Problem Statement
A particle of mass $m_1$ elastically collides with a stationary particle of mass $m_2$. Find the maximum angle of deflection of the first particle.
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. In CM frame, $m_1$ scatters at any angle. Transform to lab. Maximum deflection angle: sin $\theta$ $_max$ = $m_2/m_1$ (when $m_1$ > $m_2$).
- $\sigma = E\,\varepsilon$ — 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. In CM frame, both particles scatter symmetrically
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. Lab angle of $m_1$: tan $\theta$ = $(m_2)$ sin $\phi$ $)/(m_1 + m_2$ cos $\phi$ ) where $\phi$ is CM angle
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. For $m_1$ > $m_2$: $\theta$ $_max$ = $\arcsin (m_2/m_1)$)
$$Maximize d \theta /d \phi = 0$$
Step 4 — Solve for the required quantity: With the equations specialised, we isolate and solve for the unknown the problem asks us to find. For $m_1$ ≤ $m_2$: $\theta$ $_max$ = $\pi$ $/2$ (or 180° — full backscatter possible)
Worked Calculation
$\theta$ $_max$ = $\arcsin (m_2/m_1)$) for $m_1$ > $m_2$
Answer
$$\boxed{\theta _max = \arcsin (m_2/m_1) for m_1 > m_2}$$
This is the quantity the problem asked for, expressed in terms of the given data: $\theta _max = \arcsin (m_2/m_1) for m_1 > m_2$.
Physical Interpretation
A heavy particle cannot be deflected by more than $\arcsin (m_2/m_1)$) in an elastic collision with a lighter target — it’s geometrically constrained by the momentum triangle. 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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