A) \[\frac{{{S}_{1}}}{{{S}_{2}}}=\frac{{{m}_{2}}}{{{m}_{1}}}\]
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) \[\frac{{{S}_{1}}}{{{S}_{2}}}=-{{\left( \frac{{{m}_{2}}}{{{m}_{1}}} \right)}^{2}}\]
Correct Answer: D
Solution :
Initially both the toy carts are at rest. So the initial momentum of the system is zero. If \[{{u}_{1}}\] and \[{{u}_{2}}\]be the final velocities of both the carts after the spring is released, then by the principle of conservation of momentum, we have \[{{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}=0\]or, considering only the magnitude, we have\[u\propto \frac{1}{m}\] ? (i) Now, each cart after being released travels a certain distance before friction reduces its velocity to zero. Thus in the equation\[,\]\[{{v}^{2}}={{u}^{2}}-2as,\,\,v=0\]. Again frictional retardation in both is same\[(\mu g)\]. \[\therefore \] \[s\propto {{u}^{2}}\] From equation (i), we have \[s\propto {{u}^{2}}\propto \frac{1}{m}\] or, \[\frac{{{s}_{1}}}{{{s}_{2}}}={{\left( \frac{{{m}_{1}}}{{{m}_{2}}} \right)}^{2}}\] Taking into consideration the signs of the displacement as well, the correct option is [d].
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