Perfectly inelastic collisions — maximum KE loss

WHAT is happening?

WHY is it "maximum" loss? Momentum conservation is a constraint we cannot break. Among all final states allowed by that constraint, the one where both masses share one velocity sits at the bottom of the kinetic-energy "valley." Any other allowed final state has more KE. We will prove this below.

HOW to derive the final velocity (from scratch)

Take masses $m_1$ (velocity $u_1$ ) and $m_2$ (velocity $u_2$ ). After sticking, both move at $v$ .

Step 1 — Conserve momentum. No external horizontal force ⇒ total $p$ is unchanged. $m_1 u_1 + m_2 u_2 = (m_1 + m_2)\,v$ Why this step? This is the only conservation law that survives; energy is allowed to leak away as heat/sound/deformation.

Step 2 — Solve for the common velocity. $\boxed{\,v = \dfrac{m_1 u_1 + m_2 u_2}{m_1 + m_2}\,}$ Why this step? It's just the momentum-weighted average of the two velocities — the centre-of-mass velocity, which never changes in any collision.

HOW much KE is lost? (Derivation)

Step 1 — Write KE before and after. $KE_i=\tfrac12 m_1u_1^2+\tfrac12 m_2u_2^2,\qquad KE_f=\tfrac12(m_1+m_2)v^2$

Step 2 — Substitute $v$ and simplify. Let me carry it out for the cleanest case and then the general one.

The KE lost is $\Delta KE = KE_i-KE_f$ After algebra (substituting $v=\frac{m_1u_1+m_2u_2}{m_1+m_2}$ ): $\boxed{\;\Delta KE=\dfrac{1}{2}\,\dfrac{m_1 m_2}{m_1+m_2}\,(u_1-u_2)^2\;}$ Why this step? The combination $\mu=\dfrac{m_1m_2}{m_1+m_2}$ is the reduced mass, and $(u_1-u_2)$ is the relative velocity. The lost energy is exactly the KE of the relative motion — which is destroyed when the bodies stop moving relative to each other.

Notice: $\Delta KE$ depends only on $(u_1-u_2)^2$ . If the COM frame velocity were anything else, momentum would be violated — so this is the only allowed sticking outcome, and it kills all the relative-motion KE. That is why it's the maximum.

WHY is this the MAXIMUM loss? (the key proof)

Figure — Perfectly inelastic collisions — maximum KE loss

Worked Examples

Worked example 2. Bullet into block (ballistic pendulum)

Bullet $m=10$ g at $u=400$ m/s embeds in block $M=1.99$ kg at rest. Step 1 — $v=\frac{(0.01)(400)}{0.01+1.99}=\frac{4}{2}=2$ m/s. Why? Common velocity = momentum / total mass. Step 2 — $KE_i=\tfrac12(0.01)(400)^2=800$ J. Step 3 — $KE_f=\tfrac12(2)(2)^2=4$ J. So $\Delta KE=796$ J → 99.5% lost! Why so much? A tiny mass shares momentum with a huge mass ⇒ $v$ tiny ⇒ almost all KE goes to heat/deformation. Reduced mass $\mu\approx0.00995$ kg, $\Delta KE=\tfrac12(0.00995)(400)^2\approx796$ J ✓.

Common Mistakes

Recall Feynman: explain to a 12-year-old

Imagine two lumps of clay sliding toward each other. When they hit, they squish and stick into one bigger lump. They have to move off together at one speed (you can't have the front going one way and back going another — they're glued!). Because they're forced to agree, a lot of their "moving energy" gets used up squishing the clay and making heat, instead of staying as motion. If they were running straight at each other equally hard, the blob just stops — all the motion energy turned into a squish. That "stuck together" rule is what makes them lose the most motion energy possible, while the total "push" (momentum) stays the same.

Flashcards

What is conserved in a perfectly inelastic collision?

Momentum (always); kinetic energy is NOT conserved.

What is the final velocity after two masses stick?

v=\dfrac{m_1u_1+m_2u_2}{m_1+m_2}

(the centre-of-mass velocity).

Formula for KE lost in a perfectly inelastic collision?

\Delta KE=\dfrac12\dfrac{m_1m_2}{m_1+m_2}(u_1-u_2)^2=\tfrac12\mu\,u_{rel}^2

Why is KE loss maximum in a perfectly inelastic collision?

Sticking forces both bodies to rest in the COM frame, destroying all relative-motion KE — the lowest final KE allowed by momentum conservation.

What is the reduced mass?

\mu=\dfrac{m_1m_2}{m_1+m_2}

Equal masses, one at rest, perfectly inelastic — fraction of KE lost?

Exactly

1/2

(half).

When is 100% of KE lost in a sticking collision?

When total momentum is zero (COM at rest), e.g. equal and opposite momenta.

Where does the lost KE go?

Heat, sound, and permanent deformation of the bodies.

Why can't you use energy conservation to find

v

here?

Because KE is partly lost to non-conservative work; only momentum is conserved.

The KE lost equals the kinetic energy of what motion?

The relative motion of the two bodies (

\tfrac12\mu u_{rel}^2

Connections

Conservation of Linear Momentum — the law that survives every collision.
Elastic collisions — opposite extreme: KE fully conserved, $e=1$ .
Coefficient of restitution — perfectly inelastic means $e=0$ .
Centre of mass motion — $v$ here is the COM velocity.
Reduced mass — appears in the KE-loss formula and two-body problems.
Ballistic pendulum — classic application (Example 2).
Work–Energy theorem — where the "lost" KE actually goes.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, perfectly inelastic collision ka matlab hai do cheezein takkar ke baad chipak jaati hain aur ek hi velocity se chalti hain — jaise do clay ke lumps. Yahan ek important baat: momentum hamesha conserve hota hai (kyunki bahar se koi force nahi), lekin kinetic energy conserve NAHI hoti. Common velocity nikalne ke liye sirf momentum equation use karo: $v=\frac{m_1u_1+m_2u_2}{m_1+m_2}$ . Ye actually centre of mass ki velocity hai jo kabhi change nahi hoti.

Ab KE loss kitna hota hai? Formula hai $\Delta KE=\tfrac12\frac{m_1m_2}{m_1+m_2}(u_1-u_2)^2$ . Yahan $\frac{m_1m_2}{m_1+m_2}$ ko reduced mass kehte hain, aur $(u_1-u_2)$ relative velocity hai. Matlab jo bhi relative motion energy thi, wo poori destroy ho jaati hai — heat, sound aur deformation mein chali jaati hai. Isliye is collision mein maximum possible KE loss hota hai: chipakne ke kaaran dono COM frame mein rukk jaate hain, aur isse kam KE rakhna momentum law todega.

Galtiyan kahan hoti hain? Bahut students energy conservation use karke $v$ nikaalne lagte hain — galat! Pehle momentum se $v$ nikaalo, phir KE loss calculate karo. Aur signs ka dhyaan rakho — velocity vector hai, opposite direction pe minus lagao. Agar total momentum zero ho (equal-opposite), to dono ekdum ruk jaate hain aur 100% KE loss hota hai. Ballistic pendulum (bullet block mein ghusna) iska classic example hai — wahan chhoti bullet ki lagbhag saari KE loss ho jaati hai.

Perfectly inelastic collisions — maximum KE loss

WHAT is happening?

HOW to derive the final velocity (from scratch)

HOW much KE is lost? (Derivation)

WHY is this the MAXIMUM loss? (the key proof)

Worked Examples

Common Mistakes

Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Momentum & Collisions

Connections