BITSAT2015PhysicsLaws of MotionActual
The masses of blocks A and B are m and M respectively. Between A and B , there is a constant frictional force F and B can slide on a smooth horizontal surface. A is set in motion with velocity while B is at rest. What is the distance moved by A relative to B before they move with the same velocity?
Options
- AmMv ₀² ~F ( ~m - M )
- BmMv ₀² 2 ~F ( ~m - M )
- CmMv ₀² ~F ( ~m + M )
- DmMv ₀² 2 ~F ( M + m )
Correct answer
D. mMv ₀² 2 ~F ( M + m )
Step-by-step solution
For the blocks A and B FBD as shown below Equations of motion array l a _ A = F M ( in - x direction ) a _ B = F M ( in + x direction) array Relative acceleration, of A w.r.t. B, aligned a _ A , B &= a _ A - a _ B =- F m - F M &=- F ( M + m Mm )( along - x direction ) aligned Initial relative velocity of A w.r.t. B, u _ AB = v ₀ using equation v²=u²+2 as 0=v₀²- 2 ~F ( ~m + M ) S Mm S = Mmv ₀² 2 ~F ( ~m + M ) i.e., Distance moved by A relative to B S _ AB = Mmv ₀² 2 ~F ( ~m + M )