Question

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.

Answer 1

m = 8.81*10^7 kg

Step-by-step explanation:

relative momentum

`P = \frac{mv}{\sqrt{1 - ( v^2)/(c^2)}}`

momentum prior to collision is given as

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

Pi = 4.553 *10^7 c

let mass of asteroid is m, then final momentum is

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

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