contestada

A space probe with a rest mass of 8.1×10^7 kg and a speed of 0.49c smashes into an asteroid at rest and becomes embedded within it. If the speed of the probe-asteroid system is 0.26c after the collision, what is the rest mass of the asteroid? Answer in kg.

Respuesta :

Answer:

m = 8.81*10^7 kg

Explanation:

relative momentum

[tex]P = \frac{mv}{\sqrt{1 - \frac{ v^2}{c^2}}}[/tex]

momentum prior to collision is given as

[tex]Pi =\frac{8.1*10^7*0.49c}{\sqrt{1 - \frac{ (0.49c)^2}{c^2}}}[/tex]

Pi = 4.553 *10^7 c

let mass of asteroid is m, then final momentum is

[tex]Pf = \frac{8.1*10^7*0.26c}{\sqrt{1 - \frac{(0.26c)^2}{c^2}}}[/tex]

Pf = 0.26926 (8.1*10^7 + m) * c

from conservation of momentum

Pi = Pf

4.553 *10^7 c = 0.26926 (8.1*10^7 + m) * c

m = 8.81*10^7 kg

Otras preguntas