Question

Two particles a and b separated by a distance 2r are moving counter clock wise along the same circular path of radius r each with uniform speed v.At time t=0 a is given a acceleration of magnitude a =72 v2/25rby

Answer 1

Answer:

The answer will be   π + 5√π/6

Explanation:

According to the problem the distance between the particles is 2r and the radius of the circular path is r where each of the particle is moving with a speed of v

Now a started moving and at t = 0 a gets the acceleration .

Now b will also starts . Therefore the place at where they will collide a needs to cover that,

Now for a at time t,

s = ut +1/2at^2

   = vt +1/2 x 72v^2/25r x t^2

Now b has cover πr distance , therefore

vt +1/2 x 72v^2/25r x t^2 = vt +πr

=> 36 v^2t^2/25= πr^2

=>  t^2 = πr^2 x 25/36 v^2

=> t = 5√πr/6v

Now the angle covered by a is π extra than b

now the angle of b = vt/r

therefore angle covered by a is = π + vt/r = π + v/r x 5√πr/6v =  π + 5√π/6

Answer 2

Given that,

First particle = a

Second particle = b

Distance = 2r

Radius of circular path = r

Uniform speed = v

Acceleration $a=\dfrac{72v^2}{25 \pi r}$

Suppose , we need to find (a). The time lapse two particle to collide.

(b). The angle covered by a

The particle B covers the distance is πr.

We know that,

Distance :

The distance is equal to the product of velocity and time.

$s = vt$

We need to calculate the time lapse

Using equation of motion

$s=ut+\dfrac{1}{2}at^2$

Put the value of a, u and s

$vt+\pi r=vt+\dfrac{1}{2}\times\dfrac{72v^2}{25\pi r}t^2$

$\pi r=\dfrac{36v^2}{25\pi r}t^2$

$t^2=\dfrac{25\pi^2r^2}{36v^2}$

$t=\dfrac{5\pi r}{6v}$

(b). Angle covered by a is equal to the angle covered by b plus π.

The angular velocity of b is constant.

(b). We need to calculate the angle covered by a

Using formula for covered angle

$\theta=\dfrac{s}{r}$

$\theta=\dfrac{vt}{R}+\pi$

Put the value of t

$\theta=\dfrac{v}{r}\times\dfrac{5\pi r}{6v}+\pi$

$\theta=\dfrac{11\pi}{6}$

Hence, The time lapse two particle to collide is $\dfrac{5\pi r}{6v}$

The angle covered by a is $\dfrac{11\pi}{6}$