# Energy Transfer in Collisions

> Fraction transferred 4m1m2/(m1+m2)², perfectly inelastic loss m2/(m1+m2), neutron moderation and the ballistic pendulum — JEE Physics.

- Canonical URL: https://prepelephant.com/topics/jee/physics/energy-transfer-collision
- Exam / course: JEE · Subject: Physics
- Publisher: PrepElephant (https://prepelephant.com) — Prepared and reviewed by the PrepElephant Academic Review Team
- First published: 2026-10-02
- Last updated: 2026-10-02
- How to cite: "Energy Transfer in Collisions", PrepElephant, https://prepelephant.com/topics/jee/physics/energy-transfer-collision

## Direct answer

In a head-on elastic collision with the target at rest, the fraction of kinetic energy transferred from mass m1 to mass m2 is 4m1m2/(m1 + m2)² — exactly 1 (complete transfer) for equal masses, and small when the masses are mismatched. For a perfectly inelastic collision with a resting target, the fraction of initial kinetic energy lost is m2/(m1 + m2), approaching unity when a light projectile buries itself in a heavy block. These two fractions govern neutron moderation (hydrogen and other light nuclei sap a fast neutron's energy fastest), the ballistic pendulum (momentum conserved in capture, energy only afterwards), and the equal-mass 90-degree scattering rule of two-dimensional elastic collisions. The unifying expression for any one-dimensional collision is ΔKE = ½ μ v_rel²(1 − e²), with μ the reduced mass and e the coefficient of restitution.

## What you must remember

- **Elastic transfer fraction:** f = 4m1m2/(m1 + m2)² to a stationary target; equal masses give f = 1 (velocities exchange), m1 = 12m2 gives 48/169 ≈ 0.284, and m1 >> m2 leaves the projectile barely slowed.
- **Perfectly inelastic loss fraction:** with m2 at rest, the fraction of initial KE dissipated is m2/(m1 + m2); a heavy projectile loses little (m1 >> m2), a light one loses nearly everything.
- **Universal loss formula:** ΔKE = ½ μ v_rel²(1 − e²), μ = m1m2/(m1 + m2); elastic e = 1 loses nothing, perfectly inelastic e = 0 loses the whole COM-frame energy.
- **Moderator physics:** a neutron against hydrogen transfers up to 100 percent per head-on hit, against carbon-12 about 28, against lead under 2 — light nuclei make moderators, heavy nuclei make reflectors.
- **Ballistic pendulum:** the capture conserves momentum, v = ((M + m)/m)√(2gh); almost all the bullet's kinetic energy becomes heat and deformation, only the fraction m/(M + m) survives as the swing's motion.
- **Equal-mass 2D elastic rule:** two identical masses colliding elastically with one at rest fly apart at right angles — vector momentum plus energy conservation force the 90 degrees.
- **Momentum is not negotiable:** in every collision, however inelastic, linear momentum survives; only energy may be reclassified as heat, sound and deformation.

## A neutron picks its moderator

Fire a neutron at stationary nuclei and count the energy handed over per head-on elastic hit. Hydrogen (m2 = m1): f = 4 × 1 × 1/4 = 1 — one collision can strip everything. Carbon-12: f = 4 × 12/169 = 48/169 ≈ 0.284, about 28 percent per hit; reaching thermal energies takes some dozens of collisions. Lead-208: f = 4 × 208/209² ≈ 0.019 — you could ricochet all day. This is why reactor cores surround fuel with water, heavy water or graphite: the moderator must be light enough to sap the neutron's energy without absorbing the neutron itself, and hydrogen's near-perfect transfer is exactly what D2O trades away (deuterium, f = 0.89) in exchange for not capturing the neutron — the design tension an NCERT-level nuclear chapter expects you to articulate.

The ballistic pendulum is the inelastic showcase. A 10 g bullet buries itself in a 2 kg block hanging at rest; the block rises 5 cm. With g = 10, √(2gh) = 1 m/s, so the capture speed is v = (2.01/0.010) × 1 = 201 m/s. The bullet's arrival energy was ½(0.010)(201)² ≈ 202 J; the swing's peak energy is (M + m)gh ≈ 1 J. Ninety-nine and a half percent became heat and deformation — the numbers teach why one applies momentum first and energy only after the capture.

## Where students slip

JEE Main's version is the direct fraction, and the trap is algebraic: quoting 2m1/(m1 + m2) (the velocity-transfer ratio) where the energy fraction is asked, since energy squares the velocity result. JEE Advanced prefers layered situations: the ballistic pendulum, a neutron thermalising through N collisions (energy falls by (1 − f)^N), or the two-dimensional equal-mass rule where students forget it applies only when one body starts at rest. The conceptual trap spanning both papers: assuming kinetic energy conservation in inelastic capture — the sequence is always momentum through the collision, energy after it. Watch also the inelastic loss fraction's direction: a light projectile hitting a heavy wall loses almost all its energy, which feels backwards to students who anchor on "heavy things barely move" — they barely move precisely because they took the momentum while the energy died.

## Frequently asked questions

### What fraction of kinetic energy transfers in a head-on elastic collision?

f = 4m1m2/(m1 + m2)² to a target at rest — complete transfer for equal masses, about 28 percent from a neutron to carbon-12, and under 2 percent to a very heavy nucleus.

### Why do reactor moderators use light nuclei?

Energy transfer per elastic collision scales as 4m1m2/(m1 + m2)², maximised near equal masses, so hydrogen-rich water or graphite slows neutrons in the fewest collisions.

### What fraction of energy is lost when a bullet embeds in a block?

The fraction m2/(m1 + m2) of the bullet-block system's post-capture kinetic budget — practically, nearly all the bullet's kinetic energy becomes heat and deformation, with only (m1/(m1 + m2)) surviving as motion.

### What is the equal-mass two-dimensional elastic scattering rule?

Two identical masses, one initially at rest, depart along perpendicular directions after a non-head-on elastic collision — a direct consequence of momentum and energy conservation together.

### Which quantity is conserved in every collision?

Linear momentum, always; kinetic energy survives only when e = 1, and ΔKE = ½μv_rel²(1 − e²) quantifies exactly what did not survive.