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Mechanics

Conservation of Momentum

The law of conservation of momentum states that the total linear momentum of an isolated system of interacting particles remains constant in time. It is a fundamental conservation law directly linked to the homogeneity of space, as articulated by Noether's theorem.

UP TGT ScienceUP PGT PhysicsStudy Notes + Practice

Scope and prerequisites

TGT core

Application of linear momentum conservation in one and two-dimensional collisions, recoil of guns, and direct central impacts.

PGT extension

Variable mass systems (rocket propulsion), generalized momentum in Lagrangian mechanics, system of particles with central forces, and relativistic momentum.

Prerequisites

  • Newton's laws of motion
  • Concept of center of mass
  • Vector addition and resolution

Law of Conservation of Linear Momentum

  • If the vector sum of external forces acting on a system is zero, the total linear momentum of the system remains conserved.
  • Mathematically expressed as dp/d$t = F_ext; therefore$, when $F_ext = 0$, p = constant.
  • It holds true for macroscopic as well as microscopic systems, independent of the nature of internal forces (whether conservative or non-conservative).

Internal vs External Forces

  • Internal forces are forces exerted by particles of the system on one another, which always occur in action-reaction pairs according to Newton's third law.
  • By Newton's third law, the vector sum of all internal forces within a closed system is identically zero.
  • Only external forces from outside the system can alter the total linear momentum of the system.

Momentum Conservation in Collisions

  • During collisions, impulsive internal forces are extremely large compared to external forces, making external forces negligible during the impact duration.
  • Thus, total momentum is strictly conserved immediately before, during, and immediately after any collision.
  • Collisions are categorized into elastic (kinetic energy conserved) and inelastic (kinetic energy not conserved), but momentum is conserved in both types.

Application to Variable Mass Systems (Rocket Propulsion)

  • For systems like rockets where mass changes due to fuel consumption, the system must include both the rocket and the ejected exhaust gas.
  • The instantaneous thrust force on the rocket is given by the product of the relative exhaust velocity and the rate of mass ejection.
  • Conservation of momentum allows integration of velocity changes over time to yield Tsiolkovsky's rocket equation.

Derivations

Conservation of Momentum from Newton's Second Law

Conditions: System of n interacting particles with zero net external force.

  • Let the total momentum of the system be $P = \sum(p_i) = \sum(m_i v_i)$.
  • The time rate of change of total momentum is dP/d$t = \sum(dp_i/dt) = \sum(F_i)$.
  • The total force F_i on the i-th particle includes external force F_i^(ext) and internal forces F_ij from other particles.
  • By Newton's third law, internal forces cancel out in pairs since $F_ij = -F_ji$.
  • Therefore, dP/d$t = \sum(F_i^(ext))$. If the net external force is zero, dP/dt = 0, which yields P = constant.

Result: $P = \sum(m_i v_i) = \text{constant}$

Tsiolkovsky Rocket Equation

Conditions: Vertical or free-space motion of a rocket in the absence of external drag and gravitational forces.

  • At time t, mass of the rocket is m and velocity is v. Momentum is p(t) = mv.
  • In time interval dt, mass decreases by dm (dm is negative) and exhaust is ejected with relative velocity v_rel.
  • At time t + dt, rocket mass is m + dm and velocity is v + dv, while ejected mass dm has velocity v - v_rel.
  • Equating total momentum at t and t + dt: m$v = (m + dm)(v + dv) + dm(v - v_rel)$.
  • Expanding and dropping second-order small terms (dm dv), we get m d$v = -v_rel dm$.
  • Integrating from initial mass m_0 to m and initial velocity v_0 to v yields the final velocity expression.

