As per the given figure, two blocks each of mass 250   g are connected to a spring of spring constant 2…

As per the given figure, two blocks each of mass 250 g are connected to a spring of spring constant 2 N m-1. If both are given velocity v in opposite directions, then maximum elongation of the spring is

  1. v22
  2. v2
  3. v4
  4. v2

Solution

Let the maximum elongation of the spring be x.

Using energy conservation

Loss in kinetic energy of both blocks=Gain in spring energy

12mv2×2=12kx2

0.25v2=12×2×x2

14v2=12×2×x2

x=v2

Asked in: JEE Main 2022 (26 Jul Shift 1)

Practice more Work Power Energy questions on Aicharya