<meta name="keywords" content="Savchenko Problems in Physics, Savchenko solutions, physics problems, physics olympiad preparation, IPhO, Jaan Kalda">
<meta name="description" content="Two beams of particles moving with the same modulo velocity v intersect at an angle \alpha. Particle collisions occur in a limited area. Let's move on to the reference frame, where the particle velocities are equal in modulus and opposite in direction. It would seem that now the intersection area is the entire volume of beams, and therefore the number of collisions per unit time should be greater. Explain the resulting contradiction.">
<meta property="og:title" content="Two beams of particles moving with the same modulo velocity v intersect at an angle \alpha. Particle collisions occur in a limited area. Let's move on to the reference frame, where the particle velocities are equal in modulus and opposite in direction. It would seem that now the intersection area is the entire volume of beams, and therefore the number of collisions per unit time should be greater. Explain the resulting contradiction.">
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@@ -14,9 +14,9 @@
<meta property="og:description" content="Two beams of particles moving with the same modulo velocity v intersect at an angle \alpha. Particle collisions occur in a limited area. Let's move on to the reference frame, where the particle velocities are equal in modulus and opposite in direction. It would seem that now the intersection area is the entire volume of beams, and therefore the number of collisions per unit time should be greater. Explain the resulting contradiction.">
<title>Two beams of particles moving with the same modulo velocity v intersect at an angle \alpha. Particle collisions occur in a limited area. Let's move on to the reference frame, where the particle velocities are equal in modulus and opposite in direction. It would seem that now the intersection area is the entire volume of beams, and therefore the number of collisions per unit time should be greater. Explain the resulting contradiction.</title>
$1.4.16.$ Two beams of particles moving with the same modulo velocity $v$ intersect at an angle $\alpha$. Particle collisions occur in a limited area. Let's move on to the reference frame, where the particle velocities are equal in modulus and opposite in direction. It would seem that now the intersection area is the entire volume of beams, and therefore the number of collisions per unit time should be greater. Explain the resulting contradiction.
<p>In the new reference frame, the geometry of the beams, and therefore the area of their intersection, are the same as before. The particle velocity is not necessarily directed along the beam.
</p>
<p>
When switching to another reference frame, the particles will also form a cross. Their velocities will simply be horizontal
<small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> alex@savchenkosolutions.com <br></small>
<meta name="keywords" content="Savchenko Problems in Physics, Savchenko solutions, physics problems, physics olympiad preparation, IPhO, Jaan Kalda">
<meta name="keywords" content="Savchenko Problems in Physics, Savchenko solutions, physics problems, physics olympiad preparation, IPhO, Jaan Kalda">
<meta name="description" content="Two beams of particles moving with the same modulo velocity v intersect at an angle \alpha. Particle collisions occur in a limited area. Let's move on to the reference frame, where the particle velocities are equal in modulus and opposite in direction. It would seem that now the intersection area is the entire volume of beams, and therefore the number of collisions per unit time should be greater. Explain the resulting contradiction.">
<meta name="description" content="Two beams of particles moving with the same modulo velocity v intersect at an angle \alpha. Particle collisions occur in a limited area. Let's move on to the reference frame, where the particle velocities are equal in modulus and opposite in direction. It would seem that now the intersection area is the entire volume of beams, and therefore the number of collisions per unit time should be greater. Explain the resulting contradiction.">
<meta property="og:title" content="Two beams of particles moving with the same modulo velocity v intersect at an angle \alpha. Particle collisions occur in a limited area. Let's move on to the reference frame, where the particle velocities are equal in modulus and opposite in direction. It would seem that now the intersection area is the entire volume of beams, and therefore the number of collisions per unit time should be greater. Explain the resulting contradiction.">
<meta property="og:title" content="Two beams of particles moving with the same modulo velocity v intersect at an angle \alpha. Particle collisions occur in a limited area. Let's move on to the reference frame, where the particle velocities are equal in modulus and opposite in direction. It would seem that now the intersection area is the entire volume of beams, and therefore the number of collisions per unit time should be greater. Explain the resulting contradiction.">
<meta property="og:image" content="img/logo.png">
<meta property="og:image" content="img/logo.png">
@@ -14,9 +14,9 @@
<meta property="og:description" content="Two beams of particles moving with the same modulo velocity v intersect at an angle \alpha. Particle collisions occur in a limited area. Let's move on to the reference frame, where the particle velocities are equal in modulus and opposite in direction. It would seem that now the intersection area is the entire volume of beams, and therefore the number of collisions per unit time should be greater. Explain the resulting contradiction.">
<meta property="og:description" content="Two beams of particles moving with the same modulo velocity v intersect at an angle \alpha. Particle collisions occur in a limited area. Let's move on to the reference frame, where the particle velocities are equal in modulus and opposite in direction. It would seem that now the intersection area is the entire volume of beams, and therefore the number of collisions per unit time should be greater. Explain the resulting contradiction.">
<title>Two beams of particles moving with the same modulo velocity v intersect at an angle \alpha. Particle collisions occur in a limited area. Let's move on to the reference frame, where the particle velocities are equal in modulus and opposite in direction. It would seem that now the intersection area is the entire volume of beams, and therefore the number of collisions per unit time should be greater. Explain the resulting contradiction.</title>
<title>Two beams of particles moving with the same modulo velocity v intersect at an angle \alpha. Particle collisions occur in a limited area. Let's move on to the reference frame, where the particle velocities are equal in modulus and opposite in direction. It would seem that now the intersection area is the entire volume of beams, and therefore the number of collisions per unit time should be greater. Explain the resulting contradiction.</title>
$1.4.16.$ Two beams of particles moving with the same modulo velocity $v$ intersect at an angle $\alpha$. Particle collisions occur in a limited area. Let's move on to the reference frame, where the particle velocities are equal in modulus and opposite in direction. It would seem that now the intersection area is the entire volume of beams, and therefore the number of collisions per unit time should be greater. Explain the resulting contradiction.
$1.4.16.$ Two beams of particles moving with the same modulo velocity $v$ intersect at an angle $\alpha$. Particle collisions occur in a limited area. Let's move on to the reference frame, where the particle velocities are equal in modulus and opposite in direction. It would seem that now the intersection area is the entire volume of beams, and therefore the number of collisions per unit time should be greater. Explain the resulting contradiction.
<p>In the new reference frame, the geometry of the beams, and therefore the area of their intersection, are the same as before. The particle velocity is not necessarily directed along the beam.
<p>In the new reference frame, the geometry of the beams, and therefore the area of their intersection, are the same as before. The particle velocity is not necessarily directed along the beam.
</p>
</p>
<p>
<p>
When switching to another reference frame, the particles will also form a cross. Their velocities will simply be horizontal
When switching to another reference frame, the particles will also form a cross. Their velocities will simply be horizontal
<small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> alex@savchenkosolutions.com <br></small>
<small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> alex@savchenkosolutions.com <br></small>
