Irodov Solutions: Relativistic Mechanics- 1 Notes | EduRev

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Q.340. A rod moves lengthwise with a constant velocity v relative to the inertial reference frame K. At what value of v will the length of the rod in this frame be η = 0.5% less than its proper length? 

Solution. 340. From the formula for length contraction

Irodov Solutions: Relativistic Mechanics- 1 Notes | EduRev

So,  Irodov Solutions: Relativistic Mechanics- 1 Notes | EduRev


Q.341. In a triangle the proper length of each side equals α. Find the perimeter of this triangle in the reference frame moving relative to it with a constant velocity V along one of its
 (a) bisectors;
 (b) sides.

Solution. 341. (a) In the frame in which the triangle is at rest the space coordinates of the vertices are  Irodov Solutions: Relativistic Mechanics- 1 Notes | EduRev all measured at the same time t. In the moving frame the corresponding coordinates at time t' are

Irodov Solutions: Relativistic Mechanics- 1 Notes | EduRevIrodov Solutions: Relativistic Mechanics- 1 Notes | EduRev

The perimeter P is then

Irodov Solutions: Relativistic Mechanics- 1 Notes | EduRev

(b) The coordinates in the first frame are shown at time t. The coordinates in the moving frame are,

Irodov Solutions: Relativistic Mechanics- 1 Notes | EduRevIrodov Solutions: Relativistic Mechanics- 1 Notes | EduRev

Irodov Solutions: Relativistic Mechanics- 1 Notes | EduRev

The perimeter P is then

Irodov Solutions: Relativistic Mechanics- 1 Notes | EduRevIrodov Solutions: Relativistic Mechanics- 1 Notes | EduRev


Q.342. Find the proper length of a rod if in the laboratory frame of reference its velocity is v = c/2, the length l = 1.00 m, and the angle between the rod and its direction of motion is θ = 45°. 

Solution. 342. In the rest frame, the coordinates of the ends of the rod in terms of proper length l0

Irodov Solutions: Relativistic Mechanics- 1 Notes | EduRev

at time t. In the laboratory frame the coordinates at time t' are

Irodov Solutions: Relativistic Mechanics- 1 Notes | EduRev

Irodov Solutions: Relativistic Mechanics- 1 Notes | EduRev

Irodov Solutions: Relativistic Mechanics- 1 Notes | EduRev


Q.343.  A stationary upright cone has a taper angle θ =. 45°, and the area of the lateral surface So  = 4.0 m2. Find: (a) its taper angle; (b) its lateral surface area, in the reference frame moving with a velocity v = (4/5)c along the axis of the cone. 

Solution. 343. In the frame K in which the cone is at rest the coordinates of A are (0,0,0) and of B are (h, h tan θ, 0) . In the frame K , which is moving with velocity v along the axis o f the cone, the coordinates of A and B at time t' are

Irodov Solutions: Relativistic Mechanics- 1 Notes | EduRev

Thus the taper angle in the frame K is 

Irodov Solutions: Relativistic Mechanics- 1 Notes | EduRev

and the lateral surface area is, 

Irodov Solutions: Relativistic Mechanics- 1 Notes | EduRev

Irodov Solutions: Relativistic Mechanics- 1 Notes | EduRev

Here S0 - πh2 secθ tanθ is the lateral surface area in the rest frame and 

Irodov Solutions: Relativistic Mechanics- 1 Notes | EduRev

 
 Q.344. With what velocity (relative to the reference frame K) did the clock move, if during the time interval t = 5.0 s, measured by the clock of the frame K, it became slow by Δt = 0.10 s?

Solution. 344. Because of time dilation, a moving clock reads less time. We write,

Irodov Solutions: Relativistic Mechanics- 1 Notes | EduRev

Thus,   Irodov Solutions: Relativistic Mechanics- 1 Notes | EduRev

or, Irodov Solutions: Relativistic Mechanics- 1 Notes | EduRev


Q.345. A rod flies with constant velocity past a mark which is stationary in the reference frame K. In the frame K it takes Δt = 20 ns for the rod to fly past the mark. In the reference frame fixed to the rod the mark moves past the rod for Δt' = 25 ns. Find the proper length of the rod. 

Solution. 345. In the frame K the length l of the rod is related to the time of flight Δt by

l = v Δt

In the reference frame fixed to the rod (frame K')the proper length l0 of the rod is given by

Irodov Solutions: Relativistic Mechanics- 1 Notes | EduRev

But    Irodov Solutions: Relativistic Mechanics- 1 Notes | EduRev

Thus,     Irodov Solutions: Relativistic Mechanics- 1 Notes | EduRev

SoIrodov Solutions: Relativistic Mechanics- 1 Notes | EduRev

and   Irodov Solutions: Relativistic Mechanics- 1 Notes | EduRev


Q.346.  The proper lifetime of an unstable particle is equal to Δt0  = 10 ns. Find the distance this particle will traverse till its decay in the laboratory fraine of reference, where its lifetime is equal to Δt = 20 ns. 

