Two uniform solid spheres of equal radii R, but mass M and 4M have a centre to centre separation 6R, as shown in the figure. The two spheres are held fixed. A projectile of mass m is projected with minimum velocity from the surface of the sphere of mass M directly towards the centre of the second sphere so that it reaches the surface of the second sphere. If the speed with which the projectile reaches the surface of the sphere of mass 4M is 3√[xGM/(5R)], find x.
Physics · Mixed Chapter
JEE Mains

Question

Two uniform solid spheres of equal radii R, but mass M and 4M have a centre to centre separation 6R, as shown in the figure. The two spheres are held fixed. A projectile of mass m is projected with minimum velocity from the surface of the sphere of mass M directly towards the centre of the second sphere so that it reaches the surface of the second sphere. If the speed with which the projectile reaches the surface of the sphere of mass 4M is 3√[xGM/(5R)], find x.

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1: Let the first sphere's centre be x=0 and the second centre x=6R. Between them U/m=−GM/x−4GM/(6R−x). Step 2: The potential-energy barrier occurs where dU/dx=0: 1/x²=4/(6R−x)², giving x=2R. There U at barrier/m=−3GM/(2R). Step 3: The projectile is launched with just enough energy to reach this barrier; its conserved total energy per unit mass is −3GM/(2R). Step 4: At the second surface x=5R, Uᶠ/m=−GM/(5R)−4GM/R=−21GM/(5R). Hence ½vᶠ²=(21/5−3/2)GM/R=27GM/(10R). Step 5: vᶠ=√[27GM/(5R)]=3√[3…