Vector Algebra
Resolution of Vectors
Grade None
Question:
<p>A velocity 1/4 m/s is resolved into two components along <span>\(OA\)</span> and <span>\(OB\)</span> making angles 30° and 45°, respectively, with the given velocity. Then the component along <span>\(OB\)</span> is</p>
<p>\(\dfrac{1}{8}\) m/s</p>
<p>\(\dfrac{1}{4}(\sqrt{3}-1)\) m/s</p>
<p>\(\dfrac{1}{4}\) m/s</p>
<p>\(\dfrac{1}{8}(\sqrt{6}-\sqrt{2})\) m/s</p>
Step-by-Step Solution
Key Concept: Use the sine rule for vector resolution: when a velocity vector is resolved into two components along directions making known angles with the original vector, the components are proportional to the sines of the opposite angles in the velocity triangle.
Step 1: Let the velocity vector v = 1/4 m/s be resolved into components v_1 along OA and v_2 along OB. Step 2: The angle between v and OA is 30°, and between v and OB is 45°. The angle between OA and OB is 30° + 45° = 75°. Step 3: Using the sine rule for vector resolution: $\frac{v}{\sin(AOB)} = \frac{v_2}{\sin(30°)} = \frac{v_1}{\sin(45°)}$ Step 4: Applying the sine rule: $\frac{1/4}{\sin(75°)} = \frac{v_2}{\sin(30°)}$ Step 5: Calculate sin(75°) = sin(45° + 30°) = (√6 + √2)/4 Step 6: Therefore: $v_2 = \frac{(1/4) \times \sin(30°)}{\sin(75°)} = \frac{(1/4) \times (1/2)}{(\sqrt{6}+\sqrt{2})/4} = \frac{1}{\sqrt{6}+\sqrt{2}}$ Step 7: Rationalize: $v_2 = \frac{\sqrt{6}-\sqrt{2}}{4}$ m/s ∴ Answer: D
Correct Answer: D