A stationary body explodes into four identical fragments such that...

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 energy of the explosion will be

K

times of

{E}_{0},

then the value of

K

Correct answer

Step-by-step solution

\overset{\to}{{P}_{1}} = m{v}_{0} \overset{^}{i}

\overset{\to}{{P}_{2}} = m{v}_{0} \overset{^}{j}

\overset{\to}{{P}_{3}} = m{v}_{0}\overset{^}{j}

\overset{\to}{{P}_{4}} = m\overset{\to}{v}

0 = \overset{\to}{{P}_{1}} + \overset{\to}{{P}_{2}} + \overset{\to}{{P}_{3}} + \overset{\to}{{P}_{4}}

\overset{\to}{V} = - {v}_{0} ( \overset{^}{i} + \overset{^}{j} + \overset{^}{k} )

v = {v}_{0}\sqrt{3}

Total energy

= 3 \times \dfrac{1}{2}m{v}_{0}^{2} + \dfrac{1}{2}m{v}^{2} = 3{E}_{0} + 3{E}_{0}

= 6{E}_{0}

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