Question

Particle A moves along X-axis with a uniform velocity of magnitude 10 m/s. Particle B moves with uniform velocity 20 m/s along a direction making an angle of ${60}^o$ with the positive direction of X-axis as shown in the figure. The relative velocity of B with respect to that of A is

• A. 10 m/s along x-axis
• B. $10\sqrt{3}m/s$ along y-axis (perpendicular to X-axis)
• C. $\sqrt{5}m/s$ along the bisection of the velocities of A and B
• D. 30 m/s along negative X-axis

Answer 1

Answer: (B)

$10\sqrt{3}m/s$ along y-axis (perpendicular to X-axis)

The component of velocity of B along x and y-directions:
$V_{BX}=20cos{60}^o=20\times \frac{1}{2}=10m/s$

$V_{BY}=20\sin 60=20\frac{\sqrt{3}}{2}=20\sqrt{3}m/s$
And, $V_A=10\hat{i}$
$V_B=10\hat{i}+10\sqrt{3}\hat{j}$
$V_{BA}=V_B-V_A=10\hat{i}+10\sqrt{3}\hat{j}-10\hat{i}=10\sqrt{3}\hat{j}$