InterviewSolution
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InterviewSolution
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Two bodies A and B of masses m and 2mrespectively are placed on a smooth floor. They are connected by a light spring of stiffness k.A third body C of mass m moves with velocity v_(0) along the line joining A and B and collides elastically with A. If l_(o) be the natural length of the spring then find the minimum separation between the blocks. |
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Answer» Solution :Initially there will be collision between C and A which is elastic, therefore by using conservation of momentum we obtain, `mv_(0)=mv_(A)+mv_(c ), "" v_(0)=v_(A)+v_(C )` Since `e=1, v_(0)=v_(A)-v_(C )` Solving the above two EQUATIONS, `v_(A)=v_(0)` and `v_(c )=0`. Now A will move and compress the spring which in turn accelerate B and retard A and finally.both A and B will move with same VELOCITY v. (a) Since net EXTERNAL force is zero, therefore momentum of the system (A and B) is conserved. `rArr v=v_(0)//3` (b)If `x_(0)` is the maximum compression, then using energy conservation `(1)/(2)mv_(0)^(2)=(1)/(2)(m+2m)v^(2)+(1)/(2)kx_(0)^(2)` `rArr (1)/(2)mv_(0)^(2)=(1)/(2)(3m)(v_(0)^(2))/(9)+(1)/(2)kx_(0)^(2)` `rArr x_(0)=v_(0)sqrt((2m)/(3k))` Hence minimum distance `D=l_(0)=x_(0)==l_(0)-v_(0)sqrt((2m)/(3k))` |
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