Result: v - $v_0 = v_{rel} \ln\left(\frac{m_0}{m}\right)$

Formula sheet

Formula / lawEquationVariables and unitsConditions
Linear Momentum Definition$$p = m v$$p = momentum, m = mass, v = velocity vector
kg m s^-1
Non-relativistic mechanics
Newton's Second Law in Momentum Form$$F_{ext} = \frac{dp}{dt}$$$F_ext = net external force$, p = total linear momentum
N (kg m s^-2)
Inertial frame of reference
Conservation Principle$$p_{initial} = p_{final}$$$p_initial = total momentum before event$, $p_final = total momentum after event$
kg m s^-1
$F_ext = 0$
Recoil Velocity of a Gun$$V = -\frac{m v}{M}$$V = recoil velocity of gun, M = mass of gun, m = mass of bullet, v = velocity of bullet
m s^-1
Initial system at rest
Rocket Thrust Force$$F_{thrust} = v_{rel} \left(-\frac{dm}{dt}\right)$$$F_thrust = thrust force$, $v_rel = relative exhaust velocity$, dm/dt = rate of fuel consumption
N
No external drag
Tsiolkovsky Equation$$v = v_0 + v_{rel} \ln\left(\frac{m_0}{m}\right)$$v = final velocity, $v_0 = initial velocity$, $m_0 = initial mass$, m = final mass
m s^-1
No external forces
Center of Mass Velocity$$V_{cm} = \frac{\sum m_i v_i}{\sum m_i} = \frac{P}{M_{total}}$$$V_cm = velocity of center of mass$, P = total momentum, $M_total = total mass$
m s^-1
Standard multi-particle system
1D Elastic Collision Final Velocities$$v_1 = \frac{m_1 - m_2}{m_1 + m_2}u_1 + \frac{2m_2}{m_1 + m_2}u_2$$u_1, $u_2 = initial velocities$, $v_1 = final velocity of mass 1$, m_1, $m_2 = masses$
m s^-1
Head-on elastic collision in 1D

Graphs and experiments

Force vs Time Graph during Collision

Draw: A bell-shaped curve plotting impulsive force F(t) on the y-axis against time t on the x-axis.

Interpret: The area under the force-time curve represents the linear impulse delivered to the body, which equals the total change in linear momentum.

Exam use: Tested numerically to find momentum change from graphical areas or average force calculations.

Linear Momentum vs Time for an Isolated System

Draw: A straight horizontal line parallel to the time axis indicating constant total momentum.

Interpret: Since no external force acts on the isolated system ($F_ext = 0)$, the rate of change of momentum is zero.

Exam use: Conceptual identification questions in TGT/PGT examinations regarding isolated vs non-isolated systems.

Ballistic Pendulum Experiment

Draw: Schematic of a bullet of mass m embedding into a suspended wooden block of mass M, causing a vertical rise h.

Interpret: Combines momentum conservation during the instantaneous collision with mechanical energy conservation during the subsequent swing.

Exam use: Standard apparatus-based problem solving for finding bullet velocity.

Solved numericals

TGT core

A shell of mass 20 kg moving with a velocity of 10 m/s explodes into two fragments of masses 5 kg and 15 kg. If the 5 kg fragment comes to rest, find the velocity of the 15 kg fragment.

  • Initial momentum of the shell: $P_i = M * u = 20 kg * 10 m/s = 200 kg m/s$.
  • Let the velocity of the 15 kg fragment be v. Mass $m_1 = 5 kg with v_1 = 0 m/s$, mass $m_2 = 15 kg with velocity v$.
  • By conservation of momentum: $P_i = m_1 * v_1 + m_2 * v$.
  • $200 = (5 * 0) + (15 * v) => 15v = 200$.
  • $v = 200 / 15 = 40 / 3 m/s$.

Answer: 13.33 m/s

PGT extension

A rocket has an initial mass of 1000 kg (including 800 kg fuel). It ejects gas at a constant relative velocity of 2000 m/s and at a fuel consumption rate of 10 kg/s. Find the thrust force and the acceleration of the rocket at the start of its motion (neglect gravity).

  • Thrust force formula: $F_thrust = v_rel * (dm/dt) = 2000 m/s * 10 kg/s = 20000 N$.
  • Initial mass of the rocket $m_0 = 1000 kg$.
  • Acceleration at the start: $a = F_thrust / m_0 = 20000 N / 1000 kg = 20 m/s^2$.

