a stationary bomb explodes into two parts of masses 3kg and 1kg.the total kinetic energy of the two parts after the explosion is 2400J.the kinetic energy of smaller part is

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Solution

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$m_{1} = 3 kg m_{2} = 1 kg So by conservation of momentum : m_{1} v_{1} + m_{2} v_{2} = (m_{1} + m_{2}) 0 m_{1} v_{1} + m_{2} v_{2} = 0 m_{1} v_{1} = - m_{2} v_{2} 3 \times v_{1} = - v_{2} - 3 v_{1} = v_{2} Total kinetic energy : \frac{1}{2} m_{1} {v_{1}}^{2} + \frac{1}{2} m_{2} {v_{2}}^{2} = 2400 \frac{1}{2} (3 \times {v_{1}}^{2} + {v_{2}}^{2}) = 2400 3 \times {v_{1}}^{2} + (- 3 v_{1})^{2} = 4800 3 {v_{1}}^{2} + 9 {v_{1}}^{2} = 4800 12 {v_{1}}^{2} = 4800 {v_{1}}^{2} = 400 v_{1} = 20 m / s and v_{2} = - 3 \times 20 = - 60 m / s Here, Negative sign indicate the opposite direction of the smaller fragment . Kinetic energy of smaller part, KE = \frac{1}{2} m_{2} v_{_{2}}^{2} KE = \frac{1}{2} \times 1 \times {(60)}^{2} = 1800 J$

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