SATHEE: Kinematics Question 653

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 $α = \frac{77}{v^{2}} 25 π R$ then

Options:

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

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

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

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

Show Answer

Answer:

Correct Answer: B

Solution:

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

$∴ t = \frac{5 π R}{6v}$

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

Put t, $∴ angle covered by A= \frac{11 π}{6}$

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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 α=77v225πR then

Options:

Answer:

Solution:

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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 $α = \frac{77}{v^{2}} 25 π R$ then