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A bomb of 1...

Question

A bomb of $12 k g$ divides in two parts whose ratio of masses is $1 : 3$ . If kinetic energy of smaller part is $216 J$ , then momentum of bigger part in $k g - m / s e c$ will be

36

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72

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108

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Data is incomplete

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Solution

The correct option is A $36$
Mass of smaller part $m_{1} = \frac{1}{5} \times 12 = 2.4 k g$
Kinetic energy smaller part $E_{1} = 288 J$ (Given)
Thus momentum of bigger part $P_{1} = \sqrt{2 m_{1} E_{1}}$
$⟹ P_{1} = \sqrt{2 (2.4) (288)} = 37 k g m / s$
Since the linear momentum of the bomb before division was zero. Thus according to law of conservation of linear momentum, sum of linear momentum of both parts is also zero. So, both parts move with equal linear momentum but in opposite direction. i.e. $| P_{1} | = | P_{2} |$
Thus momentum of bigger part $| P_{2} | = | P_{1} | = 37 k g m / s$

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A bomb of 12kg divides in two parts whose ratio of masses is 1:3. If kinetic energy of smaller part is 216J, then momentum of bigger part in kg−m/sec will be

A bomb of $12 k g$ divides in two parts whose ratio of masses is $1 : 3$ . If kinetic energy of smaller part is $216 J$ , then momentum of bigger part in $k g - m / s e c$ will be