Question

A 1kg mass is attached to a spring of force constant 600N/m and rests on a smooth horizontal surface with other end of the spring tied to a wall as shown in the figure. A second mass of 0.5kg slides on the surface and hits the first at 3m/s. If the masses make a perfectly inelastic collision, then find the amplitude of oscillation of the combined mass and time period of oscillation. 

• A. \(5\,\text{cm},\; \dfrac{\pi}{10}\,\text{s}\)
• B. \(5\,\text{cm},\; \dfrac{\pi}{5}\,\text{s}\)
• C. \(4\,\text{cm},\; \dfrac{2\pi}{5}\,\text{s}\)
• D. 4cm, (π)/(3)s

Hint:

After a perfectly inelastic collision: A=(v)/(ω), ω=√((k)/(m)) Use momentum conservation first, then SHM formulas.

Answer 1

The Correct Option is A

Step 1: Using conservation of momentum for perfectly inelastic collision: v = (m₂ u)/(m₁+m₂) = (0.5×3)/(1.5) = 1m/s
Step 2: Total mass after collision: M = 1 + 0.5 = 1.5kg
Step 3: Angular frequency of oscillation: ω = √((k)/(M)) = √((600)/(1.5)) = √(400) = 20rad/s
Step 4: Amplitude of oscillation: A = (v)/(ω) = (1)/(20) = 0.05m = 5cm
Step 5: Time period: T = (2π)/(ω) = (2π)/(20) = (π)/(10)s