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1. Suppose that the position of one particle at time t is given by x1= 3 sint yı = 2 cost 0<1<21 and the position of a second
0 0
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Answer #1

Here, we are given the position of the particles, Particle I and Particle II.

We first of all eliminate the parameter t

2 0 Posetion of one particle at time t: = 3 sint, y, - 2 costi Ost <2t here, h - sinti sint; y, cost by tu = sent tcost =) si2 0 Posetion of one particle at time t: = 3 sint, y, - 2 costi Ost <2t here, h - sinti sint; y, cost by tu = sent tcost =) si

(a)

After elimination, we get to know that the particle I is moving on elliptical path and particle II is moving on circular path as shown in the figure:

@graph of the path of the parterly, Yz bartice I 2 partile I А ( R 22 . ti -31B -2

Here, t is the angle subtended by the particle.

For partiilio tis the angle, the parterle makes with the positive Y-asust For partuile t is the angle, the partite makes with

From the figure, it is clear that there are two point of intersection of particles I and II e.g, Points A and B

(b)

Coordinates of the posetion Partule IT Particle time TV2 ( ر2-) (2ر3) (ا را- (5ر3 ) ( ر2 - (2 ره) هر 3) (2 -,ه) رہ ر3) TC 31/Coordinates of the posetion Partule IT Particle time TV2 ( ر2-) (2ر3) (ا را- (5ر3 ) ( ر2 - (2 ره) هر 3) (2 -,ه) رہ ر3) TC 31/

From the table and figure, there is only one collision point of the two particles, i.e, there is only one time, t= 3π/2 , such that the particles are at the same place.

The collision point is B(-3,0)

(c)

If path of the second particle is changed, the two particles never collide at any point of time as shown in the table and figure below:

If the path of the second particle is: x2 = 3+Lost : Y 2 = 1 tsent; ost<em => (42-3)? +42-1)=1 is a urcle. Nowo 2 paretice I

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