Problem Statement
A proton (mass m) elastically scatters off a stationary proton. After collision, one proton moves at angle $\theta$ $_1$ and the other at $\theta$ $_2$ to the original direction. Show that $\theta$ $_1 + \theta _2$ = 90°.
Given Information
- All numerical data are stated in the problem above; symbols are defined as they appear.
Physical Concepts & Formulas
These problems use Hooke’s law and the elastic moduli that relate stress and strain in a deformed solid. For equal-mass elastic collisions in $2D$: the velocity vectors of the two particles after collision are always perpendicular. This follows from conservation of both momentum and KE with $m_1$ = $m_2$.
- $\sigma = E\,\varepsilon$ — Hooke’s law (Young’s modulus)
- $\tau = G\,\gamma$ — shear stress and strain
- $U = \tfrac{1}{2}\dfrac{\sigma^2}{E}$ — elastic energy density
Step-by-Step Solution
Step 1 — Identify the governing principle: We begin by recognising which physical law controls the situation and why it is the correct starting point for this problem.
$$Conservation: mv_0 = mv_1’ + mv_2’ (vectors), (1/2)mv_0^{2}= (1/2)mv_1’^{2}+ (1/2)mv_2’^{2}$$
Step 2 — Set up the relevant equations: Next we write down the equations that follow from that principle, introducing the symbols we will carry through the algebra. $v_0^{2}$ = v_1’$^{2}$ + v_2’$^{2}$ + 2v_1’ * v_2’ (from momentum squared) and $v_0^{2}$ = v_1’$^{2}$ + v_2’$^{2}$ (from KE)
Step 3 — Apply the given conditions: We now substitute the specific conditions and constraints given in the problem so the equations describe this particular situation. Therefore 2v_1’ * v_2’ = 0 → v_1’ ⊥ v_2’ → $\theta$ $_1 + \theta _2$ = 90°
Worked Calculation
$$v_1’ \cdot v_2’ = 0 → \theta _1 + \theta _2 = 90^{\circ}$$
Answer
$$\boxed{\theta _1 + \theta _2 = 90^{\circ} (for equal-mass elastic collision)}$$
This is the quantity the problem asked for, expressed in terms of the given data: $\theta _1 + \theta _2 = 90^{\circ} (for equal-mass elastic collision)$.
Physical Interpretation
This is the famous 90° rule for equal-mass elastic scattering — billiard balls always part at right angles, which snooker players know intuitively. The magnitude of the answer is consistent with everyday physical experience for this class of problem in Irodov’s Part 1 — the result shows how the answer scales with the given quantities. If we doubled the dominant input, the boxed formula tells us exactly how the output would respond, and that scaling is the key physical insight this problem trains. Comparing the answer with the appropriate limiting cases (very small or very large values of the dominant parameter) recovers the familiar Newtonian or intuitive expectation, which is a useful sanity check.
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