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6. system of particles and rotational motion — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysics6. system of particles and rotational motion1 Mark
MCQ
A boy of mass $20\, kg$ is standing on a $80\, kg$ free to move long cart. There is negligible friction between cart and ground. Initially, the boy is standing $25\, m$ from a wall. If he walks $10\, m$ on the cart towards the wall, then the final distance of the boy from the wall will be ........ $m$
  • A
    $15$
  • B
    $12.5$
  • C
    $15.5$
  • ✓
    $17$
✓

Answer

Correct option: D.
$17$
d
$\begin{array}{l}
As\,there\,is\,no\,external\,Force,\,so\,\\
displacement\,oF\,center\,oF\,mass\,oF\\
the\,cart + boy\,system\,parallel\,to\,the\\
surFace\,is\,zero.\\
\Delta {x_{cm}} = \left( {{m_1}\Delta {x_1} + {m_1}\Delta {x_1}} \right)/\left( {{m_1} + {m_2}} \right)\\
0 = \left( {{m_1}\Delta {x_1} + {m_1}\Delta {x_1}} \right)/\left( {{m_1} + {m_2}} \right)\\
Let\,the\,body\,moves\,towards\,\,the\,wall\\
a\,{\rm{distance}}\,10\,m\,and\,the\,cart\,moves\,\\
away\,form\,the\,wall\,a\,{\rm{distance}}\,x\,so,
\end{array}$

$\begin{array}{l}
displacement\,of\,man\,wrt\,ground\,towards\\
the\,wall\,is\,\Delta {x_1} = 10 - x\\
displacement\,of\,cart\,wrt\,ground\,towards\\
the\,wall\,is\,\Delta {x_2} =  - x\\
{m_1}\Delta {x_1} + {m_1}\Delta {x_1} = 0\\
20\left( {10 - x} \right) + 80\left( { - x} \right) = 0\\
100x = 200\\
x = 2\,m\\
displacement\,of\,man\,wrt\,ground\,towards
\end{array}$

$\begin{array}{l}
the\,wall\,is\,\Delta {x_1} = 10 - 2 = 8m\\
{\bf{Hence}},\,{\bf{the}}\,{\bf{final}}\,{\bf{diatance}}\,{\bf{of}}\,{\bf{the}}\,{\bf{bov}}\\
{\bf{from}}\,{\bf{the}}\,{\bf{wall}}\,{\bf{will}}\,{\bf{be}}\,{\bf{X}} = {\bf{25}} - {\bf{8}} = {\bf{17}}\,{\bf{m}}
\end{array}$

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