A ring of radius R and mass 5M is initially at rest and free to move. A point mass M is initially on its axis at distance √3R from the ring centre O. Starting from rest, find the speed of M when it crosses O.

Question

G1 Physics · Accepted Answer

Step 1: Initially the point mass M is at z=√3R on the axis of a ring of mass 5M and radius R. Their gravitational potential energy is Uᵢ=-G(5M)(M)/√(R²+3R²)=-5GM²/(2R). At crossing O, U𝒇=-5GM²/R. Thus the gain in kinetic energy is 5GM²/(2R). Step 2: Initial total momentum is zero. If the point mass speed is v, ring speed is V=v/5 in the opposite direction. Total kinetic energy is (1/2)Mv²+(1/2)(5M)(v/5)²=(3/5)Mv². Step 3: (3/5)Mv²=5GM²/(2R) ⇒ v²=25GM/(6R). Hence v=√(25GM/6R). Final: A.