In the figure, mₐ = mᵦ = 1 kg, block A is uncharged and is attached to a spring of constant 18 N m⁻¹; block B carries qᵦ = −1 C. Both are on a smooth horizontal surface in an electric field E = 10 N C⁻¹ to the right. B starts from rest 1.8 m to the right of A. The spring is initially unstretched and the collision is perfectly inelastic. If the subsequent amplitude is √(53x)/9 m, find x. [The source equation omits the printed parameter x; refer to PDF page 51.]
Physics · Mixed Chapter
JEE Mains

Question

In the figure, mₐ = mᵦ = 1 kg, block A is uncharged and is attached to a spring of constant 18 N m⁻¹; block B carries qᵦ = −1 C. Both are on a smooth horizontal surface in an electric field E = 10 N C⁻¹ to the right. B starts from rest 1.8 m to the right of A. The spring is initially unstretched and the collision is perfectly inelastic. If the subsequent amplitude is √(53x)/9 m, find x. [The source equation omits the printed parameter x; refer to PDF page 51.]

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Explanation: Block B accelerates toward A at 10 m/s² over 1.8 m, giving impact speed √(2×10×1.8)=6 m/s. After perfectly inelastic collision the combined mass is 2 kg and speed 3 m/s. The spring gives ω=√(18/2)=3 rad/s; the electric force shifts equilibrium by |xₑᵩ|=10/18=5/9 m. Amplitude A=√[(5/9)²+(3/3)²]=√106/9=√(53×2)/9, hence x=2. The question’s printed radical is partly incomplete.