JEE Advanced2006PhysicsLaws of MotionActual
System shown in figure is in equilibrium and at rest. The spring and string are massless, now the string is cut. The acceleration of mass 2 m and m just after the string is cut will be
Options
- Ag / 2 upwards, g downwards
- Bg upwards, g / 2 downwards
- Cg upwards, 2 g downwards
- D2 g upwards, g downwards
Correct answer
A. g / 2 upwards, g downwards
Step-by-step solution
Initially under equilibrium of mass ' m ' T=m g Now, the string is cut. Therefore, T=m g force is decreased on mass m upwards and downwards on mass 2 m . a_m= m g m =g (downwards) and a_ 2 m = m g 2 m = g 2 (upwards)