Conservation of Momentum — Revision Notes

⚡ 30-Second Revision

Momentum: — $vec{p} = mvec{v}$ (vector quantity, SI unit: kg·m/s)
Conservation of Momentum: — For an isolated system ($vec{F}_{\text{net, ext}} = 0$), total momentum is constant: $sum vec{p}_{\text{initial}} = sum vec{p}_{\text{final}}$
Elastic Collision: — Momentum conserved, Kinetic Energy conserved. $e=1$.
Inelastic Collision: — Momentum conserved, Kinetic Energy NOT conserved. $0 < e < 1$.
Perfectly Inelastic Collision: — Objects stick together. Momentum conserved, max KE loss. $e=0$. Formula: $m_1u_1 + m_2u_2 = (m_1+m_2)V$.
Impulse: — $vec{J} = Delta vec{p} = vec{F}_{\text{avg}}Delta t$. Also, $vec{J} = int vec{F} dt$.
Recoil/Explosion: — Initial momentum (often zero) = sum of final momenta of fragments.

2-Minute Revision

The core concept of Conservation of Momentum states that for an isolated system, the total linear momentum remains constant. An isolated system is one where no net external forces act on it. Momentum ($vec{p}$) is a vector quantity, calculated as mass times velocity ($mvec{v}$). This principle is a direct consequence of Newton's third law. When objects collide or interact, their individual momenta may change, but the vector sum of their momenta for the entire system stays the same.

Collisions are categorized based on kinetic energy conservation. In elastic collisions, both momentum and kinetic energy are conserved, and the coefficient of restitution ($e$) is 1. In inelastic collisions, momentum is conserved, but kinetic energy is lost (converted to heat, sound, deformation), with $0 < e < 1$.

A special case is perfectly inelastic collisions, where objects stick together ($e=0$), resulting in maximum kinetic energy loss. Applications include recoil of guns and rocket propulsion. Remember to treat momentum as a vector, assigning appropriate signs for direction in 1D problems and resolving into components for 2D problems.

Impulse, defined as the change in momentum, is also a key related concept.

5-Minute Revision

To master Conservation of Momentum for NEET, start with the fundamental definition: the total linear momentum of an isolated system remains constant. An isolated system is crucial – it means the net external force acting on the system is zero. Momentum, $vec{p} = mvec{v}$, is a vector, so direction is paramount. For 1D problems, use positive/negative signs; for 2D, resolve into components (x and y) and apply conservation independently for each.

Types of Collisions:

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Elastic: — Both momentum and kinetic energy are conserved. The coefficient of restitution ($e$) is 1. For a 1D elastic collision between $m_1$ (initial $u_1$) and $m_2$ (initial $u_2$), final velocities $v_1, v_2$ can be found using:

$m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$ (Momentum) $rac{1}{2}m_1u_1^2 + \frac{1}{2}m_2u_2^2 = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2$ (Kinetic Energy) A useful derived relation for 1D elastic collisions: $u_1 - u_2 = -(v_1 - v_2)$ (relative speed of approach equals relative speed of separation).

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Inelastic: — Momentum is conserved, but kinetic energy is not (some is lost). $0 < e < 1$.

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Perfectly Inelastic: — Objects stick together after collision. Momentum is conserved, kinetic energy loss is maximum. $e=0$. The combined mass moves with a common final velocity $V$: $m_1u_1 + m_2u_2 = (m_1+m_2)V$.

Applications:

Recoil of a gun: — If a gun (mass $m_g$) fires a bullet (mass $m_b$) with velocity $v_b$, the gun recoils with velocity $v_g$. If initially at rest: $0 = m_b v_b + m_g v_g$.
Explosions: — If an object at rest explodes into fragments, the vector sum of the momenta of all fragments is zero.

Impulse-Momentum Theorem: Impulse ($vec{J}$) is the change in momentum: $vec{J} = Delta vec{p} = vec{p}_{\text{final}} - vec{p}_{\text{initial}}$. It's also the average force multiplied by the time interval: $vec{J} = vec{F}_{\text{avg}}Delta t$. For a variable force, $vec{J} = int vec{F} dt$.

Worked Example: A $0.1,\text{kg}$ ball moving at $2,\text{m/s}$ collides head-on with a stationary $0.2,\text{kg}$ ball. If they stick together, find their common velocity.

$m_1 = 0.1,\text{kg}$, $u_1 = 2,\text{m/s}$
$m_2 = 0.2,\text{kg}$, $u_2 = 0,\text{m/s}$
Perfectly inelastic collision: $(m_1+m_2)V = m_1u_1 + m_2u_2$
$(0.1+0.2)V = (0.1)(2) + (0.2)(0)$
$0.3V = 0.2$
$V = \frac{0.2}{0.3} = \frac{2}{3} approx 0.67,\text{m/s}$.

Always double-check units and vector directions. This topic is highly testable, so thorough practice is key.

Prelims Revision Notes

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Definition of Momentum: — Linear momentum $vec{p}$ is a vector quantity, $vec{p} = mvec{v}$. SI unit is kg·m/s. Direction is same as velocity.
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Conservation Principle: — For an isolated system (no net external force), total linear momentum is conserved: $sum vec{p}_{\text{initial}} = sum vec{p}_{\text{final}}$. This means $m_1vec{u}_1 + m_2vec{u}_2 + dots = m_1vec{v}_1 + m_2vec{v}_2 + dots$.
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Isolated System: — Crucial condition. Internal forces (e.g., collision forces) do not change total system momentum. External forces (e.g., friction, gravity) do.
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Types of Collisions:

* Elastic: Momentum conserved, Kinetic Energy conserved. $e=1$. Relative speed of approach = relative speed of separation. * Inelastic: Momentum conserved, Kinetic Energy NOT conserved (some lost to heat, sound, deformation). $0 < e < 1$. * Perfectly Inelastic: Objects stick together. Momentum conserved, maximum KE loss. $e=0$. Combined mass $(m_1+m_2)$ moves with common velocity $V$: $m_1u_1 + m_2u_2 = (m_1+m_2)V$.

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Coefficient of Restitution ($e$): — $e = \frac{\text{relative speed of separation}}{\text{relative speed of approach}} = \frac{|v_2 - v_1|}{|u_1 - u_2|}$.
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Impulse-Momentum Theorem: — Impulse $vec{J}$ is the change in momentum: $vec{J} = Delta vec{p} = vec{p}_{\text{final}} - vec{p}_{\text{initial}}$. Also, $vec{J} = vec{F}_{\text{avg}}Delta t$. For variable force, $vec{J} = int vec{F} dt$.
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Applications:

* Recoil: Gun-bullet system. Initial momentum (at rest) is zero. $0 = m_{\text{bullet}}v_{\text{bullet}} + m_{\text{gun}}v_{\text{gun}}$. * Explosions: If an object at rest explodes, the vector sum of momenta of all fragments is zero.

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Key Problem-Solving Steps:

* Identify system and check if isolated. * Define positive direction (for 1D) or coordinate axes (for 2D). * Write down initial total momentum. * Write down final total momentum. * Equate initial and final total momenta (vectorially). * If elastic, also conserve kinetic energy or use relative velocity relation. * Solve for unknowns.

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Common Traps: — Forgetting vector nature, confusing momentum with kinetic energy, incorrect unit conversions (g to kg), algebraic errors with signs.

Vyyuha Quick Recall

MICE: Momentum Is Conserved Everywhere (in an isolated system)!