Olympiad workbookNSEPLaws of Motion
Two equal blocks, each of mass M , hang on either side of a frictionless light pulley with a light string. A rider of mass m is placed on one of the blocks (as shown). When the system is released, the block with rider descends a distance H till the rider is caught by a ring that allows the block to pass through. The system moves a further distance D taking time t . In such a situation, the acceleration due to gravity
Options
- Ag= (2 M+m) D^2 2 m H t^2
- Bg= (M+m) D^2 2 m H t^2
- Cg= (2 M+m) D m H t^2
- Dg= (M+2 m) D^2 m H t^2
Correct answer
A. g= (2 M+m) D^2 2 m H t^2
Step-by-step solution
No Solution