A stationary body explodes into four identical fragments such that three of them fly off mutually perpendicular to each other, each with same kinetic energy, $E_{0}$ . The minimum energy of explosion will be

6 E_{0}

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\frac{4 E_{0}}{3}

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4 E_{0}

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8 E_{0}

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Solution

The correct option is C $6 E_{0}$
Let the three fragments move along x, y, z direction with velocities $V_{a}^i, V_{b}^j, V_{c}^k$

hence, $m_{a} V_{a}^i + m_{b} V_{b}^j + m_{c} V_{c}^j + m_{d}_{d} = 0$

As $m_{a} = m_{b} = m_{c} = m_{d}$

so, $_{d} = - V_{0} (V_{a}^i + V_{b}^j + V_{c}^k)$

Again as K.E are same and mass are same

so, $V_{a} = V_{b} = V_{c} = V_{0}$

Hence, $_{d} = - V_{0} (^i +^j +^k)$

The energy of explosion $= K . E_{f} - K . E_{i}$

$= \frac{1}{2} m {V_{d}}^{2} + [\frac{1}{2} m {V_{0}}^{2} X^{3}]$

$= \frac{1}{2} m {V_{0}}^{2} (3) + \frac{1}{2} m {V_{0}}^{2} (3)$

$= 3 m {V_{0}}^{2}$

$^P (i n i t i a l) =^P (f i n a l) = 6 E_{0}$

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