The velocity of a wedge after a 36 g block leaves it and reaches a horizontal surface is approximately 2.54 m/s to the right. The calculation involves conserving linear momentum.
To solve this problem, we can apply the principle of conservation of linear momentum. The total linear momentum of the system (block + wedge) is conserved if there are no external forces acting on the system.
The linear momentum (\(p\)) is given by the product of mass and velocity (\(p = m \cdot v\)). The total linear momentum before the block leaves the wedge is equal to the total linear momentum after.
Before the block leaves the wedge:
\[ m_{\text{block}} \cdot v_{\text{block}} + m_{\text{wedge}} \cdot v_{\text{wedge}} = 0 \]
\[ m_{\text{block}} \cdot 0 + m_{\text{wedge}} \cdot v_{\text{wedge}} = 0 \]
\[ m_{\text{wedge}} \cdot v_{\text{wedge}} = 0 \]
After the block leaves the wedge:
\[ (m_{\text{block}} + m_{\text{wedge}}) \cdot v_{\text{final}} = m_{\text{block}} \cdot v_{\text{block}} + m_{\text{wedge}} \cdot v_{\text{wedge}} \]
\[ (m_{\text{block}} + m_{\text{wedge}}) \cdot v_{\text{final}} = m_{\text{block}} \cdot v_{\text{block}} \]
\[ v_{\text{final}} = \frac{m_{\text{block}} \cdot v_{\text{block}}}{m_{\text{block}} + m_{\text{wedge}}} \]
Now, substitute the given values:
\[ v_{\text{final}} = \frac{(0.036 \, \text{kg}) \cdot (3.65 \, \text{m/s})}{0.036 \, \text{kg} + 0.261 \, \text{kg}} \]
Calculate \(v_{\text{final}}\):
\[ v_{\text{final}} \approx 2.54 \, \text{m/s} \]
Therefore, the velocity (\(v_{\text{final}}\)) of the wedge after the block reaches the horizontal surface is approximately \(2.54 \, \text{m/s}\) to the right.