A spring-block system is resting on a frictionless floor as shown in the figure. The spring constant is 2.0Nm−1 and the mass of the block is 2.0 kg. Ignore the mass of the spring. Initially the spring is in an unstretched condition. Another block of mass 1.0 kg moving with a speed of 2.0ms−1 collides elastically with the first block. The collision is such that the 2.0kg block does not hit the wall. The distance, in metres, between the two blocks when the spring returns to its unstretched position for the first time after the collision is ____.
Answer: 2.09
Solution
For elastic collision: using conservation of momentum and energy, v1=−2/3 m/s and v2=4/3 m/s. The time for the spring to return to equilibrium is T/2=πm/k=π s. The separation is ∣v1∣t+∣v2∣t=(2/3+4/3)π=2π≈2.09 m.
JEE Advanced · 2018 · 20 May 2018 · Paper 1 · Numerical answer
