Two blocks of masses 3 kg and 6 kg rest on a horizontal frictionless surface. The 3 kg block is attached to a spring of force constant K = 900 N m⁻¹, compressed 2 m initially from equilibrium. When the 3 kg mass is released, it strikes the 6 kg mass and the two stick together.

Question

G1 Physics · Accepted Answer

Before collision, the 3 kg block has A=2 m and ω=√(900/3)=10√3 rad s⁻¹. At x=1 m, v=ω√(A²-x²)=10√3·√3=30 m s⁻¹. Perfectly inelastic collision with 6 kg gives v′=(3×30)/9=10 m s⁻¹, so Sub 1=A. After collision ω′=√(900/9)=10 rad s⁻¹ and A′=√[x²+(v′/ω′)²]=√(1+1)=√2 m, so Sub 2=C.