A spring is compressed between two toy carts of masses $$m1$$ and $$m2$$. When the toy carts are released the spring exerts on each toy cart equal and opposite forces for the same time t. If the coefficients of friction μ between the ground and the toy carts are equal, then the displacements of the toy carts are in the ratio :

Question

A spring is compressed between two toy carts of masses $$m1$$ and $$m2$$. When the toy carts are released the spring exerts on each toy cart equal and opposite forces for the same time t. If the coefficients of friction μ between the ground and the toy carts are equal, then the displacements of the toy carts are in the ratio :

Infinity Learn · Accepted Answer

Let the minimum stopping distance be s. The force of friction would be p m g. Hence work done against $$frictionW=$$μmgs  Initial K.E. of toy cart $$=p2/2m$$ ∴ μ$$mgs=p2/2m$$ or  $$s=p2/2$$μ$$gm2$$ so s∝$$1/m2$$ [.' p and μ are same for two toy carts. Further, the momentum is numerically $$same)$$ $$s1s2=m2m12$$  As $$s1$$ and $$s2$$ are in opposite directions, so  $$s1s2=$$−$$m2m12$$

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