A bomb of 12 kg divides in two parts whose ra | vedclass

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Question

(a)The bomb of mass $ 12kg$  divides into two masses $m_1$  and $m_2$ then ${m_1} + {m_2} = 12$&hellip;(i)

and $\frac{{{m_1}}}{{{m_2}}} =

Vedclass · Accepted Answer

$36$

Vedclass · Answer

(a)The bomb of mass $ 12kg$  divides into two masses $m_1$  and $m_2$ then ${m_1} + {m_2} = 12$&hellip;(i)

and $\frac{{{m_1}}}{{{m_2}}} = \frac{1}{3}$&hellip;(ii) 

by solving we get ${m_1} = 3kg$ and ${m_2} = 9kg$ 

Kinetic energy of smaller part = $\frac{1}{2}{m_1}v_1^2 = 216J$

$v_1^2 = \frac{{216 	imes 2}}{3}$

==> ${v_1} = 12m/s$ 

So its momentum = ${m_1}{v_1} = 3 	imes 12 = 36\;kg{m{ - }}{m{m}}/s$ 

As both parts possess same momentum therefore momentum of each part is $36\;kg{m{ - }}m/s$

A bomb of $12 kg$ 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 kg-m/sec will be

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