NEET  >  Irodov Solutions: Relativistic Mechanics - 2

# Irodov Solutions: Relativistic Mechanics - 2 - Notes | Study Physics Class 11 - NEET

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Q. 356. At two points of the reference frame K two events occurred separated by a time interval Δt. Demonstrate that if these events obey the cause-and-effect relationship in the frame K (e.g. a shot fired and a bullet hitting a target), they obey that relationship in any other inertial reference frame K'.

Solution. 356. We can take the coordinates of the two events to be

For B to be the effect and A to be cause we must have

In the moving frame the coordinates of A and B become

Since

Q. 357. The space-time diagram of Fig. 1.93 shows three events A, B, and C which occurred on the x axis of some inertial reference frame. Find:
(a) the time interval between the events A and B in the reference frame where the two events occurred at the same point;
(b) the distance between the points at which the events A and C occurred in the reference frame where these two events are simultaneous.

Solution. 357. (a) The four-dimensional interval between A and B (assuming Δy = Δz = 0) is :

52 - 32 bs 16 units

Therefore the time interval between these two events in the reference frame in which the events occurred at the same place is

(b) The four dimensional interval between A and C is (assuming Δy = Δz = 0)

32 - 52 = - 16

So the distance between the two events in the fram in which they are simultaneous is 4 units = 4m.

Q. 358. The velocity components of a particle moving in the xy plane of the reference frame K are equal to vx  and vy. Find the velocity v' of this particle in the frame K' which moves with the velocity V relative to the frame K in the positive direction of its x axis.

Solution. 358. By the velocity addition formula

and

Q. 359. Two particles move toward each other with velocities v1 = 0.50c and v2  = 0.75c relative to a laboratory frame of reference.
Find:
(a) the approach velocity of the particles in the laboratory frame of reference;
(b) their relative velocity.

Solution. 359. (a) By definition the velocity of apporach is

in the reference frame K .

(b) The relative velocity is obtained by the transformation law

Q. 360. Two rods having the same proper length l0  move lengthwise toward each other parallel to a common axis with the same velocity v relative to the laboratory frame of reference. What is the length of each rod in the reference frame fixed to the other rod?

Solution. 360. The velocity of one of the rods in the reference frame fixed to the other rod is

The length of the moving rod in this frame is

Q. 361. Two relativistic particles move at right angles to each other in a laboratory frame of reference, one 'with the velocity v1 and the other with the velocity v2. Find their relative velocity.

Solution. 361. The approach velocity is defined by

in the laboratory frame.

On the other hand, the relative velocity can be obtained by using the velocity addition formula and has the components

Q. 362. An unstable particle moves in the reference frame K' along its y' axis with a velocity v'. In its turn, the frame K' moves relative to the frame K in the positive direction of its x axis with a velocity V. The x' and x axes of the two reference frames coincide, the y' and y axes are parallel. Find the distance which the particle traverses in the frame K, if its proper lifetime is equal to Δt0

Solution. 362. The components of the velocity of the unstable particle in the frame K are

so the velocity relative to K is

The life time in this frame dilates to

and the distance traversed is

Q. 363. A particle moves in the frame K with a velocity v at an angle θ to the x axis. Find the corresponding angle in the frame K' moving with a velocity V relative to the frame K in the positive direction of its x axis, if the x and x' axes of the two frames coincide.

Solution. 363. In the frame K' the components of the velocity of the particle are

Q. 364. The rod AB oriented parallel to the x' axis of the reference frame K' moves in this frame with a velocity v' along its y' axis. In its turn, the frame K' moves with a velocity V relative to the frame K as shown in Fig. 1.94. Find the angle θ between the rod and the x axis in the frame K.

Solution. 364. In K' the coordinates of A and B are

After performing Lorentz transformation to the frame K we get

By translating   we can write

the coordinates of B as B : t = γ t'

Thus

Hence

Q. 365. The frame K' moves with a constant velocity V relative to the frame K. Find the acceleration w' of a particle in the frame K', if in the frame K this particle moves with a velocity v and acceleration w along a straight line
(a) in the direction of the vector V;
(b) perpendicular to the vector V.

