advanced34 min studyLesson 582 in Physics

Two-body nuclear reaction kinematics

Transform relative motion into laboratory observables without assigning a dynamical angular distribution.

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What you’ll understand

  • Separate centre-of-mass translation and relative kinetic energy.
  • Partition exit-channel kinetic energy between unequal masses.
  • Check a laboratory energy-angle result against momentum conservation.

A two-body exit channel is constrained strongly by energy and momentum conservation, yet those laws do not specify which emission angles are likely. The centre-of-mass frame separates the allowed relative motion from the translation of the whole pair. Transforming back to the laboratory explains why the same centre-of-mass energy can produce different observed particle energies. It also prevents a detector angle from being mistaken for a centre-of-mass angle.

Work with the relative energy first

Assume nonrelativistic motion and a target initially at rest. The projectile kinetic energy splits into centre-of-mass translation plus the kinetic energy of relative motion. The target-to-total-mass ratio multiplies the laboratory beam energy to give the initial relative energy. Adding the channel Q-value gives the final relative energy, which must be nonnegative. Treat reaction mass differences in the rest-energy balance while neglecting their small effect on inertial mass sums only when that low-energy approximation is adequate.

In the final centre-of-mass frame, the two momenta are equal and opposite. Consequently the lighter product has greater speed and a larger share of kinetic energy. The reduced mass conveniently packages this partition into a single relative momentum. If a residual nucleus is left excited, lower the Q-value before calculating that momentum. A discrete set of final excitations can therefore produce distinct energy groups even with the same incident beam and nominal reaction channel.

Ei=mAma+mATa,Ef=Ei+Q=p∗22μf,Tb∗=mBmb+mBEf,μf=mbmBmb+mBE_i=\frac{m_A}{m_a+m_A}T_a,\qquad E_f=E_i+Q=\frac{p_*^2}{2\mu_f},\qquad T_b^*=\frac{m_B}{m_b+m_B}E_f,\quad\mu_f=\frac{m_bm_B}{m_b+m_B}
All E and T variables are kinetic energies, p* is the exit centre-of-mass momentum, and μf is the exit reduced mass. The star denotes the centre-of-mass frame. These expressions assume nonrelativistic two-body kinematics and consistent treatment of small rest-mass defects.↔ Scroll wide equations horizontally.

The laboratory adds a velocity vector

For elastic scattering, the centre-of-mass velocity is constant and the projectile’s relative speed has the same magnitude before and after collision. Its laboratory exit velocity is the vector sum of the centre-of-mass velocity and the rotated relative velocity. Squaring that sum produces an angle-dependent laboratory energy. The cross term can add or subtract depending on direction, so a single centre-of-mass speed corresponds to a range of laboratory energies.

For unequal masses, the laboratory angle requires the direction of that vector sum, preferably using a two-argument arctangent so the correct quadrant is retained. Endothermic reactions near threshold can even have two allowed laboratory energies at one laboratory angle. Conservation laws determine such branches but not their relative intensities. Cross sections require the transition amplitude and, when transforming angular densities between frames, the appropriate solid-angle Jacobian rather than only a relabeling of the angle.

vb=Vcm+vb∗,Tb=mb2(Vcm2+vb∗2+2Vcmvb∗cos⁡θ∗),θlab=atan2⁡(vb∗sin⁡θ∗,Vcm+vb∗cos⁡θ∗)\mathbf v_b=\mathbf V_{\rm cm}+\mathbf v_b^*,\qquad T_b=\frac{m_b}{2}\left(V_{\rm cm}^2+v_b^{*2}+2V_{\rm cm}v_b^*\cos\theta_*\right),\qquad\theta_{\rm lab}=\operatorname{atan2}(v_b^*\sin\theta_*,V_{\rm cm}+v_b^*\cos\theta_*)
Velocities are in m/s, mb in kilograms, and Tb in joules when using SI. θ* is measured from the incident beam. Galilean velocity addition is valid at nonrelativistic speeds; relativistic beams require Lorentz transformations.↔ Scroll wide equations horizontally.

