Work, Energy and Power · Lesson 14 of 16

Head-On Elastic and Inelastic Collisions

“Derive final velocities, sticking losses and energy transfer in one dimension.”

Learning Objectives

• Use signed velocities consistently. • Derive the elastic velocity formulas. • Calculate completely inelastic energy loss and neutron moderation.

Start with momentum and a second condition

One-dimensional momentum balanceLaTeX

m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}

All velocities are signed along one chosen axis in the same inertial frame. External impulse is negligible.

This single equation contains two unknown final velocities. To solve, specify more physics. An elastic collision adds kinetic-energy conservation. A completely inelastic collision adds the shared final velocity condition v1 = v2. Momentum alone cannot tell how much kinetic energy is transformed or released. For two approaching objects in one dimension, a restitution relation can also supply the missing condition, but it must be stated rather than assumed.

Completely inelastic collision

Common velocity and kinetic-energy lossLaTeX

v = \frac{m _{1} u _{1} + m _{2} u _{2}}{m _{1} + m _{2}}; K_{i} - K_{f} = \frac{1}{2} \frac{m _{1} m _{2}}{m _{1} + m _{2}} (u_{1} - u_{2})^{2}

This loss formula assumes no additional energy release, fixed masses and a final shared velocity. The reduced-mass factor has units of mass.

Derive the velocity by replacing both final velocities in the momentum equation by v. Substitute that result into Kf = ½(m1+m2)v² and subtract from Ki = ½m1u1²+½m2u2². Expanding and collecting terms leaves m1m2(u1−u2)²/[2(m1+m2)]. The nonnegative square shows why sticking removes the relative-motion energy. Some kinetic energy can remain because the centre of mass still moves.

A sticking pair

Problem
A 2 kg trolley moving at +3 m/s catches a stationary 1 kg trolley and they stick. Find final velocity and energy loss.

1.Total momentum is 6 kg m/s and combined mass is 3 kg, so v = +2 m/s.
2.Initial K is 9 J; final K is ½×3×2² = 6 J. Loss is 3 J.
3.The lost relative-motion energy becomes internal energy or other forms; the 6 J of common translation remains.

Derive the elastic relative-speed rule

Elastic energy balanceLaTeX

m_{1} u_{1}^{2} + m_{2} u_{2}^{2} = m_{1} v_{1}^{2} + m_{2} v_{2}^{2}

The common factor ½ has been removed.

Rearrange momentum as m1(u1−v1)=m2(v2−u2). Factor the energy equation as m1(u1−v1)(u1+v1)=m2(v2−u2)(v2+u2). For a nontrivial collision with nonzero transferred momentum, divide the second equation by the first. This gives u1+v1=v2+u2, hence u1−u2 = v2−v1: relative approach speed equals relative separation speed. The unchanged-velocity algebraic case must not be obtained by dividing by zero and is not the interacting outcome for ordinary approaching smooth masses.

General one-dimensional elastic resultLaTeX

v_{1} = \frac{m _{1} - m _{2}}{m _{1} + m _{2}} u_{1} + \frac{2 m _{2}}{m _{1} + m _{2}} u_{2}; v_{2} = \frac{2 m _{1}}{m _{1} + m _{2}} u_{1} + \frac{m _{2} - m _{1}}{m _{1} + m _{2}} u_{2}

Solve the momentum equation together with u1−u2=v2−v1.

A stationary target and limiting masses

Mass relation, with u₂ = 0	Result
m₁ = m₂	v₁ = 0 and v₂ = u₁: velocities exchange
m₂ much greater than m₁	v₁ ≈ −u₁ and v₂ ≈ 0: light projectile rebounds
m₁ much greater than m₂	v₁ ≈ u₁ and v₂ ≈ 2u₁: light target can leave faster than projectile’s initial speed

Unequal masses

Problem
A 2 kg body moves at +6 m/s and hits a stationary 4 kg body elastically head-on. Find final velocities.

1.v1 = (2−4)/(2+4)×6 = −2 m/s.
2.v2 = 2×2/(2+4)×6 = +4 m/s.
3.Momentum after is 2×(−2)+4×4 = 12 kg m/s, matching the initial 12. K after is 4+32 = 36 J, matching the initial 36 J.

Energy transfer and neutron moderation

Stationary-target energy fractionsLaTeX

\frac{K _{1 f}}{K _{1 i}} = (\frac{m _{1} - m _{2}}{m _{1} + m _{2}})^{2}; \frac{K _{2 f}}{K _{1 i}} = \frac{4 m _{1} m _{2}}{( m _{1} + m _{2} ) ^{2}}

The first fraction is retained by the projectile, not the fraction lost. These are ideal head-on elastic results.

For a neutron hitting a stationary deuterium nucleus of about twice its mass, the retained fraction is 1/9 and the transferred fraction is 8/9, about 88.9%. For a carbon nucleus of about 12 neutron masses, the retained fraction is 121/169 ≈ 71.6%, and transferred fraction is 48/169 ≈ 28.4%. Light nuclei can therefore slow neutrons through repeated elastic encounters. Real glancing encounters transfer less than this head-on maximum on average. The source calls the retained fraction “lost” in one sentence; the equations and these labels distinguish them correctly.

Check your understanding

Quick check

Equal masses in a one-dimensional elastic collision

Explanation: The general formulas reduce to v1=u2 and v2=u1.

Quick check

A completely inelastic collision imposes

Explanation: The bodies share a final velocity.

Quick check

For an elastic head-on encounter, approach and separation speeds are

Explanation: It follows from momentum and kinetic-energy conservation.

Quick check

A light projectile striking a much heavier stationary target elastically

Explanation: The velocity factor tends to −1.

Quick check

For a target of twice the projectile mass, retained projectile energy fraction is

Explanation: [(1−2)/(1+2)]²=1/9.

Key Takeaways

• Momentum requires an additional collision condition to fix both velocities. • Use signed velocity formulas and verify momentum and energy. • Retained and transferred energy fractions must not be confused.

Previous · Lesson 13

Collisions, Impulse and Energy Changes

Next · Lesson 15

Oblique Collisions and Multi-Stage Energy Problems

Start with momentum and a second condition

Completely inelastic collision

Derive the elastic relative-speed rule

A stationary target and limiting masses

Energy transfer and neutron moderation

Check your understanding

Quiz

Key Takeaways