Block A of mass m is hanging from a vertical spring of constant k and is at rest. The value of m/k = 1/150 in SI units. Block B also of mass m strikes block A with velocity v and sticks to it. Then find the value of v in m/s for which the spring just attains natural length. Take g = 10 m s⁻².
Physics · Mixed Chapter
JEE Mains

Question

Block A of mass m is hanging from a vertical spring of constant k and is at rest. The value of m/k = 1/150 in SI units. Block B also of mass m strikes block A with velocity v and sticks to it. Then find the value of v in m/s for which the spring just attains natural length. Take g = 10 m s⁻².

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1: Initially the spring extension is x₀=mg/k. During the brief perfectly inelastic collision, momentum conservation gives the pair’s immediate upward speed v/2. Step 2: To just reach the natural length with zero speed, use energy from x=x₀ to x=0: ½(2m)(v/2)² + ½kx₀² = (2mg)x₀. Step 3: Substitute x₀=mg/k: mv²/4 + m²g²/(2k)=2m²g²/k. Thus v²=6mg²/k. Step 4: With m/k=1/150 and g=10, v²=4 and v=2 m s⁻¹.