Discussion
A particle is constrained to move at a constant speed on an inclined plane (ABCD) along the curved path shown in the figure. Edges AD and BC are parallel to the \(y\) axis. The inclined plane makes an angle \(\theta\) with the \(xy\)-plane. The velocity vector of the particle makes an angle \(\phi\) with the dotted line which is parallel to edge \(AB\). If the speed of the particle is 2 m/s, \(\phi = 30^{\circ}\), and \(\theta = 40^{\circ}\), then the \(z\)-component of the velocity of the particle in m/s is __________
Answer: C
The z-component is \(-1.11\) m/s.
The velocity lies in the plane. Split it along the dotted line parallel to AB, which is the line of steepest slope at angle \(\theta\) to the xy-plane, and along the y direction, which is horizontal. Only the AB component has a vertical part, and the particle moves down the slope:
\[ \begin{aligned} v_{z} &= -v\cos\phi\sin\theta \\ &= -2 \times \cos 30^{\circ} \times \sin 40^{\circ} \\ &= -2 \times 0.866 \times 0.643 = -1.11 \end{aligned} \]
\(-1.32\) uses \(\cos 40^{\circ}\) in place of \(\sin 40^{\circ}\), \(-1.00\) is \(-v\sin\phi\), which ignores the slope altogether, and \(-1.50\) does not follow from these angles.
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