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A projectile is launched vertically upward. It explodes into two pieces at the top point of its trajectory. One piece has twice the mass of the other. Immediately after the explosion, the more massive piece has kinetic energy $E$ . What is the total kinetic energy of both pieces immediately after the explosion?

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$1.5 E$

$2 E$

$3 E$

not enough information to answer

answer is C.

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Detailed Solution

Let velocities of fragments be $V_{1} and V_{2}$ and of the system is zero.

$\Rightarrow \frac{2 m}{3} V_{1} - \frac{m}{3} V_{2} = 0 \Rightarrow \frac{V_{1}}{V_{2}} = \frac{1}{2}$

Ratio of $K E$

$\frac{K_{2}}{K_{1}} = \frac{\frac{1}{2} (\frac{1}{3} m) V_{2}^{2}}{\frac{1}{2} (\frac{2 m}{3}) V_{1}^{2}} = \frac{1}{2} (2)^{2} = 2 (K_{1} = E given) K_{2} = 2 K_{1} = 2 E$

Total $KE = K_{1} + K_{2} = 3 E$

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