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 :

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 ∴ μmgs=p2/2m or  s=p2/2μgm2 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