Exams › JEE Main › Physics
Two blocks of mass 3 kg and 6 kg lie on a smooth horizontal surface, joined by a light spring of force constant k = 200 N/m that is initially unstretched. The blocks are given velocities (as shown) such that they approach each other with a relative speed of 3 m/s along the line of the spring. Find the maximum extension (or compression) of the spring.
- 30 cm
- 10 cm
- 15*sqrt(2) cm
- 15 cm
Correct answer: 30 cm
Solution
At maximum extension the relative velocity is zero, so the kinetic energy of relative motion converts to spring PE: (1/2)*mu*v_rel² = (1/2)*k*x² with mu = 2 kg and v_rel = 3 m/s gives x = 0.30 m = 30 cm.
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