A) \[\frac{\sqrt{19-4\sqrt{3}}}{2}\text{ m/s}\]
B) \[\frac{\sqrt{13}}{2}\text{ m/s}\]
C) \[\sqrt{3}\text{m/s}\]
D) \[\sqrt{7}\text{ m/s}\]
Correct Answer: D
Solution :
[d] Supposing block is fixed and wedge moves with velocity 3m/s relative to block, and so \[3\sin 30{}^\circ ={{v}_{y}}\cos 30{}^\circ ,\text{ }\therefore {{\text{v}}_{y}}=\sqrt{3}m/s.\] \[\text{and }{{u}_{x}}=\frac{1}{\sin 30{}^\circ }=2m/s\] \[\therefore u=\sqrt{u_{x}^{2}+u_{y}^{2}}=\sqrt{4+3}=\sqrt{7}m/s.\]You need to login to perform this action.
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