After the hanging block hits the table, what is the distance s along the surface of the plane which the top block continues to slide before stopping? (Assume that the plane is long enough that the blo

Question

After the hanging block hits the table, what is the distance s along the surface of the plane which the top block continues to slide before stopping? (Assume that the plane is long enough that the block does not fall off)

TestHub · Accepted Answer

When the lower block hits the table, the string becomes slack, and the tension in the string becomes zero. So, the acceleration of the block on the wedge becomes:

$a = g sin θ - μg cos θ = g \times \frac{5}{13} - \frac{1}{2} \times g \times \frac{12}{13} = - \frac{g}{13}$

Let $u = v_{0}$ and $v = 0$ (finally at rest). Using the equation $v^{2} - u^{2} = 2 a s$ :

$0^{2} - v_{0}^{2} = 2 \times - \frac{g}{13} \times s$

$\Rightarrow s = \frac{13 v _{0}^{2}}{2 g}$

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