Use elastic limiting cases as diagnostics

For a head-on elastic collision, solve the one-dimensional momentum and kinetic-energy equations together. A light projectile on a heavier stationary target can reverse direction while retaining part of its initial energy. For equal masses, a head-on collision transfers all kinetic energy to the target. In the infinitely heavy-target limit, the projectile speed is unchanged while its momentum direction can change, because a vanishing target speed can still carry finite recoil momentum.

These limits distinguish energy transfer from momentum transfer. A large scattering angle does not necessarily imply a large energy loss for a very heavy target. Conversely, the equal-mass case permits complete transfer even when no internal excitation occurs. In practice, reaction identification also needs isotope masses, excitation energies, energy loss, angular acceptance, and calibration. An ideal kinematic curve is an allowed locus for specified states, not a full detector response or proof that the scattering was elastic.

A light projectile backscatters from a heavier target

In a toy nonrelativistic elastic collision, a projectile of mass m and laboratory energy 8 MeV hits a stationary target of mass 3m. Its centre-of-mass scattering angle is 180°. Find its laboratory direction and energy, and the target recoil energy.

  1. If the incoming laboratory speed is v, the centre-of-mass speed is v/4v/4. The projectile’s initial relative velocity is 3v/43v/4 along the beam, while the target’s is −v/4-v/4.

  2. A 180° centre-of-mass deflection reverses the projectile relative velocity to −3v/4-3v/4. Adding the centre-of-mass velocity gives laboratory velocity v/4−3v/4=−v/2v/4-3v/4=-v/2: it travels backward at half its incident speed.

  3. Its kinetic energy is therefore 8(1/2)2=2 MeV8(1/2)^2=2\,\mathrm{MeV}. Elastic energy conservation leaves 6 MeV for the target. Momentum gives target speed v/2v/2, whose energy 12(3m)(v/2)2\tfrac12(3m)(v/2)^2 is indeed three times the projectile’s final energy.

The result

The projectile emerges backward with 2 MeV and the target recoils forward with 6 MeV. Energy and momentum conservation agree; the 180° angle used in the construction was a centre-of-mass angle.

Explore the model

Interactive lab

Interactive lab · vectors

Check the direction of an outgoing relative velocity

Reset and set both controls to angle 180° and magnitude 1.5; predict x projection −1.5 and y projection 0. Reset and set both to angle 90° and magnitude 1.5; predict x = 0 and y = 1.5. These represent two possible normalized relative-velocity directions. The lab displays one vector only; adding the centre-of-mass velocity and calculating laboratory energies remain separate calculations.

Check the direction of an outgoing relative velocityPlayback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Current readings: x component: 1.393; y component: 0.975; Magnitude: 1.7. All parameters are adjustable with the labeled controls below.−7.46−3.7303.737.46−3.5−1.7501.753.5xyvvₓ = 1.39
Vectorx / y projectionsTrace marker
x component
1.393
y component
0.975
Magnitude
1.7
t = 0 s

Playback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forRelative velocity rotates in the centre-of-mass frame before the laboratory translation is added.

Check your understanding

For a two-body exit channel in its centre-of-mass frame, which product has more kinetic energy?

Try it yourself

An 8 MeV projectile elastically scatters from a stationary target of the same mass. The projectile centre-of-mass deflection is 90°. Find both laboratory kinetic energies and the projectile laboratory angle.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

The centre-of-mass velocity and projectile relative speed are both v/2; add perpendicular vectors.

Compare with the solution

The projectile laboratory velocity has components (v/2,v/2)(v/2,v/2), so its squared speed is v2/2v^2/2 and its kinetic energy is 4 MeV. Its laboratory angle is 45°, not 90°. The target velocity is (v/2,−v/2)(v/2,-v/2), giving another 4 MeV. The transverse momenta cancel, the longitudinal momenta sum to the incoming value, and the total kinetic energy remains 8 MeV.

Take it with you

  • Relative energy is the beam energy after removing centre-of-mass translation.
  • Two-body products have equal momentum magnitudes, not generally equal energies.
  • A laboratory kinematic locus does not determine angular cross sections.

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