Solution. 365.

In K the velocities at time t and t + dt are respectively v and v + wdt along x - axis which is parallel to the vector  In the frame K' moving with velocity  with respect to K, the velocities are respectively,

The latter velocity is written as

Also by Lorentz transformation

Thus the acceleration in the K' frame is

(b) In the K frame the velocities of the particle at the time t and t + di are repectively

where    is along jt-axis. In the K frame the velocities are

and

Thus the acceleration

Q. 366. An imaginary space rocket launched from the Earth moves with an acceleration w' = 10g which is the same in every instantaneous co-moving inertial reference frame. The boost stage lasted τ =1.0 year of terrestrial time. Find how much (in per cent) does the rocket velocity differ from the velocity of light at the end of the boost stage. What distance does the rocket cover by that moment?

Solution. 366. In the instantaneous rest frame v = V and

So,

w' is constant by assumption. Thus integration gives

Integrating once again

Q. 367. From the conditions of the foregoing problem determine the boost time τ0 in the reference frame fixed to the rocket. Remember that this time is defined by the formula where dt is the time in the geocentric reference frame.

Solution. 367. The boost time τ0 in the reference frame fixed to the rocket is related to the time τ elapsed on the earth by

Q. 368. How many times does the relativistic mass of a particle whose velocity differs from the velocity of light by 0.010% exceed its rest mass?

Solution. 368.

For

Q. 369. The density of a stationary body is equal to p0. Find the velocity (relative to the body) of the reference frame in which the density of the body is η = 25% greater than p0

Solution. 369. We define the density p in the frame K in such a way that p dx dy dz is the rest mass dm0 of the element That is p dx dy dz = p0 dx0 dy0 dz0, where p0 is the proper density dx0, dy0 , dz0 are the dimensions of the element in the rest frame K0. Now

if the frame K is moving with velocity, v relative to the frame K0. Thus

Defining

We get

or

Q. 370. A proton moves with a momentum p = 10.0 GeV/c, where c is the velocity of light. How much (in per cent) does the proton velocity differ from the velocity of light?

Solution. 370. We have

or

or

Q.371. Find the velocity at which the relativistic momentum of a particle exceeds its Newtonian momentum η = 2 times.

Solution. 371. By definition of η,

or

Q.372. What work has to be performed in order to increase the velocity of a particle of rest mass mo  from 0.60 c to 0.80 c? Compare the result obtained with the value calculated from the classical formula.

Solution. 372. The work done is equal to change in kinetic energy which is different in the two cases Classically i.e. in nonrelativistic mechanics, the change in kinetic energy is

Q.373. Find the velocity at which the kinetic energy of a particle equals its rest energy.

Solution. 373.

or

or

Q.374.  At what values of the ratio of the kinetic energy to rest energy can the velocity of a particle be calculated from the classical formula with the relative error less than ε = 0.010?

Solution. 374. Relativistically

So

Thus

But Classically,

Hence if

the velocity β is given by the classical formula with an error less than ε.

Q.375. Find how the momentum of a particle of rest mass m0 depends on its kinetic energy. Calculate the momentum of a proton whose kinetic energy equals 500 MeV.

Solution. 375. From the formula

we find

or

Q.376. A beam of relativistic particles with kinetic energy T strikes against an absorbing target. The beam current equals I, the charge and rest mass of each particle are equal to e and m0 respectively. Find the pressure developed by the beam on the target surface, and the power liberated there.

Solution. 376. Let the total force exerted by the beam on the target surface be .F and the power liberated there be P. Then, using the result of the previous problem we see

since I = Ne, N being the number of particles striking the target per second. Also,

These will be, respectively, equal to the pressure and power developed per unit area of the target if I is current density.

Q.377. A sphere moves with a relativistic velocity v through a gas whose unit volume contains n slowly moving particles, each of mass m. Find the pressure p exerted by the gas on a spherical surface element perpendicular to the velocity of the sphere, provided that the particles scatter elastically. Show that the pressure is the same both in the reference frame fixed to the sphere and in the reference frame fixed to the gas.

