Irodov Problem 1.13

Figure shows an typical instantaneous state of the two particles during their motion as particle B is continuously aimed towards particle A. Let x be the horizontal distance between the two particles and y be the vertical distance between the two particles. From figure 2 we can write the motion of the particle B given by the following equations,






Now, integrating both side of (1a) from the t=0 to the time when A and B meet we obtain,

The integral is from 0 to 0 in the above equation because the horizontal distance between A and B initially is 0 and finally when they meet also it will again be 0. This can seen from the figure, from the definition of x - as the horizontal distance between A and B.

In the figure, we resolve all velocities along and perpendicular to the line connecting the two bodies.
If r is the distance between the two bodies then, as seen from the figure r decreases as at any given instant of time. Thus,




The same result can of course also be derived by transforming (1a,1b) to polar coordinates. Now if r is the vector connecting particles A and B then,








Using (4) can rewrite (1a,1b) as,










Integrating both side of (3), it would mean that,









Now we can use (2) to solve for from (5) as,

Irodov Problem 1.12

While a very mathematical and rigorous treatment of the problem is possible however, it can be solved by simply exploiting the inherent symmetry in the problem. The first step is to understand that because of the inherent symmetry in the positions and motions of the three particles, their positions at any given time will always form equilateral triangle as shown in the figure. Thus, the motion of particles will depict a shrinking equilateral triangle that is also rotating about the centroid as it shrinks. This is depicted in the figure.

The problem becomes quite simple if one solves it from the point of view of an observer sitting on one of the particles. To an observer sitting on one the particles, the rotational motion of the triangle will not be present (since the triangle is always an equilateral triangle at any given point of time) and he only perceive a shrinking equilateral triangle as shown in the second figure.

To an observer located on one of the particles the components of velocities of its adjacent particle at any given point of time are
depicted in the third figure. Thus, any given particles feel that
while it is moving towards its adjacent particle at a speed v, the adjacent particle is moving towards it at a speed v/2. Thus the particles seem to approach each other at a constant rate of v + v/2 = 3v/2. Since the initial distance was a, the time takes to meet is given by, a/(3v/2) = 2a/3v.

Irodov Problem 1.11

Let us refer to the particle that was shot with v1 as P1 and the one that was shot with v2 as P2. Since v1 and v2 are supposed to be in opposite directions we arbitrarily choose v1 as the positive directions and v2 as the negative direction. Further, we assume that gravity acting downwards is the negative direction.

Since, gravity acts vertically downwards, it does not effect the horizontal components of velocities of P1 and P2. However, both P1 and P2 accelerate at equal rates of g downwards from zero initial velocity component downwards.

The velocity of P1 after time t is given by and that of P2 is given by . When the two velocity vectors are perpendicular to each other, the dot product of the vectors will be null, i.e.








The vertical distance traveled by the P1 and P2 will be identical, thus at any time t the distance between the two particles is only a function of the horizontal components of their positions given by, (v1 + v2)t. Thus, the distance between the two particles when their velocity vectors are mutually perpendicular are given by,

Irodov Problem 1.10

The figure shows the components of the velocities of the two bodies along the horizontal and vertical components. The position of the vertically thrown body after time t will be where i and j are the unit vectors along the horizontal and vertical directions respectively. The position of the other body at a function of time will be . Now the distance between the two is simply the euclidean distance given by,