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
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
$36$
$72$
$108$
Data is incomplete
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
The bomb of mass $ 12kg$ divides into two masses $m_1$ and $m_2$ then ${m_1} + {m_2} = 12$…(i)
and $\frac{{{m_1}}}{{{m_2}}} = \frac{1}{3}$…(ii)
by solving we get ${m_1} = 3kg$ and ${m_2} = 9kg$
Kinetic energy of smaller part = $\frac{1}{2}{m_1}v_1^2 = 216J$
$v_1^2 = \frac{{216 \times 2}}{3}$
==> ${v_1} = 12m/s$
So its momentum = ${m_1}{v_1} = 3 \times 12 = 36\;kg{\rm{ - }}{\rm{m}}/s$
As both parts possess same momentum therefore momentum of each part is $36\;kg{\rm{ - }}m/s$
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