A particle of mass $4M$ at rest splits into two particles of mass $M$ and $3M$. The ratio of the kinetic energ

Question

A particle of mass $4M$ at rest splits into two particles of mass $M$ and $3M$. The ratio of the kinetic energies of mass $M$ and $3M$ would be:

Accepted Answer

1. **Conservation of Momentum:** Since the initial particle is at rest, the initial total momentum is zero. In the absence of external forces, the total momentum of the system is conserved during the split (explosion/decay) [Class 11 Physics, Ch 6, Sec 6.4, Eq 6.18a].
   $$ \vec{P}_{initial} = \vec{P}_{final} = 0 $$
   $$ \vec{p}_1 + \vec{p}_2 = 0 \implies \vec{p}_1 = -\vec{p}_2 $$
   The magnitudes of momentum for both particles are equal: $p_1 = p_2 = p$.
2. **Relate Kinetic Energy to Momentum:** Kinetic energy ($K$) is related to momentum ($p$) and mass ($m$) by the formula:
   $$ K = \frac{p^2}{2m} $$
3. **Calculate Ratio:**
   For the first particle (mass $M$): $K_1 = \frac{p^2}{2M}$
   For the second particle (mass $3M$): $K_2 = \frac{p^2}{2(3M)}$
   $$ 	ext{Ratio } \frac{K_1}{K_2} = \frac{\frac{p^2}{2M}}{\frac{p^2}{6M}} = \frac{p^2}{2M} 	imes \frac{6M}{p^2} = 3 $$
   Thus, the ratio is $3:1$.

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