Question

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

• A. $-1.32$
• B. $-1.00$
• C. $-1.11$
• D. $-1.50$

Hint:

Always break velocity components into directions defined by the geometry. For inclined planes, vertical components arise from the plane’s tilt and the projection of motion within the plane.

Answer 1

The Correct Option is C

The particle moves on a plane inclined at angle $\theta = 40^\circ$ to the $xy$-plane. The velocity vector has magnitude 2 m/s and makes an angle $\phi = 30^\circ$ with the dotted reference line that lies in the inclined plane.
The z-component of velocity is obtained by projecting the velocity onto the vertical direction. Since the motion is constrained to the inclined plane, the vertical contribution comes from the tilt of the plane:
\[ v_z = -v \cos(\phi)\sin(\theta) \] Substituting values:
\[ v_z = -2 \cos(30^\circ)\sin(40^\circ) \] \[ \cos(30^\circ) = \frac{\sqrt{3}}{2} \approx 0.866,\quad \sin(40^\circ) \approx 0.643 \] \[ v_z = -2(0.866)(0.643) \approx -1.11 \ \text{m/s} \] Thus, the z-component of velocity is \(-1.11\) m/s.