Amit & Deepali Sinha Foundation

Reduced Mass Absorbs the Lost Energy

Irodov 1.167DifficultyUniqueness

Problem Irodov 1.167

Two spheres of masses m1m_1 and m2m_2 form an isolated system: no external force acts on the pair. Moving with initial velocities v1\vect v_1 and v2\vect v_2, they collide perfectly inelastically. Find the resulting change ΔT\Delta T in the system’s total kinetic energy.

Before: the spheres approach with v_1 and v_2 at a general angle. After: they move off together at the common velocity V. The drawn angle is illustrative; the derivation assumes nothing about it.

Before: the spheres approach with v1\vect v_1 and v2\vect v_2 at a general angle. After: they move off together at the common velocity V\vect V. The drawn angle is illustrative; the derivation assumes nothing about it.

Given
  • m1m_1
  • m2m_2
  • v1\vect v_1
  • v2\vect v_2
Find
ΔT\Delta T

Solution

Step 1Common velocity from momentum conservation

The system is closed, so no external force acts during the collision and the total momentum is the same immediately before and after it. Equating the momentum of the two separate spheres to that of the single combined body of mass m1+m2m_1+m_2 gives their shared velocity V\vect V after impact.

m1v1+m2v2=(m1+m2)Vm_1\vect v_1+m_2\vect v_2 = (m_1+m_2)\vect V
V=m1v1+m2v2m1+m2\vect V = \dfrac{m_1\vect v_1+m_2\vect v_2}{m_1+m_2}
(1)
Step 2Kinetic energy before and after

The kinetic energy before the collision is the sum of the two spheres’ separate kinetic energies, while afterward the combined body of mass m1+m2m_1+m_2 carries only (1)’s common velocity. Their difference is the sought increment.

ΔT=12(m1+m2)V2−(12m1v12+12m2v22)=(m1v1+m2v2)22(m1+m2)−12m1v12−12m2v22\begin{align*}\Delta T &= \frac12(m_1+m_2)V^2-\left(\frac12m_1v_1^2+\frac12m_2v_2^2\right) \\ &= \frac{(m_1\vect v_1+m_2\vect v_2)^2}{2(m_1+m_2)}-\frac12m_1v_1^2-\frac12m_2v_2^2\end{align*}
Step 3Reducing the difference to the relative velocity

Clear the denominator first: multiplying the last line above by 2(m1+m2)2(m_1+m_2), expanding the square and expanding the product, the four squared-speed terms cancel in pairs.

2(m1+m2) ΔT=(m1v1+m2v2)2−(m1+m2)(m1v12+m2v22)=m12v12+2m1m2 v1⋅v2+m22v22−m12v12−m1m2v22−m1m2v12−m22v22=−m1m2(v12−2v1⋅v2+v22)=−m1m2(v1−v2)2\begin{align*}2(m_1+m_2)\,\Delta T &= (m_1\vect v_1+m_2\vect v_2)^2-(m_1+m_2)(m_1v_1^2+m_2v_2^2) \\ &= m_1^2v_1^2+2m_1m_2\,\vect v_1\cdot\vect v_2+m_2^2v_2^2 \\ &\qquad -m_1^2v_1^2-m_1m_2v_2^2-m_1m_2v_1^2-m_2^2v_2^2 \\ &= -m_1m_2\bigl(v_1^2-2\vect v_1\cdot\vect v_2+v_2^2\bigr) \\ &= -m_1m_2(\vect v_1-\vect v_2)^2\end{align*}

What survives depends on the two velocities only through their difference. Dividing back by 2(m1+m2)2(m_1+m_2),

ΔT=−m1m2m1+m2⋅(v1−v2)22\Delta T = -\frac{m_1m_2}{m_1+m_2}\cdot\frac{(\vect v_1-\vect v_2)^2}{2}

The mass combination standing in front is the system’s reduced mass; naming it once puts the result in its standard form.

μ=m1m2m1+m2\mu = \dfrac{m_1m_2}{m_1+m_2}
(2)
ΔT=−μ(v1−v2)22\Delta T = -\dfrac{\mu(\vect v_1-\vect v_2)^2}{2}
(3)
Check

Since μ>0\mu>0 and (v1−v2)2≥0(\vect v_1-\vect v_2)^2\ge0 always, the boxed ΔT\Delta T is never positive, consistent with a perfectly inelastic collision only ever dissipating kinetic energy. Equality ΔT=0\Delta T=0 holds exactly when v1=v2\vect v_1=\vect v_2, when the two spheres already share one velocity and no collision actually occurs.

Remark

The numerator −m1m2(v1−v2)2-m_1m_2(\vect v_1-\vect v_2)^2 that survived the cancellation above is the same combination that splits the total kinetic energy into a center-of-mass part (m1+m2)V2/2(m_1+m_2)V^2/2 and a relative part μ(v1−v2)2/2\mu(\vect v_1-\vect v_2)^2/2. A perfectly inelastic collision leaves the center-of-mass part untouched, since the combined body still carries the whole momentum, and destroys exactly the relative part, which is why ΔT\Delta T depends only on the relative velocity and never on v1\vect v_1, v2\vect v_2 separately.