Two astronauts, of masses 60 kg and 80 kg, are initially right next to each other and at rest in outer space. They suddenly push each other apart. How far has the heavier astronaut moved when the lighter astronaut has moved 12 m

21 m

Explanation:

Given

Mass of the first astronaut, m1 = 60 kg

Mass of the second astronaut, m2 = 80 kg

Distance moved by the first astronaut, d1 = 12 m

Distance moved by the second astronaut, d2 = ?

using the law of conservation of momentum, we can say

m1d1 = m2d2

So that, d2 = m1d1 / m2

d2 = (60 * 12) / 80

d2 = 720 / 80

d2 = 9 m

If we then take into account the distance of 12 m moved by the first astronaut. We have the total distance moved by the second astronaut to be 9 + 12 = 21 m

Thus, the distance moved by the heavier astronaut when the lighter astronaut has moved 12 m, is 21 m

v=ir = 3x15 =45 v