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A block of mass m is placed on a triangu...

A block of mass `m` is placed on a triangular block of mass `M(M = 2m)` , as shown. All surfaces are smooth. Calculate the velocity of triangular block when the smaller block reaches at bottom.

Text Solution

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When the smaller block reaches at bottom
`V : v_(M//g)`
`v : v_(m//M)`
`v_(m//g)` in horizontal direction `= v cos 30^(@) - V`
`v_(m//g)` in vertical direction `= v sin 30^(@)`
By the momentum conservation in the horizontal direction
`m(v cos 30^(@) - V) = MV = 2 mV`
`3 V = v cos 30^(@) = (sqrt(3) v)/(2) rArr v = 2 sqrt(3) V`
By the energy conservation
`mgH = (1)/(2) m(v cos 30^(@) - V)^(2)`
`+ (1)/(2)m ( v sin 30^(@))^(2) + (1)/(2) MV^(2)`
`= (1)/(2) m( 2V)^(2) + (1)/(2) m ( 2 V sqrt(3) xx (1)/(2))^(2) + (1)/(2) xx 2mV^(2)`
`= 2mV^(2) + (3 mV^(2))/(2) + mV^(2) = (9)/(2) mV^(2)`
`V = sqrt((2 gH)/(9)) = sqrt(2 gH)/(3)`
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Knowledge Check

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