Solution. 377. In the tome fixed to the sphere The momentum transferred to the eastically scattered particle is

The density of the moving element is, from 1.369,

and the momentum transferred per unit time per unit area is

In the frame fixed to the gas When the sphere hits a stationary particle, the latter recoils with a velocity

The momentum transferred is

and the pressure

Q.378. A particle of rest mass mo  starts moving at a moment t = 0 due to a constant force F. Find the time dependence of the particle's velocity and of the distance covered.

Solution. 378. The equation of motion is

Integrating

or using r = 0 at r - 0, we get,

Q.379. A particle of rest mass m0 moves along the x axis of the frame K in accordance with the law  where a is a constant, c is the velocity of light, and t is time. Find the force acting on the particle in this reference frame.

Solution. 379.

Q.380. Proceeding from the fundamental equation of relativistic dynamics, find:
(a) under what circumstances the acceleration of a particle coincides in direction with the force F acting on it;
(b) the proportionality factors relating the force F and the acceleration w in the cases when F ⊥ and F II v, where v is the velocity of the particle.

Solution. 380.

Thus

Q.381. A relativistic particle with momentum p and total energy E moves along the x axis of the frame K. Demonstrate that in the frame K' moving with a constant velocity V relative to the frame K in the positive direction of its axis x the momentum and the total energy of the given particle are defined by the formulas:

Solution. 381. By definition,

where   i.s the invariant interval (dy = dz - 0)

Thus,

Q.382. The photon energy in the frame K is equal to ε. Making use of the transformation formulas cited in the foregoing problem, find the energy ε' of this photon in the frame K' moving with a velocity V relative to the frame K in the photon's motion direction. At what value of V is the energy of the photon equal to ε' = ε/2?

Solution. 382. For a photon moving in the x direction

In the moving frame,

Note that

Q.383. Demonstrate that the quantity E2 — p2c2  for a particle is an invariant, i.e. it has the same magnitude in all inertial reference frames. What is the magnitude of this invariant?

Solution. 383. As before

Similarly

Then

Q.384. A neutron with kinetic energy T = 2m0c2, where m0 is its rest mass, strikes another, stationary, neutron. Determine:
(a) the combined kinetic energy
of both neutrons in the frame of their centre of inertia and the momentu  each neutron in that frame;
(b) the velocity of the centre of inertia of this system of particles. Instruction. Make use of the invariant E2  — p2c2  remaining constant on transition from one inertial reference frame to another (E is the total energy of the system, p is its composite momentum).

Solution. 384. (b) & (a) In the CM frame, the total momentum is zero, Thus

where wc have used the result of problem (Q.375) Then

Total energy in the CM frame is

So

Q.385. A particle of rest mass m0 with kinetic energy T strikes a stationary particle of the same rest mass. Find the rest mass and the velocity of the compound particle formed as a result of the collision.

Solution. 385.

Also

Q.386. How high must be the kinetic energy of a proton striking another, stationary, proton for their combined kinetic energy in the frame of the centre of inertia to be equal to the total kinetic energy of two protons moving toward each other with individual kinetic energies T = 25.0 GeV?

Solution. 386. Let T = kinetic energy of a proton striking another stationary particle of the same rest mass. Then, combined kinetic energy in the CM frame

Q.387. A stationary particle of rest mass m0 disintegrates into three particles with rest masses m1, m2, and m3. Find the maximum total energy that, for example, the particle m1 may possess.

Solution. 387. We have

Hence

The L.H.S.

The R.H.S. is an invariant We can evaluate it in any frame. Choose the CM frame of the particles 2 and 3.

In this frame

Q.388. A relativistic rocket emits a gas jet with non-relativistic velocity u constant relative to the rocket. Find how the velocity v of the rocket depends on its rest mass m if the initial rest mass of the rocket equals m0.

Solution. 388. The velocity of ejected gases is u realtive to the rocket. In an earth centred frame it is

in the direction of the rocket The momentum conservation equation then reads

or

Here - dm is the mass of the ejected gases, so

Integrating

The constant

Thus

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