 2\ \text{m/s}$. - Problem 5: Masses $3\ \text{kg}$ and $5\ \text{kg}$ approach each other at $4\ \text{m/s}$ and $2\ \text{m/s}$. If perfectly inelastic, find common velocity.
Solution: $v = \frac{(3)(4) + (5)(-2)}{3 + 5} = \frac{2}{8} = 0.25\ \text{m/s}$. - Problem 6: A ball dropped from $20\ \text{m}$ rebounds to $5\ \text{m}$. Find coefficient of restitution $e$.
Solution: $e = \sqrt{\frac{5}{20}} = \sqrt{0.25} = 0.5$. - Problem 7: A bullet ($0.05\ \text{kg}$, $200\ \text{m/s}$) embeds into a stationary block ($1.95\ \text{kg}$). Find common velocity.
Solution: $v = \frac{(0.05)(200)}{0.05 + 1.95} = \frac{10}{2.0} = 5\ \text{m/s}$. - Problem 8: Elastic collision between equal masses where $u_1 = 8\ \text{m/s}$ and $u_2 = 2\ \text{m/s}$ in same direction. Find final velocities.
Solution: Velocities exchange: $v_1 = 2\ \text{m/s}$, $v_2 = 8\ \text{m/s}$. - Problem 9: A $2\ \text{kg}$ body at $10\ \text{m/s}$ hits a stationary $2\ \text{kg}$ body with $e = 0.8$. Find separation velocity.
Solution: $v_2 – v_1 = e(u_1 – u_2) = (0.8)(10) = 8\ \text{m/s}$. - Problem 10: Calculate kinetic energy lost when a $4\ \text{kg}$ mass at $6\ \text{m/s}$ hits a stationary $4\ \text{kg}$ mass and sticks.
Solution: $\Delta K = \frac{1}{2} \left(\frac{(4)(4)}{4 + 4}\right) (6)^2 = (1)(18) = 18\ \text{J}$.
