A bomb at rest explodes into two fragments of masses 3.0 kg and 1.0 kg. The total kinetic energy of the fragment is 6x10^4 joules. Calculate the kinetic energy of the bigger fragment. Also, in which type of collision, elastic or inelastic, momentum is conserved?

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Solution

$m_{1} = 3 k g m_{2} = 1 k g S o b y c o n s e r v a t i o n o f m o m e n t u m : m_{1} v_{1} + m_{2} v_{2} = (m_{1} + m_{2}) 0 m_{1} v_{1} + m_{2} v_{2} = 0 m_{1} v_{1} = - m_{2} v_{2} 3 v_{1} = - 1 v_{2} - 3 v_{1} = v_{2} T o t a l k i n e t i c e n e r g y : 1 / 2 * m_{1} {v_{1}}^{2} + 1 / 2 * m_{2} {v_{2}}^{2} = 6 * 10^{4} 1 / 2 (3 * {v_{1}}^{2} + {v_{2}}^{2}) = 6 * 10^{4} 3 * {v_{1}}^{2} + (- 3 v_{1})^{2} = 12 * 10^{4} 3 * {v_{1}}^{2} + 9 {v_{1}}^{2} = 12 * 10^{4} 12 {v_{1}}^{2} = 12 * 10^{4} {v_{1}}^{2} = 10^{4} v_{1} = 100 m / s s o v_{2} = - 300 m / s$
In elastic collision the momentum is conserved as there is no loss in the body deterioration!

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