A block of mass m is pushed towards a movable wedge of mass 2m and height h with initial speed u. All surfaces are smooth. What is the minimum value of u for which the block just reaches the top of the wedge?

sqrt(2*g*h)
2*sqrt(2*g*h)
sqrt(3*g*h)
sqrt(g*h)

Correct answer: sqrt(3*g*h)

Solution

At minimum u, block and wedge move together at the top. Momentum gives v = u/3. Energy equation: mu²/2 = 3m*(u/3)²/2 + mgh => u²/2 - u²/6 = gh => u²/3 = gh => u = sqrt(3gh).

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