Solution. 346. The distance travelled in the laboratory frame of reference is vΔ t where v is the velocity of the particle. But by time dilation

Irodov Solutions: Relativistic Mechanics- 1 Notes | EduRev

Thus the distance traversed is

Irodov Solutions: Relativistic Mechanics- 1 Notes | EduRev


Q.347. In the reference frame K a muon moving with a velocity v = 0.990c travelled a distance l = 3.0 km from its birthplace to the point where it decayed. Find:
 (a) the proper lifetime of this muon;
 (b) the distance travelled by the muon in the frame K "from the muon's standpoint". 

Solution. 347. (a) If τ0 is the proper life time of the muon the life time in the moving frame is

Irodov Solutions: Relativistic Mechanics- 1 Notes | EduRev

Thus   Irodov Solutions: Relativistic Mechanics- 1 Notes | EduRev

(The words "from the muon’s stand point" are not part of any standard terminology)


Q.348. Two particles moving in a laboratory frame of reference along the same straight line with the same velocity v = (3/4)c strike against a stationary target with the time interval Δt = 50 ns. Find the proper distance between the particles prior to their hitting the target. 

Solution. 348. In the frame K in which the particles are at rest, their positions are A and B whose coordinates may be taken as,

Irodov Solutions: Relativistic Mechanics- 1 Notes | EduRev

In the frame K' with respect to which K is moving with a velocity v the coordinates of A and B at time t' in the moving frame are

Irodov Solutions: Relativistic Mechanics- 1 Notes | EduRev

Suppose B hits a stationary target in K' after time t'B while A hits it after time tB + Δt. Then,

Irodov Solutions: Relativistic Mechanics- 1 Notes | EduRev

Irodov Solutions: Relativistic Mechanics- 1 Notes | EduRev


Q.349. A rod moves along a ruler with a constant velocity. When the positions of both ends of the rod are marked simultaneously in the reference frame fixed to the ruler, the difference of readings on the ruler is equal to Δx1  = 4.0 m. But when the positions of the rod's ends are marked simultaneously in the reference frame fixed to the rod, the difference of readings on the same ruler is equal to Δx2 = 9.0 m. Find the proper length of the rod and its velocity relative to the ruler. 

Solution. 349. In the reference frame fixed to the ruler the rod is moving with a velocity v and suffers Lorentz contraction. If l0 is the proper length of the rod, its measured length will be

Irodov Solutions: Relativistic Mechanics- 1 Notes | EduRev

In the reference frame fixed to the rod the ruler suffers Lorentz contraction and we must have

Irodov Solutions: Relativistic Mechanics- 1 Notes | EduRev

and  Irodov Solutions: Relativistic Mechanics- 1 Notes | EduRev


Q.350. Two rods of the same proper length l0  move toward each other parallel to a common horizontal axis. In the reference frame fixed to one of the rods the time interval between the moments, when the right and left ends of the rods coincide, is equal to Δt. What is the velocity of one rod relative to the other?

Solution. 350. The coordinates of the ends of the rods in the frame fixed to the left rod are shown. The points B and D coincides when

Irodov Solutions: Relativistic Mechanics- 1 Notes | EduRev

Irodov Solutions: Relativistic Mechanics- 1 Notes | EduRev

The points A and E coincide when

Irodov Solutions: Relativistic Mechanics- 1 Notes | EduRev

From this    Irodov Solutions: Relativistic Mechanics- 1 Notes | EduRev


Q.351. Two unstable particles move in the reference frame K along a straight line in the same direction with a velocity v = 0.990c. The distance between them in this reference frame is equal to l = 120 m. At a certain moment both particles decay simultaneously in the reference frame fixed to them. What time interval between the moments of decay of the two particles will be observed in the frame K? Which particle decays later in the frame K? 

