Question

Particle A moves along X-axis with a uniform velocity of magnitude 10 m/s. Particle B moves with uniform velocity 20m/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. $10\sqrt{5}$ a long the bisection of the velocities of A and B
• D. 30 m/s along negative X-axis

Answer 1

The correct option is B $10\sqrt{3}$ m/s along Y-axis (perpendicular to X-axis)

$\overrightarrow{R}=\overrightarrow{A}+\overrightarrow{B}$

ACCORDING TO THE GIVEN PROBLEM, OBJECT B IS MOVING WITH VELCITY $V_B$ MAKING ANGLE WITH OBJECT A WHICH IS MOVING WITH VELCITY $V_A$ AND THE ANGLE BETWEEN THEM IS ${60}^0$,

NOW THE CONCEPT OF RELATIVE VELOCITY IMPLIES THAT THE VELOCITY OF B WITH RESPECT TO A IS GIVEN BY THE FOOLOWING :

$V_B^{\prime }=V_B-V_A$

AND ACCORDING TO PARLLELOGARM LAW OFVECTOR ADDITION :

$(V_B^{\prime }{)}^2=(V_A{)}^2+\left(V_B^2\right)-2V_A{V}_B\ast \cos \theta$

WHERE $\theta$ IS THE ANGLE BETWEEN THE VELOCITIES,

BY PUTTING THE VALUE $V_A=10,V_B=20AND\theta ={60}_0$ THEN,

$V_B^{\prime }=10\sqrt{3}m/s$