Law of conservation of momentum and it's applications

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A shell of mass 20kg at rest explodes into two fragments whose masses are in the ratio 2: 3. The smaller fragment moves with a velocity of 6 m/s. The kinetic energy (in joule) of the larger fragment is

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Solution

$\begin{array}{l} m_{1} + m_{2} = 20 k g and \frac{m_{1}}{m_{2}} = \frac{2}{3} \\ \Rightarrow m_{1} = 8 k g ; m_{2} = 12 k g \\ Conservation of momentum, \\ M v = m_{1} v_{1} + m_{2} v_{2} \\ \Rightarrow 0 = 8 (6) + 12 (v_{2}) \\ \Rightarrow v_{2} = - 4 m / s \\ So the kinetic energy of the larger fragment is \\ \Rightarrow K . E = \frac{1}{2} m_{2} v_{2}^{2} = \frac{1}{2} (12) {(- 4)}^{2} = 6 (16) = 96 J \end{array}$

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Law of conservation of momentum and it's applications

Remember concepts with our Masterclasses.

A shell of mass 20kg at rest explodes into two fragments whose masses are in the ratio 2: 3. The smaller fragment moves with a velocity of 6 m/s. The kinetic energy (in joule) of the larger fragment is

m1+m2=20kg and m1m2=23⇒m1=8kg ; m2=12kgConservation of momentum, M v=m1v1+m2v2⇒0=8(6)+12(v2)⇒v2=−4 m/s So the kinetic energy of the larger fragment is ⇒K.E=12m2v22=12(12)(−4)2=6(16)=96J

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