Solution. 351. In K0, the rest frame of the particles, the events corresponding to the decay of the particles are,

Irodov Solutions: Relativistic Mechanics- 1 Notes | EduRev

In the reference frame Kf the corresponding coordintes are by Lorentz transformation

Irodov Solutions: Relativistic Mechanics- 1 Notes | EduRev

Now  Irodov Solutions: Relativistic Mechanics- 1 Notes | EduRev

by Lorentz Fitzgerald contraction formula. Thus the time lag of the decay time of B is 

Irodov Solutions: Relativistic Mechanics- 1 Notes | EduRev

B decays later (B is the forward particle in the direction of motion)


Q.352. A rod AB oriented along the x axis of the reference frame K moves in the positive direction of the x axis with a constant velocity v. The point A is the forward end of the rod, and the point B its rear end. Find:
 (a) the proper length of the rod, if at the moment tA the coordinate of the point A is equal to xA, and at the moment tB the coordinate of the point B is equal to xB;
 (b) what time interval should separate the markings of coordinates of the rod's ends in the frame K for the difference of coordinates to become equal to the proper length of the rod. 

Solution. 352. (a) In the reference frame K with respect to which the rod is moving with velocity v, the coordinates of A and B are

Irodov Solutions: Relativistic Mechanics- 1 Notes | EduRev

Irodov Solutions: Relativistic Mechanics- 1 Notes | EduRev

Irodov Solutions: Relativistic Mechanics- 1 Notes | EduRev
Irodov Solutions: Relativistic Mechanics- 1 Notes | EduRev


Q.353. The rod A'B' moves with a constant velocity v relative to the rod AB (Fig. 1.91). Both rods have the same proper length l0 and at the ends of each of them clocks are mounted, which are synchronized pairwise: A with B and A' with B'. Suppose the moment when the clock B' gets opposite the clock A is taken for the beginning of the time count in the reference frames fixed to each of the rods. Determine: 

(a) the readings of the clocks B and B' at the moment when they are opposite each other;
 (b) the same for the clocks A and A'. 

Irodov Solutions: Relativistic Mechanics- 1 Notes | EduRev

Solution. 353.  At the instant the picture is taken the coordintes of A, B, A', B' in the rest frame of AB are

Irodov Solutions: Relativistic Mechanics- 1 Notes | EduRev

Irodov Solutions: Relativistic Mechanics- 1 Notes | EduRev

In this frame the coordinates o f B ' at other times are B': (t, vt, 0, 0). So B ' is opposite to B at time Irodov Solutions: Relativistic Mechanics- 1 Notes | EduRev In the frame in which B', A' is at rest the time corresponding this is by Lorentz tranformation.

Irodov Solutions: Relativistic Mechanics- 1 Notes | EduRev

Similarly in the rest frame of A, B, te coordinates of A at other times are

Irodov Solutions: Relativistic Mechanics- 1 Notes | EduRev

A' is opposite to A at time  Irodov Solutions: Relativistic Mechanics- 1 Notes | EduRev

The corresponding time in the frame in which A', B' are at rest is

Irodov Solutions: Relativistic Mechanics- 1 Notes | EduRev


Q.354. There are two groups of mutually synchronized clocks K and K' moving relative to each other with a velocity v as shown in Fig. 1.92. The moment when the clock A' gets opposite the clock A is taken for the beginning of the time count. Draw the approximate position of hands of all the clocks at this moment "in terms of the K clocks"; "in terms of the K' clocks".

Irodov Solutions: Relativistic Mechanics- 1 Notes | EduRev

Solution. 354. By Lorentz transformation  Irodov Solutions: Relativistic Mechanics- 1 Notes | EduRev

So at time   Irodov Solutions: Relativistic Mechanics- 1 Notes | EduRev

Irodov Solutions: Relativistic Mechanics- 1 Notes | EduRev and we get the diagram given below "in terms o f th eif-clock ".

Irodov Solutions: Relativistic Mechanics- 1 Notes | EduRev

The situation in terms of the K' clock is reversed.


Q.355. The reference frame K' moves in the positive direction of the x axis of the frame K with a relative velocity V. Suppose that at the moment when the origins of coordinates O and O' coincide, the clock readings at these points are equal to zero in both frames. Find the displacement velocity Irodov Solutions: Relativistic Mechanics- 1 Notes | EduRev of the point (in the frame K) at which the readings of the clocks of both reference frames will be permanent- ly identical. Demonstrate that Irodov Solutions: Relativistic Mechanics- 1 Notes | EduRev

Solution. 355. Suppose x (t) is the locus of points in the frame K at which the readings of the clocks of both reference system are permanently identical then by Lorentz transformation

Irodov Solutions: Relativistic Mechanics- 1 Notes | EduRev

So differentiatin  Irodov Solutions: Relativistic Mechanics- 1 Notes | EduRev

Let  Irodov Solutions: Relativistic Mechanics- 1 Notes | EduRev

Irodov Solutions: Relativistic Mechanics- 1 Notes | EduRev

Irodov Solutions: Relativistic Mechanics- 1 Notes | EduRev

 (tan h θ is a monotonically increasing function of θ)

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