A) \[\frac{{{S}_{1}}}{{{S}_{2}}}=\frac{{{m}_{2}}}{{{m}_{2}}}\]
B) \[\frac{{{S}_{1}}}{{{S}_{2}}}=\frac{{{m}_{1}}}{{{m}_{2}}}\]
C) \[\frac{{{S}_{1}}}{{{S}_{2}}}={{\left( \frac{{{m}_{2}}}{{{m}_{1}}} \right)}^{2}}\]
D) \[\frac{{{S}_{1}}}{{{S}_{2}}}={{\left( \frac{{{m}_{1}}}{{{m}_{2}}} \right)}^{2}}\]
Correct Answer: C
Solution :
[c] Minimum stopping distance =s Work done against the friction \[=W=\mu mgs\] Initial momentum gained by both toy carts will be same because same force acts for same time initial kinetic energy of the toy cart =\[\left( \frac{{{p}^{2}}}{2m} \right)\] Therefore, \[\mu mgs=\frac{{{p}^{2}}}{2m}\]or \[s=\left( \frac{{{p}^{2}}}{2\mu g{{m}^{2}}} \right)\] For the two toy carts, momentum is numerically the same. Further \[\mu \]and g are the same for the toy carts. So, \[\frac{{{S}_{1}}}{{{S}_{2}}}={{\left( \frac{{{m}_{2}}}{{{m}_{1}}} \right)}^{2}}\]
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