Problem Statement
A nucleus of mass M at rest decays into two fragments. Fragment $1 has$ mass $m_1$ and kinetic energy $T_1$. Find the kinetic energy of fragment $2$
Given Information
- All numerical data are stated in the problem above; symbols are defined as they appear.
Physical Concepts & Formulas
These problems exploit conservation laws. Identify the system, check which quantities (energy, linear momentum, angular momentum) are conserved, and equate initial and final states. From $m_1v_1$ = $m_2v_2$: $p_1$ = $p_2$ = p. So $T_1/T_2$ = $m_2/m_1$.
- $E_k = \tfrac{1}{2}mv^2$ — kinetic energy
- $\vec{p} = m\vec{v}$ — linear momentum
- $\vec{L} = \vec{r}\times m\vec{v}$ — angular momentum
- $A = \Delta E$ — work–energy theorem
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.
$$p_1 = p_2 (momentum conservation, nucleus at rest)$$
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.
$$T = p^{2}/(2m) → T_1 = p^{2}/(2m_1), T_2 = p^{2}/(2m_2)$$
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.
$$T_2 = T_1 \cdot m_1/m_2 = T_1 \cdot m_1/(M-m_1)$$
Worked Calculation
$$T_2 = T_1 \cdot m_1/(M-m_1)$$
Answer
$$\boxed{T_2 = T_1 \cdot m_1/(M – m_1)}$$
This is the quantity the problem asked for, expressed in terms of the given data: $T_2 = T_1 \cdot m_1/(M – m_1)$.
Physical Interpretation
In any two-body decay from rest, both fragments carry equal and opposite momenta, so the lighter fragment always has more kinetic energy. 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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