Question:
A body which is initially at rest breaks into 2 pieces of masses $4M$ and $6M$ respectively, together having a total kinetic energy $E$. The piece with mass $4M$, after breaking has a kinetic energy of:
A body which is initially at rest breaks into 2 pieces of masses $4M$ and $6M$ respectively, together having a total kinetic energy $E$. The piece with mass $4M$, after breaking has a kinetic energy of:
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When a body breaks from rest into two pieces, their kinetic energies are inversely proportional to their masses because their momentum magnitudes are equal ($K \propto \frac{1}{m}$):
$$\frac{K_1}{K_2} = \frac{m_2}{m_1} = \frac{6M}{4M} = \frac{3}{2}$$
Thus, the fractional energy of the first piece is:
$$K_1 = \frac{3}{3+2}E = \frac{3}{5}E = 0.6E$$
Updated On: Jun 10, 2026
- $0.6\text{ E}$
- $0.4\text{ E}$
- $0.2\text{ E}$
- $0.8\text{ E}$
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The Correct Option is A
Solution and Explanation
Concept:
During an internal explosion or breakage of an isolated system, no external forces act on the system. Therefore, the total linear momentum of the system must remain conserved.
The relationship between the linear momentum ($p$) of a particle and its corresponding kinetic energy ($K$) is given by:
$$K = \frac{p^2}{2m}$$
Step 1: The body is initially at rest, meaning its initial linear momentum is zero: $$\vec{P}_{\text{initial}} = 0$$ After splitting into two pieces of masses $m_1 = 4M$ and $m_2 = 6M$, let their respective final velocities be $\vec{v}_1$ and $\vec{v}_2$. Applying the law of conservation of linear momentum: $$\vec{P}_{\text{final}} = \vec{P}_{\text{initial}}$$ $$m_1\vec{v}_1 + m_2\vec{v}_2 = 0$$ $$4M\vec{v}_1 + 6M\vec{v}_2 = 0 \implies 4M\vec{v}_1 = -6M\vec{v}_2$$ Taking the magnitude of momentum on both sides shows that both fragments acquire equal and opposite momentum vectors: $$|p_1| = |p_2| = p$$
Step 2: Since both pieces share the exact same momentum magnitude $p$, we can write their individual kinetic energies using the momentum-energy relation: $$K_1 = \frac{p^2}{2m_1} = \frac{p^2}{2(4M)} = \frac{p^2}{8M}$$ $$K_2 = \frac{p^2}{2m_2} = \frac{p^2}{2(6M)} = \frac{p^2}{12M}$$
Step 3: We are given that the total combined kinetic energy after the separation is $E$: $$E = K_1 + K_2$$ $$E = \frac{p^2}{8M} + \frac{p^2}{12M}$$ Finding a common denominator (which is $24M$) to combine the terms: $$E = p^2 \cdot \left( \frac{3}{24M} + \frac{2}{24M} \right) = \frac{5p^2}{24M}$$ From this expression, we can isolate the term $\frac{p^2}{M}$: $$\frac{p^2}{M} = \frac{24E}{5} \quad \cdots (1)$$
Step 4: We need to determine the value of $K_1$ (the kinetic energy of the $4M$ mass): $$K_1 = \frac{p^2}{8M} = \frac{1}{8} \cdot \left(\frac{p^2}{M}\right)$$ Substituting the isolated value from equation (1) into this equation: $$K_1 = \frac{1}{8} \cdot \left( \frac{24E}{5} \right)$$ $$K_1 = \frac{3E}{5} = 0.6\text{ E}$$ Thus, the kinetic energy of the $4M$ fragment is $0.6\text{ E}$, matching Option (A).
Step 1: The body is initially at rest, meaning its initial linear momentum is zero: $$\vec{P}_{\text{initial}} = 0$$ After splitting into two pieces of masses $m_1 = 4M$ and $m_2 = 6M$, let their respective final velocities be $\vec{v}_1$ and $\vec{v}_2$. Applying the law of conservation of linear momentum: $$\vec{P}_{\text{final}} = \vec{P}_{\text{initial}}$$ $$m_1\vec{v}_1 + m_2\vec{v}_2 = 0$$ $$4M\vec{v}_1 + 6M\vec{v}_2 = 0 \implies 4M\vec{v}_1 = -6M\vec{v}_2$$ Taking the magnitude of momentum on both sides shows that both fragments acquire equal and opposite momentum vectors: $$|p_1| = |p_2| = p$$
Step 2: Since both pieces share the exact same momentum magnitude $p$, we can write their individual kinetic energies using the momentum-energy relation: $$K_1 = \frac{p^2}{2m_1} = \frac{p^2}{2(4M)} = \frac{p^2}{8M}$$ $$K_2 = \frac{p^2}{2m_2} = \frac{p^2}{2(6M)} = \frac{p^2}{12M}$$
Step 3: We are given that the total combined kinetic energy after the separation is $E$: $$E = K_1 + K_2$$ $$E = \frac{p^2}{8M} + \frac{p^2}{12M}$$ Finding a common denominator (which is $24M$) to combine the terms: $$E = p^2 \cdot \left( \frac{3}{24M} + \frac{2}{24M} \right) = \frac{5p^2}{24M}$$ From this expression, we can isolate the term $\frac{p^2}{M}$: $$\frac{p^2}{M} = \frac{24E}{5} \quad \cdots (1)$$
Step 4: We need to determine the value of $K_1$ (the kinetic energy of the $4M$ mass): $$K_1 = \frac{p^2}{8M} = \frac{1}{8} \cdot \left(\frac{p^2}{M}\right)$$ Substituting the isolated value from equation (1) into this equation: $$K_1 = \frac{1}{8} \cdot \left( \frac{24E}{5} \right)$$ $$K_1 = \frac{3E}{5} = 0.6\text{ E}$$ Thus, the kinetic energy of the $4M$ fragment is $0.6\text{ E}$, matching Option (A).
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