Answer: Thrust force = 20000 N, Acceleratio$n = 20 m/s^2$

PGT extension

A ball of mass 0.5 kg moving at 6 m/s strikes a smooth vertical wall normally and bounces back with the same speed of 6 m/s. What is the magnitude of the impulse delivered by the wall to the ball?

  • Choose direction away from the wall as positive. Initial velocity $u = -6 m/s$, final velocity $v = +6 m/s$.
  • Impulse J equals change in momentum: Delta $p = m(v - u)$.
  • Delta $p = 0.5 kg * (6 - (-6)) m/s = 0.5 * 12 = 6 kg m/s$.

Answer: 6 Ns (or 6 kg m/s)

Assumptions and limitations

  • Conservation of linear momentum applies strictly in inertial frames of reference unless pseudo-forces are explicitly introduced in non-inertial frames.
  • At relativistic speeds (v comparable to c), classical momentum p = mv must be replaced by relativistic momentum $p = m_0 v / \sqrt(1 - v^2/c^2)$.

Common misconceptions

Common error: Internal forces can change the total momentum of a system if the bodies have different masses.

Correct: Internal forces can never change the total momentum of a system.

By Newton's third law, internal forces occur in equal and opposite pairs, canceling out completely for the system as a whole.

Common error: Kinetic energy is always conserved whenever momentum is conserved.

Correct: Momentum is conserved in inelastic collisions, but kinetic energy is not.

Kinetic energy can be transformed into heat, sound, or deformation energy during interactions, whereas momentum remains invariant in the absence of external forces.

Common error: Momentum conservation depends on the type of forces (conservative or non-conservative) acting internally.

Correct: Momentum conservation depends only on the absence of net external forces.

Internal conservative and non-conservative forces both satisfy Newton's third law pairs of action and reaction.

Quick revision

  • Total linear momentum is conserved if and only if the net external force acting on the system is zero.
  • Linear Momentum Definition: p = m v
  • Newton's Second Law in Momentum Form: $F_{ext} = \frac{dp}{dt}$

Must remember

  • The total linear momentum of a system remains conserved if and only if the vector sum of external forces acting on it is zero ($F_{ext} = 0$).
  • Internal forces, acting in action-reaction pairs via Newton's third law, cancel out and cannot alter the system's total momentum.
  • In both elastic and inelastic collisions, linear momentum is conserved because impulsive internal forces far exceed external forces during impact.
  • For a rocket in free space, the instantaneous thrust force is given by $F_{thrust} = v_{rel} \left(-\frac{dm}{dt}\right)$.
  • Tsiolkovsky's rocket equation is expressed as $v = v_0 + v_{rel} \ln\left(\frac{m_0}{m}\right)$.

Glossary

Linear Momentum ($p$)The product of an object's mass and its velocity vector, given by $p = m v$.
Impulsive Internal ForcesExtremely large forces exerted internally between colliding bodies over a very short duration of impact.
Relative Exhaust Velocity ($v_{rel}$)The velocity of the ejected gas relative to the rocket itself.
Center of Mass Velocity ($V_{cm}$)The velocity of the center of mass of a system, defined as $V_{cm} = \frac{\sum m_i v_i}{\sum m_i}$.
Elastic CollisionA collision in which both linear momentum and total kinetic energy are conserved.

Elastic vs Inelastic Collisions

ParameterElastic CollisionInelastic Collision
Linear MomentumConservedConserved
Total Kinetic EnergyConservedNot Conserved
Nature of ForcesConservative internal forcesNon-conservative forces (heat, deformation)

Internal vs External Forces

ParameterInternal ForcesExternal Forces
OriginWithin the system (between particles)Outside the system
Effect on Total MomentumVector sum is zero; cannot change total momentumCan alter the total linear momentum of the system
Governing LawNewton's Third Law (action-reaction pairs)Newton's Second Law ($F_{ext} = \frac{dp}{dt}$)

Practice Quiz

Choose an option to reveal the explanation.

1. What is the condition for the total linear momentum of a system to remain constant?

Explanation: According to the law of conservation of linear momentum, when the vector sum of external forces acting on a system is zero ($F_{ext} = dp/dt = 0$), momentum $p$ remains constant.

