Kinematics Question 653

Question: Two particles A and B separated by a distance 2R are moving counter clockw it’se along the same circular path of radius R each with uniform speed v. At time $ t=0 $ , a is given a tangential acceleration of magnitude $ \alpha =\frac{\text{77}}{{v^{2}}}{25\pi \text{R}} $ then

Options:

A) the time lapse for the two bodies to collide it’s $ \frac{6\pi \text{R}}{5\text{v}} $

B) the angle covered by a is 11$ \pi $ /6

C) angular velocity of a is $ \frac{11\text{v}}{5\text{R}} $

D) radial acceleration of a is $ \text{289 }{{\text{v}}^{\text{2}}}\text{/5R} $

Show Answer

Answer:

Correct Answer: B

Solution:

[b] As when they collide $ vt+\frac{1}{2}( \frac{77v^{2}}{25\pi R} )t^{2}-\pi R=vt $

$ \therefore \text{t}=\frac{\text{5}\pi \text{R}}{\text{6v}} $

Now, angle covered by $ \text{A}=\pi +\frac{\text{vt}}{\text{R}} $

Put t, $ \therefore \text{ angle covered by A=}\frac{11\pi }{6} $



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