A $$1 \mathrm{~kg}$$ ball moving at $$12 \mathrm{~ms}^{-1}$$ collides with a $$2 \mathrm{~kg}$$ ball moving in opposite direction at $$24 \mathrm{~ms}^{-1}$$. If the coefficient of restitution is $$2 / 3$$, then their velocities after the collision are
A ball hits the floor and rebounds after an inelastic collision. In this case
In figure $$E$$ and $$v_{\mathrm{cm}}$$ represent the total energy and speed of centre of mass of an object of mass $$1 \mathrm{~kg}$$ in pure rolling. The object is
Two bodies of masses $$8 \mathrm{~kg}$$ are placed at the vertices $$A$$ and $$B$$ of an equilateral triangle $$A B C$$. A third body of mass $$2 \mathrm{~kg}$$ is placed at the centroid $$G$$ of the triangle. If $$A G=B G=C G=1 \mathrm{~m}$$, where should a fourth body of mass $$4 \mathrm{~kg}$$ be placed, so that the resultant force on the $$2 \mathrm{~kg}$$ body is zero?