2. Why do internal forces not affect the total momentum of a closed system?

Explanation: By Newton's third law, internal forces occur in action-reaction pairs that are equal in magnitude and opposite in direction, making their vector sum identically zero.

3. Which of the following quantities is strictly conserved in both elastic and inelastic collisions?

Explanation: Linear momentum is conserved in all types of collisions because impulsive internal forces vastly exceed external forces during impact. Kinetic energy is only conserved in elastic collisions.

4. The formula for the recoil velocity of a gun of mass $M$ firing a bullet of mass $m$ with velocity $v$ is given by:

Explanation: Applying momentum conservation with the initial system at rest: $0 = M V + m v \implies V = -\frac{m v}{M}$.

5. What is the expression for the instantaneous thrust force acting on a variable mass rocket?

Explanation: The thrust force is given by $F_{thrust} = v_{rel} \left(-\frac{dm}{dt}\right)$, where $-\frac{dm}{dt}$ represents the positive rate of fuel consumption.

6. Tsiolkovsky's rocket equation relates the final velocity $v$ to which of the following parameters?

Explanation: Tsiolkovsky's rocket equation is $v = v_0 + v_{rel} \ln\left(\frac{m_0}{m}\right)$, involving initial velocity, relative exhaust velocity, and the mass ratio.

7. What is the velocity of the center of mass of a system in terms of total momentum $P$ and total mass $M_{total}$?

Explanation: The center of mass velocity is defined as $V_{cm} = \frac{\sum m_i v_i}{\sum m_i} = \frac{P}{M_{total}}$.

8. Which misconception regarding momentum conservation is correct according to physical principles?

Explanation: Momentum conservation depends solely on the absence of net external forces, regardless of internal force types or mass differences.

9. In a 1D head-on elastic collision, what is the final velocity $v_1$ of mass $m_1$ colliding with mass $m_2$ initially at rest ($u_2 = 0$)?

Explanation: Setting $u_2 = 0$ in the formula $v_1 = \frac{m_1 - m_2}{m_1 + m_2}u_1 + \frac{2m_2}{m_1 + m_2}u_2$ yields $v_1 = \frac{m_1 - m_2}{m_1 + m_2}u_1$.

10. What are the standard units of linear momentum ($p$)?

Explanation: Linear momentum is defined as $p = m v$, having units of mass (kg) times velocity ($\text{m s}^{-1}$), i.e., $\text{kg m s}^{-1}$.

11. Why can external forces be neglected during the brief duration of a collision?

Explanation: During impacts, impulsive internal forces are orders of magnitude larger than external forces (like gravity or friction), rendering external forces negligible during the impact duration.

12. What is the mathematical form of Newton's second law relating force to momentum?

Explanation: Newton's second law in momentum form is expressed as $F_{ext} = \frac{dp}{dt}$.

13. Does the conservation of linear momentum hold true for microscopic systems?

Explanation: The law of conservation of linear momentum holds true universally for both macroscopic and microscopic systems, independent of internal force natures.

14. If a bomb at rest explodes into two fragments of masses $m_1$ and $m_2$ with velocities $v_1$ and $v_2$, what is the ratio of their velocities $\frac{v_1}{v_2}$?

Explanation: By momentum conservation, $0 = m_1 v_1 + m_2 v_2 \implies \frac{v_1}{v_2} = -\frac{m_2}{m_1}$.

15. In Lagrangian mechanics, a generalized momentum corresponding to a cyclic coordinate is:

Explanation: According to Noether's theorem and Lagrangian mechanics, if a coordinate is cyclic (absent from the Lagrangian), its corresponding generalized momentum is conserved.

PYQ Patterns

These are pattern-based practice questions. Authentic year labels are only used when verified from a source.

1. A shell of mass $M$ at rest explodes into two fragments of masses $m$ and $M-m$. If the fragment of mass $m$ has a velocity $v$, what is the velocity of the second fragment?

Explanation: Using conservation of linear momentum: $0 = m v + (M-m) v_2 \implies v_2 = -\frac{mv}{M-m}$.

