Conservation laws overview — momentum, energy, angular momentum when to use

Question

When do we use conservation of momentum vs. conservation of energy vs. conservation of angular momentum? How do you decide which conservation law to apply in a given problem? Give examples where only one law works but the others do not.

(JEE Main + NEET — strategy + conceptual)

Solution — Step by Step

Law	Conserved When	Mathematical Form
Conservation of Linear Momentum	No external force on the system ( $\vec{F}_{ext} = 0$ )	$\vec{p}_i = \vec{p}_f$ or $m_1\vec{v}_1 + m_2\vec{v}_2 = \text{constant}$
Conservation of Energy	No non-conservative forces (or if we account for work done by them)	$KE_i + PE_i = KE_f + PE_f$ (or $+ W_{nc}$ )
Conservation of Angular Momentum	No external torque ( $\vec{\tau}_{ext} = 0$ )	$\vec{L}_i = \vec{L}_f$ or $I_1\omega_1 = I_2\omega_2$

Scenario	Best Law	Why
Collision (elastic or inelastic)	Momentum (always) + Energy (if elastic)	External forces during collision are internal to the system
Explosion/recoil (gun firing, bomb splitting)	Momentum	No external force; energy is NOT conserved (chemical energy released)
Object falling/sliding (no friction)	Energy	Gravity is conservative; momentum changes (gravity is external force)
Spring problems	Energy	Spring force is conservative
Spinning figure skater pulling arms in	Angular momentum	No external torque about vertical axis
Inelastic collision	Momentum only	Energy is lost to deformation/heat — NOT conserved

In perfectly inelastic collisions (objects stick together), kinetic energy is NOT conserved — some converts to heat, sound, and deformation. But momentum IS still conserved (no external forces during the brief collision).

Energy loss: $\Delta KE = \frac{1}{2}\frac{m_1 m_2}{m_1 + m_2}(v_1 - v_2)^2$

In explosions (reverse of collision), kinetic energy INCREASES (chemical energy converts to KE). But momentum is still conserved — total momentum before = total momentum after (usually both = 0).

A 2 kg ball moving at 3 m/s collides with a 3 kg ball at rest. They stick together. Find the velocity after collision and energy lost.

Step 1 — Momentum (always conserved in collisions):

2 \times 3 + 3 \times 0 = (2 + 3) \times v_f

v_f = \frac{6}{5} = \mathbf{1.2 \text{ m/s}}

Step 2 — Energy (to find what was lost):

KE_i = \frac{1}{2}(2)(3)^2 = 9 \text{ J}

KE_f = \frac{1}{2}(5)(1.2)^2 = 3.6 \text{ J}

\Delta KE = 9 - 3.6 = \mathbf{5.4 \text{ J lost}}

graph TD
    A["Physics Problem"] --> B{"Collision/Explosion?"}
    B -->|Yes| C["Use Momentum Conservation"]
    C --> D{"Elastic?"}
    D -->|Yes| E["Also use Energy Conservation"]
    D -->|No| F["Momentum only — energy lost"]
    B -->|No| G{"Rotation involved?"}
    G -->|Yes| H{"External torque = 0?"}
    H -->|Yes| I["Angular Momentum Conservation"]
    G -->|No| J{"Conservative forces only?"}
    J -->|Yes| K["Energy Conservation"]
    J -->|No| L["Work-Energy Theorem"]
    style A fill:#fbbf24,stroke:#000,stroke-width:2px
    style C fill:#86efac,stroke:#000
    style K fill:#93c5fd,stroke:#000

Why This Works

Conservation laws are the most powerful tools in physics because they bypass the need to track forces and accelerations moment by moment. Instead of solving $F = ma$ through a messy collision, we just compare “before” and “after” states. The laws work because they are rooted in fundamental symmetries of nature: momentum conservation comes from translational symmetry, energy conservation from time symmetry, and angular momentum conservation from rotational symmetry.

Common Mistake

The biggest error: using energy conservation in an inelastic collision. In an inelastic collision, kinetic energy is NOT conserved. Only momentum is conserved. If you set $KE_i = KE_f$ for an inelastic collision, you will get the wrong answer. Always check: “Is the collision elastic or inelastic?” before deciding whether to use energy conservation.

JEE shortcut: in problems where you need to find velocity after collision, ALWAYS start with momentum conservation — it works for ALL collisions. Then check if the collision is elastic (KE also conserved) to get a second equation if needed. Two unknowns need two equations: momentum gives one, elasticity condition gives the other.