A) \[\frac{{{p}^{2}}}{2\mu g}\]
B) \[\frac{{{p}^{2}}}{2m\mu g}\]
C) \[\frac{{{p}^{2}}}{2{{m}^{2}}\mu g}\]
D) \[\frac{{{p}^{2}}}{2mg}\]
Correct Answer: C
Solution :
If the car stops after covering distance\[s,\]then \[F\times s=\frac{{{p}^{2}}}{2m}\] where \[F=\]friction Now \[F=\mu mg\] So, \[\mu mg\times s=\frac{{{p}^{2}}}{2m}\] \[\Rightarrow \] \[s=\frac{{{p}^{2}}}{2\mu {{m}^{2}}g}\]
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