2. A rocket consumes fuel at a constant rate of $-\frac{dm}{dt} = r$ with a relative exhaust velocity $v_{rel}$. If there are no external drag forces, the thrust force on the rocket is:

Explanation: The thrust force is given by $F_{thrust} = v_{rel} \left(-\frac{dm}{dt}\right) = v_{rel} r$.

3. A bullet of mass $10\text{ g}$ is fired horizontally with a velocity of $500\text{ m s}^{-1}$ from a gun of mass $5\text{ kg}$. What is the recoil velocity of the gun?

Explanation: Using $V = -\frac{m v}{M} = -\frac{0.01 \times 500}{5} = -\frac{5}{5} = -1.0\text{ m s}^{-1}$.

4. Which of the following statements is correct regarding an inelastic collision in an isolated system?

Explanation: In any isolated system collision, linear momentum is always conserved due to zero net external force, whereas kinetic energy is generally not conserved in inelastic collisions.

5. If the net external force acting on a system of particles is zero, then which of the following is constant?

Explanation: When $F_{ext} = 0$, $\frac{dp}{dt} = 0$, which implies the total linear momentum $p$ is constant.

6. A particle of mass $m_1$ moving with velocity $u_1$ collides head-on elastically with a stationary particle of mass $m_2$. If $m_1 = m_2$, what are their final velocities?

Explanation: For equal masses ($m_1 = m_2$) in an elastic head-on collision, velocities are interchanged. Thus, the first particle comes to rest ($v_1 = 0$) and the second particle moves with $u_1$.

Timed Topic Test

Attempt these questions under a ten-minute limit using a separate timer.

1. What is the physical significance of $dp/dt = F_{ext}$ in non-relativistic mechanics?

Explanation: Newton's second law states that the rate of change of linear momentum with respect to time is equal to the net external force acting on the system.

2. Which of the following conditions must be satisfied for Tsiolkovsky's rocket equation to apply without external disturbances?

Explanation: Tsiolkovsky's equation $v = v_0 + v_{rel} \ln(m_0/m)$ is derived under the condition of zero external forces acting on the rocket-fuel system.

3. If two masses $m_1$ and $m_2$ experience only internal mutual forces, what happens to the center of mass of the system?

Explanation: Since internal forces sum to zero, the net external force is zero, meaning the center of mass acceleration is zero. Thus, the center of mass moves with constant velocity.

4. In a completely inelastic collision between two bodies, what happens to the lost kinetic energy?

Explanation: Kinetic energy is not conserved in inelastic collisions; the 'lost' kinetic energy is dissipated into thermal energy, acoustic energy, or work done in permanent deformation.

5. A rocket starts from rest with initial mass $m_0$ and burns fuel until its final mass is $m_0 / e$. If the relative exhaust velocity is $v_{rel}$, what is its final velocity (neglecting external forces)?

Explanation: Using Tsiolkovsky's equation with $v_0 = 0$ and $m = m_0 / e$: $v = v_{rel} \ln(m_0 / (m_0/e)) = v_{rel} \ln(e) = v_{rel}$. (Note: $\ln(e) = 1$).

6. Which statement best describes the nature of internal forces within a closed mechanical system?

Explanation: Internal forces are exerted between particles of the system and always occur in action-reaction pairs according to Newton's third law, summing to zero.

7. What is the velocity of the center of mass for an isolated system of particles?

Explanation: For an isolated system ($F_{ext} = 0$), total momentum $P$ is constant, hence the center of mass velocity $V_{cm} = P / M_{total}$ remains constant.

8. If a particle's mass is constant, how does the formula $F_{ext} = dp/dt$ simplify?

Explanation: For constant mass ($dm/dt = 0$), $dp/dt = d(mv)/dt = m(dv/dt) = m a$.

9. During a rocket launch, what constitutes the 'system' when applying conservation of momentum?

Explanation: Because mass changes due to fuel consumption, the closed system must include both the rocket and all the exhaust gas expelled from it.

10. Which of the following scenarios violates the conservation of linear momentum?

Explanation: A car accelerating on a highway relies on external friction between the tires and the road; hence, external forces are non-zero, and momentum of the car alone is not conserved.