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AP EAMCET201923 Apr 2019Morning ShiftPhysicsCenter of Mass, Momentum and CollisionActual

In the figure shown, the blocks have equal masses. Friction, mass of the string and the mass of the pulley are negligible. The magnitude of the acceleration of the centre of mass of the two blocks is (Acceleration due to gravity (=g )).

Options

  1. A( ( 3 -1 2 ) g )
  2. B( g 2 )
  3. C(( 3 -1) g )
  4. D( ( 3 -1 4 2 ) g )

Correct answer

D. ( ( 3 -1 4 2 ) g )

Step-by-step solution

For a pulley and block system on a smooth double inclined plane as shown below Force equation for both the blocks, ( m g 30^ -T=m a ) ...(i) ( T-m g 60^ =m a ) ...(ii) From above equation we get, (a= ( 3 -1) 4 g ) ( ) Magnitude of the acceleration of centre of mass, ( a _ C M = | m a ₁+m a ₂ m+m | ) Here, ( a _ C M = | ( 3 -1 4 ) g i + ( 3 -1 4 ) g j 2 | ) ( aligned & = g 2 ( 3 -1 4 )^2+ ( 3 -1 4 )^2 & = g 2 2 3 -1 4 & = ( 3 -1 4 2 ) g aligned ) Hence, option (d) is correct.

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