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Sergeu [11.5K]
2 years ago
5

In an "atom smasher," two particles collide head on at relativistic speeds. If the velocity of the first particle is 0.741c to t

he left, and the velocity of the second particle is 0.543c to the right (both as measured in the lab rest frame), how fast are the particles moving with respect to each other?
Physics
1 answer:
galina1969 [7]2 years ago
5 0

Answer:

W_x = 0.9156\ c

Explanation:

given,

velocity of particle 1 = 0.741 c to left

velocity of second particle = 0.543 c to right

relative velocity between the particle = ?

for the relative velocity calculation we have formula

W_x = \dfrac{|u_x - v_x|}{1-\dfrac{u_xv_x}{c^2}}

u_x = 0.543 c

v_x = - 0.741 c

W_x = \dfrac{0.543 c - (-0.741 c)}{1-\dfrac{(0.543 c)(-0.741 c)}{c^2}}

W_x = \dfrac{0.543 c +0.741 c)}{1+\dfrac{(0.543)(0.741)c^2}{c^2}}

W_x = \dfrac{1.284c}{1+0.402363}

W_x = 0.9156\ c

Relative velocity of the particle is W_x = 0.9156\ c

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<u><em>m = 0.15 kg</em></u>

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Now, we find the launch angle of the ball by using the following formula:

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