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5.Work, Energy, Power and Collision
medium
$12 kg$ નો સ્થિર બોમ્બ ફૂટતાં $1:3$ દળના $2$ ટુકડા થાય છે.નાના ટુકડાની ગતિઊર્જા $216 J$ હોય,તો મોટા ટુકડાનું વેગમાન કેટલા ............ $ kg-m/sec$ થશે?
A
$36$
B
$72 $
C
$108 $
D
$216 $
Solution
(a)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 $36\;kg{\rm{ – }}m/s$
Standard 11
Physics
