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. $0.6\text{ E}$
• B. $0.4\text{ E}$
• C. $0.2\text{ E}$
• D. $0.8\text{ E}$

Hint:

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: $$K_1 = \frac{3}{3+2}E = \frac{3}{5}E = 0.6E$$

Answer 1

The Correct Option is A

Step 1: Understand the breakup.
A body is at rest and then breaks into two pieces. One piece has mass $4M$ and the other has mass $6M$. Together they share a total kinetic energy $E$. We want the kinetic energy of the lighter piece $4M$.

Step 2: Use momentum conservation.
Before breaking, the body is at rest, so total momentum is zero. After breaking, the two pieces must have equal and opposite momenta so they still add up to zero: \[ p_1 = p_2 = p \]

Step 3: Link kinetic energy to momentum.
For any piece, kinetic energy can be written using momentum and mass: \[ K = \frac{p^2}{2m} \] Since both pieces have the same momentum size $p$, the kinetic energy depends inversely on the mass.

Step 4: Write each piece's energy.
For the $4M$ piece: \[ K_1 = \frac{p^2}{2(4M)} \] For the $6M$ piece: \[ K_2 = \frac{p^2}{2(6M)} \]

Step 5: Find the share of the lighter piece.
The energies are in the ratio of the inverse masses, so $K_1 : K_2 = \frac{1}{4} : \frac{1}{6} = 6 : 4 = 3 : 2$. The total parts are $3 + 2 = 5$. The lighter piece gets $\frac{3}{5}$ of the total energy.

Step 6: Compute the value.
So the kinetic energy of the $4M$ piece is: \[ K_1 = \frac{3}{5} E = 0.6 E \] The lighter piece carries more energy because lighter pieces move faster after a breakup. \[ \boxed{0.6\,E} \]