A small block of mass $m_{1}$ is released from rest at the top of a curve-shaped, frictionless wedge of mass $m_{2}$ which sits on a frictionless horizontal surface as shown. When the block leaves the wedge its velocity is measured to be $4.00 m / s$ to the right as shown in the figure. If the mass of the block is doubled to become $2 m_{1}$ , what can be said about the speed with which it leaves the wedge?

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Its speed is less than $4.00 m / s$

Its speed is greater than $4.00 m / s$

Its speed is equal to $4.00 m / s$

Not enough information is given.

answer is A.

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Detailed Solution

$m_{1} g h = \frac{1}{2} μ {v^{2}}_{r e l} = \frac{1}{2} \frac{m_{1} m_{2}}{m_{1} + m_{2}} {v^{2}}_{r e l} v_{r e l} = \sqrt{2 g h (1 + \frac{m_{1}}{m_{2}})} v_{1} = v_{1 c} = \frac{μ}{m_{1}} v_{r e l} = \frac{m_{2}}{m_{1} + m_{2}} \times \sqrt{2 g h (\frac{m_{2} + m_{1}}{m_{2}})} = \sqrt{2 g h (\frac{m_{2}}{m_{1} + m_{2}})}$

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