A) \[\frac{{{p}^{2}}}{2\mu {{m}^{2}}g}\]
B) \[\frac{p}{2\mu {{m}^{2}}g}\]
C) \[\frac{{{p}^{2}}}{2\mu mg}\]
D) \[\frac{p}{2\mu mg}\]
Correct Answer: A
Solution :
Key Idea: While stopping, all of the kinetic energy is dissipated as frictional energy, When a body of mass \[m\]is moving with velocity\[v\], it has kinetic energy \[K=\frac{1}{2}m{{v}^{2}}\] Also frictional force exists between the tyres and road. When car stops Kinetic energy = Frictional energy \[\frac{1}{2}m{{v}^{2}}=\mu \,\,mg\times s\] Where \[\mu \] is coefficient of friction and \[s\] is stopping distance \[\therefore \] \[s=\frac{m{{v}^{2}}}{2\mu \,\,mg}\] Multiplying and dividing RHS by\[m\], we get \[s=\frac{{{m}^{2}}{{v}^{2}}}{2\mu \,\,{{m}^{2}}g}\] Also, momentum \[p\]=mass\[\times \]velocity\[=mv\] \[\therefore \] \[s=\frac{{{p}^{2}}}{2\mu \,\,,{{m}^{2}}g}\]
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