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The blocks A and B shown in figure have masses `M_A=5kg` and `M_B=4kg`. The system is released from rest. The speed of B after A has travelled a distance `1m` along the incline is
image
A. (a) `sqrt3/2sqrtg`
B. (b) `sqrt3/4sqrtg`
C. (c) `(sqrtg)/(2sqrt3)`
D. (d) `sqrtg/2`

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Correct Answer - C
If A moves down the incline by `1m`, B shall move up by `1/2m`.
If the speed of B is v, then the speed of A will be `2v`.
From conservation of energy,
Gain in KE=loss in PE
`1/2m_A(2v)^2+1/2m_Bv^2=m_Agxx3/5-m_Bgxx1/2`
Solving, we get
`v=1/2sqrt(g/3)`

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