Irodov Problem 1.380

a) For this problem let us denote v as the magnitude of the instantaneous velocity of the particle and as the unit vector along the instantaneous direction of motion of the particle. In other words the instantaneous velocity vector of the particle is given by .

From the fundamental laws of dynamics we can write,


















Similarly, we have,







In order for (1) and (2) be parallel there are only two possible ways

1. , in which case both vectors get trivially aligned. However, this also means that the particle never changes it direction. In other words, the force has to be aligned in the instantaneous direction of motion of the particle i.e. F is parallel to velocity vector.

2. , in which case the magnitude of the velocity of the particle never changes. This can only happen if the force acting on the particle always acts perpendicular to the instantaneous direction of its motion i.e F is perpendicular to the velocity vector.

b) The solution to this part is quite evident from Eqns (1), and (2) considering each of the two cases separately.

Case 1: F is parallel to velocity vector













Case 2 : F is perpendicular to velocity vector

Irodov Problem 1.379

This problem can be solved from a very simple inspection of solution to problem 1.378 and seeing that,





The constant offset is not important since the origin could always be displaced by a constant value without effecting the dynamics.

Irodov Problem 1.378

We can solve this using the fundamental equation of dynamics as follows:

Irodov Problem 1.377

In the frame of reference of the sphere

As seen from the frame of reference of the sphere, the sphere itself is stationary. The gas particles are however moving at a speed v towards it. Let us now consider the collision of the sphere with a single particle. As asked in the question consider a section of the sphere that is normal to the direction of velocity. Assuming that the sphere is extremely large compared to the gas particles, in case of an elastic collision, the gas particle will simply bounce off with the same velocity v in the opposite direction. The net change in momentum of the gas particle is given by,








As seen from the reference frame of the particles there are n particles per unit volume. However, the same unit volume with n particles as seen by an observer moving with the sphere is shrunk by a factor of since space itself would have shrunk by this factor in the direction of motion. In other words, the density of the gas a seen from the observer on the sphere would be





Since n' particles change their momentum each second, the net force exerted on the sphere as a reaction will be





Solution in reference frame of the particles
As seen from the view of the gas particles, they were initially stationary and the sphere was moving at a velocity v. Since as seen from the sphere's reference frame, the final velocity (velocity after collision) was 2v, their final velocity as seen by a stationary observer can be obtained by using the relativistic transformation as,





This means that the change in momentum from the point of view of a stationary observer is given by,










Since the number of particles colliding with the sphere per unit time is n, the net force exerted on the sphere is given by , which evaluates exactly the same as calculated from the point of view of an observer on the sphere!