Statement A : A bomb of mass M at rest explodes into two parts of mass " $m_{1}$ " and " $m_{2}$ " . The total energy released is E. The energy carried by mass $m_{1}$ is $\frac{E m_{2}}{M}$
Statement B : The kinetic energy is not conserved during inelastic collision

A and B are true

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A is true but B is false

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A is false but B is true

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A and B are false

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Solution

The correct option is B A and B are true
The bomb was at rest initially therefore the initial momentum is zero
Now Applying conservation of momentum
$p_{i} = p_{f}$
let the velocity of parts after explosion be $v_{1}$ and $v_{2}$
$0 = m_{1} v_{1} + m_{2} v_{2}$

$v_{1} = - \frac{m_{2} v_{2}}{m_{1}}$

The energy released is in the form of K.E.
$K E_{1} + K E_{2} = E$

$K E_{1} + \frac{1}{2} m_{2} v_{2}^{2} = E$

substituting value of $v_{2}$

$\frac{1}{2} m_{1} v_{1}^{2} + \frac{1}{2} m_{2} \times (- \frac{m_{1} v_{1}}{m_{2}})^{2} = E$

$\frac{1}{2} m_{1} v_{1}^{2} (1 + \frac{m_{1}}{m_{2}}) = E$

$\frac{1}{2} m_{1} v_{1}^{2} (\frac{m_{1} + m_{2}}{m_{2}}) = E$

we know $m_{1} + m_{2} = M$

$K E_{1} \times \frac{M}{m_{2}} = E$

$K E_{1} = \frac{E m_{2}}{M}$

In case of inelastic collisions some of the kinetic energy is converted in some internal energy and hence is not conserved.

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Statement A : A bomb of mass M at rest explodes into two parts of mass " m1" and " m2" . The total energy released is E. The energy carried by mass m1 is Em2MStatement B : The kinetic energy is not conserved during inelastic collision

Statement A : A bomb of mass M at rest explodes into two parts of mass " $m_{1}$ " and " $m_{2}$ " . The total energy released is E. The energy carried by mass $m_{1}$ is $\frac{E m_{2}}{M}$
Statement B : The kinetic energy is not conserved during inelastic collision