Question

Suppose that, at t = 9.00 × 10 − 4 s, the space coordinates of a particle are x = 135 m, y = 30.0 m, and z = 55.0 m according to coordinate system S . If reference frame S ′ moves at speed 1.50 × 10 5 m/s in the + x - direction relative to frame S , compute the corresponding coordinate values as measured in frame S ′ . The reference frames start together, with their origins coincident at t = 0 .

Answer 1

Answer:

$x'=134.999983\ m$

$y'=y=30\ m$

$z'=z=55\ m$

Explanation:

Given:

• time, $t=9\times 10^{-4}\ s$

• x coordinates of a particle, $x=135\ m$

• y coordinates of a particle, $y=30\ m$

• z coordinates of a particle, $x=55\ m$

• Relative Speed of frame of reference S' in the +x direction, $v_x=1.5\times 10^{5} m.s^{-1}$

• Since the speed of the frame S' is comparable to the speed of light in vacuum therefore the observer from S' frame will observe a contracted length of the dimensions in the direction of motion.

Now from the equation of length contraction:

$x'=x\times \sqrt{1-\frac{v_x^2}{c^2} }$

$x'=135\times \sqrt{1-\frac{(1.5\times 10^{5})^2}{(3\times 10^8)^2} }$

$x'=134.999983\ m$

Rest other values will remain unaffected since they are along the axis of motion. So,

$y'=y=30\ m$

$z'=z=55\ m$