Irodov 1.177: The Right Angle Hidden in Momentum – Amit & Deepali Sinha Foundation

Problem Irodov 1.177

A moving molecule collides with a second, stationary molecule of the same mass. Show that

if the collision is ideally elastic, the angle between the two molecules’ velocities after the collision (the angle of divergence) equals $90^\circ$ ;
if the collision is inelastic, that angle differs from $90^\circ$ .

The setup: a molecule of mass m arrives with velocity v_0 at a second molecule of the same mass, initially at rest. Nothing about the two outgoing velocities is settled before the collision is classified as elastic or inelastic. — The setup: a molecule of mass $m$ arrives with velocity $\vect v_0$ at a second molecule of the same mass, initially at rest. Nothing about the two outgoing velocities is settled before the collision is classified as elastic or inelastic.

Given

two molecules, equal mass $m$
one initially at rest

Find

Demonstrate that the angle between the two final velocities is

90^\circ

for an elastic collision and not

90^\circ

for an inelastic one.

Solution

Step 1Momentum conservation removes the mass entirely

Let $\vect v_0$ be the initial velocity of the moving molecule, and $\vect v_1$ , $\vect v_2$ the final velocities of the two molecules after the collision. Since both molecules have the same mass $m$ , momentum conservation

m\vect v_0 = m\vect v_1+m\vect v_2

divides through by $m$ on both sides, leaving a relation between the velocities alone:

\vect v_0 = \vect v_1+\vect v_2

(1)

Step 2Squaring the vector relation

Squaring (1) (i.e. dotting it with itself),

v_0^2 = \left(\vect v_1+\vect v_2\right)\cdot\left(\vect v_1+\vect v_2\right) = v_1^2+v_2^2+2\vect v_1\cdot\vect v_2.

Writing $\alpha$ for the angle of divergence between $\vect v_1$ and $\vect v_2$ where they leave the collision, the dot product is $\vect v_1\cdot\vect v_2 = v_1v_2\cos\alpha$ , so substituting gives

v_0^2 = v_1^2+v_2^2+2v_1v_2\cos\alpha

(2)

Step 3Part (a): the elastic case

The velocity parallelogram for the elastic case: v_1 and v_2 compose to the initial velocity v_0, at the right angle this step establishes. The ratio drawn is v_1:v_2:v_0=3:4:5. — The velocity parallelogram for the elastic case: $\vect v_1$ and $\vect v_2$ compose to the initial velocity $\vect v_0$ , at the right angle this step establishes. The ratio drawn is $v_1:v_2:v_0=3:4:5$ .

An ideally elastic collision conserves kinetic energy as well as momentum:

\frac{1}{2}mv_0^2 = \frac{1}{2}mv_1^2+\frac{1}{2}mv_2^2 \quad\Longrightarrow\quad v_0^2 = v_1^2+v_2^2.

Comparing this with (2), the two expressions for $v_0^2$ agree only if $2v_1v_2\cos\alpha=0$ .

Since $v_1\neq0$ and $v_2\neq0$ in general,

\cos\alpha = 0 \;\Longrightarrow\; \alpha = 90^\circ \text{ (elastic collision)}

(3)

Step 4Part (b): the inelastic case

An inelastic collision loses kinetic energy, $mv_0^2/2 > mv_1^2/2+mv_2^2/2$ , so

v_0^2 > v_1^2+v_2^2.

Substituting into (2), the same comparison now gives

v_1^2+v_2^2+2v_1v_2\cos\alpha > v_1^2+v_2^2 \quad\Longrightarrow\quad \cos\alpha > 0.

\cos\alpha > 0 \;\Longrightarrow\; \alpha < 90^\circ \text{ (inelastic collision)}

(4)

Check

The figure’s own numbers are a direct check: for $v_1:v_2:v_0=3:4:5$ , substituting into $v_0^2=v_1^2+v_2^2+2v_1v_2\cos\alpha$ gives $25 = 9+16+24\cos\alpha$ , so $\cos\alpha=0$ and $\alpha=90^\circ$ exactly as part (a) claims, the same right angle that makes 3, 4, 5 a Pythagorean triple.

Remark

Part (a) is the vector version of a familiar fact: two equal masses leaving an elastic collision always separate at a right angle (billiard players rely on this), except the head-on case, where the striker stops and no divergence angle exists, while part (b) shows that same right angle closing up whenever energy is lost to internal degrees of freedom, so measuring the divergence angle alone is enough to tell an elastic collision from an inelastic one, without ever measuring speeds.