JEE Advanced2018PhysicsOscillationsActual
A spring - block system is resting on a frictionless floor as shown in the figure. The spring constant is 2.0 N m - 1 and the mass of the block is 2.0 k g . Ignore the mass of the spring. Initially the spring is in an unstretched condition. Another block of mass 1.0kg moving with a speed of 2.0 m s - 1 collides elastically with the first block. The collision is such that the 2.0kg block does not hit the wall. The dis
Correct answer
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Step-by-step solution
Given here: m = 2 . 0 kg and k = 2 . 0 N m - 1 . Time period of spring is given as T = 2 π m k = 2 π s e c Block returns to original position in T 2 = π s e c The speed of block after collision is v = 2 × 2 2 + 1 = 4 3 m / s and v ' = 2 2 + 1 = 2 3 m / s . The distance between the two blocks is d = 2 3 π = 2 3 3.14 = 2.0933 m d = 2.09 m