WB JEE Medical WB JEE Medical Solved Paper-2015

  • question_answer 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)  \[10\sqrt{5}\,m/s\]along the bisection of the velocities of A and B

    D)  30 m/s along negative X-axis

    Correct Answer: B

    Solution :

     The component of velocity of B along x-direction \[{{v}_{Bx}}=20\cos {{60}^{o}}=20\times \frac{1}{2}=10\,m/s\] \[{{v}_{A}}=10\,\hat{i}\Rightarrow {{v}_{B}}=10\hat{i}+10\sqrt{3}j\] \[{{v}_{BA}}={{v}_{B}}-{{v}_{A}}=10\hat{i}+10\sqrt{3}\hat{j}-10\hat{i}=10\sqrt{3}\hat{j}\]


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