A) 32m
B) 24m
C) 16m
D) 12m
Correct Answer: C
Solution :
Given, \[{{m}_{1}}=200kg,\] \[{{m}_{2}}=300kg,\] \[{{s}_{1}}=36m\] Using law of conservation of momentum, \[{{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}}=0\] or \[{{m}_{1}}{{v}_{1}}=-{{m}_{2}}{{v}_{2}}\] \[\frac{{{m}_{1}}}{{{m}_{2}}}=-\frac{{{v}^{2}}}{{{v}_{1}}}\] ?.(i) Kinetic energy of cart =work done against friction force. For first cart, \[\frac{1}{2}{{m}_{1}}v_{1}^{2}={{f}_{s}}\times {{s}_{1}}\] \[\frac{1}{2}{{m}_{1}}v_{1}^{2}=\mu {{m}_{1}}g\times {{s}_{1}}\] ...(ii) For second car, \[\frac{1}{2}{{m}_{2}}v_{2}^{2}={{f}_{s}}\times {{s}_{2}}=\mu {{m}_{2}}g\times {{s}_{2}}\] ?.(iii) \[\therefore \] \[\frac{\frac{1}{2}{{m}_{1}}v_{1}^{2}}{\frac{1}{2}{{m}_{2}}v_{2}^{2}}=\frac{\mu {{m}_{1}}g\times {{s}_{1}}}{\mu {{m}_{2}}f\times {{s}_{2}}}\] \[\therefore \] \[\frac{{{s}_{1}}}{{{s}_{2}}}=\frac{v_{1}^{2}}{v_{2}^{2}}\] Using Eq. (i), \[\frac{{{s}_{1}}}{{{s}_{2}}}=\frac{m_{2}^{2}}{m_{1}^{2}}={{\left( \frac{300}{200} \right)}^{2}}=\frac{9}{4}\] \[\frac{36}{{{s}_{2}}}=\frac{9}{4}\] \[{{s}_{2}}=16m\]
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