An object is projected at an initial velocity of  at angle of  with the horizontal surface. At the highest point of the path of the projectile, the object breaks into two parts of mass ratio 1:3. The lighter part comes to momentary rest just after the break. Find the separation between the points of landing of lighter and heavier parts.  

Just after the impact, the heavier part must have horizontal momentum. Hence, both the parts must have same time of flight and reach the horizontal surface simultaneously. As the parts land, centre of mass follows the origin parabolic path.
C.M. must divide the line joining in the inverse ratio of masses.
$mR/2=-3mx$ 
$\therefore x=-R/6=-\left(\frac{1}{6}\right)\cdot \left[\frac{{10}^2\times \sin \left(2\times {45}^{\circ }\right)}{10}\right]=-5/3 m$ 
  Separation between the parts $=\frac{R}{2}+\left|x\right|=\frac{10}{2}+\frac{5}{3}=\frac{20}{3}m$