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Question

(c)As the momentum of both fragments are equal therefore

$\frac{{{E_1}}}{{{E_2}}} = \frac{{{m_2}}}{{{m_1}}} = \frac{3}{1}$

i.e. ${E_1} = 3{E_2}$

Vedclass · Accepted Answer

$4.8 	imes {10^4}J$

Vedclass · Answer

(c)As the momentum of both fragments are equal therefore

$\frac{{{E_1}}}{{{E_2}}} = \frac{{{m_2}}}{{{m_1}}} = \frac{3}{1}$

i.e. ${E_1} = 3{E_2}$&hellip;(i)

According to problem ${E_1} + {E_2} = 6.4 	imes {10^4}J$ &hellip;(ii)

By solving equation (i) and (ii) we get

${E_1} = 4.8 	imes {10^4}J$ and ${E_2} = 1.6 	imes {10^4}J$

A bomb is kept stationary at a point. It suddenly explodes into two fragments of masses $1\, g$ and $3\;g$. The total K.E. of the fragments is $6.4 \times {10^4}J$. What is the K.E. of the smaller fragment

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