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Two identical particles A and B are attached to a string of length 2l A to middle and B to one of the ends. The string is whirled in a horizontal circle , with the end o fixed . If the kinetic energy of B relative to A is E , then the absolute kinetic energies of A and B are E and 2E E and 4 E E and 3E E and 5E

Two identical particles A and B are attached to a string of length 2l A to middle and B to one of the ends. The string is whirled in a horizontal circle , with  the end o fixed . If the kinetic energy of B relative to A is E , then the absolute kinetic energies of A and B are 
  1. E and 2E 
  2. E and 4 E
  3. E and 3E
  4. E and 5E

Grade:11

1 Answers

Arun
25757 Points
3 years ago
Since A and B are rotating with the same angular velocity,
The kinetic energies are
Va = wl and Vb = 2wl
 
E = ½ m (Vb2 – Va2) = E
 
hence Ea =E
 
and Eb = 4E
 
hence option 2 is correct.

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