Kinematics Question 653
Two particles A and B separated by a distance 2R are moving counter clockwise 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{77}{v^{2}} \cdot \frac{25\pi}{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 ring is $ \frac{11\text{v}}{5\text{R}} $
D) radial acceleration of a is $ \text{289 }{{\text{v}}^{\text{2}}}\text{/5R} $
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Answer:
Correct Answer: B
Solution:
As when they collide $ vt+\frac{1}{2}( \frac{77v^{2}}{25\pi R} )t^{2}-\pi R^{2}=vt $
$ \therefore \text{t}=\frac{\text{5}\pi \text{R}}{\text{6v}} $
Now, angle covered by $ \theta=\pi +\frac{\text{vt}}{\text{R}} $
Put t, $ \therefore \text{ angle covered by A} = \frac{11\pi}{6} $
BETA
