Question

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?

Answer 1

Answer:

The velocity of the two $\pi$-meson are 0.574 in positive direction and - 0.9367c in the negative direction along X-axis

Solution:

As per the question:

The speed of the $\pi$-meson along x-axis, v' = 0.828c

The speed of the K-meson along positive x-axis, v'' = 0.486c

Now, to calculate the velocities of the two $\pi$-meson:

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

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

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

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

$v_{px} = 0.574c$

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

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

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

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

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

Answer 2

Answer:

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

Explanation:

To solve this problem, we need to use the following formula

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

Givens:

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

In same direction, we have

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

Notice, that we just replaced each given value, and solve basic operations.

In different direction,

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

As you can observe, to solve this kind of problem, you just neet to consider the right sign of each velocity, when they are in the same direction, they must have the same sign, but when they are in opposite direction, they must have different signs.

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