A neutral K meson at rest decays into two π mesons, which travel in opposite directions along the x axis with speeds of 0.828c.

Question

15 days ago

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A neutral K meson at rest decays into two π mesons, which travel in opposite directions along the x axis with speeds of 0.828c.

If instead the K meson were moving in the positive x direction with a velocity of 0.486c, what would be the velocities of the two π mesons?

Physics

2 answers:

Sav [3.1K] · Answer 1 · 2026-04-06T21:45:19+03:00

Answer:

The velocities of both $\pi$ -mesons are 0.574 in the positive direction and -0.9367c in the negative direction along the X-axis

Solution:

Referring to the question:

The speed of the $\pi$ -meson is v' = 0.828c

And the K-meson's speed in the positive x direction is v'' = 0.486c

Now, to determine the speeds of the two $\pi$ -mesons:

Velocity of the first $\pi$ -meson, $v_{px}$ :

$v_{px} = \frac{v' - v''}{1 - \frac{v'v''}{c^{2}}}$

$v_{px} = \frac{0.828c - 0.486c}{1 - \frac{0.828c\times 0.486c}{c^{2}}}$

$v_{px} = 0.574c$

Velocity of the second $\pi$ -meson, $v_{px}$ :

$v_{px} = \frac{v' - v''}{1 - \frac{v'v''}{c^{2}}}$

$v_{px} = \frac{- 0.828c - 0.486c}{1 + \frac{0.828c\times 0.486c}{c^{2}}}$

$v'_{px} = \frac{- 0.828c - 0.486c}{1 + \frac{0.828c\times 0.486c}{c^{2}}}$

$v'_{px} = - 0.937c$

kicyunya [3.2K] · Answer 2 · 2026-04-06T21:47:57+03:00

Answer:

$v'_{x}=0.572c$ and $v''_{x}=-0.939c$

Explanation:

To resolve this issue, we must apply the following formula

$v'_{x}=\frac{v_{x} -u}{1-\frac{v_{x} u}{c^{2} } }$

Provided information:

$v_{x}=0.828c\\u=0.486c\\$

When in the same direction, we have

$v'_{x}=\frac{(+0.828c)-(+0.486c)}{1-\frac{(0.828c)(0.486c)}{c^{2} } }=\frac{0.342c}{1-\frac{0.402c^{2} }{c^{2} } }\\ v'_{x}=\frac{0.342c}{0.598} \\v'_{x}=0.572c$

It's worth noting that we simply substituting the given values and perform basic calculations.

In opposite directions,

$v''_{x}=\frac{-0.828c-(+0.486c)}{1-\frac{(-0.828c)(0.486c)}{c^{2} } }=\frac{-1.314c}{1+\frac{0.402c^{2} }{c^{2} } }\\ v''_{x}=\frac{-1.314c}{1.40} \\v''_{x}=-0.939c$

You can see that tackling these types of problems requires careful attention to the signs of each velocity. In the same direction, the signs must match, while in opposing directions, they must differ.

Thus, the results are $v'_{x}=0.572c$ and $v''_{x}=-0.939c$