Question

Two manned satellites approaching one another at a relative speed of 0.450 m/s intend to dock. The first has a mass of 4.50 ✕ 103 kg, and the second a mass of 7.50 ✕ 103 kg. If the two satellites collide elastically rather than dock, what is their final relative velocity? Adopt the reference frame in which the second satellite is initially at rest and assume that the positive direction is directed from the second satellite towards the first satellite.

Answer 1

Ok, so adopting that the 2nd satellite is at rest and that we're not moving anywhere near the speed of light (so no special relativity considerations), we can just add the two speed together, and say the 1st satellite is moving at 0.9m/s at the 2nd satellite. We can then set up our conservation of momentum equation, m₁v₁+m₂v₂ = m₁v₃+m₂v₄, where I'm calling v 1 and 2 the initial velocities of satellite 1 and 2 and v 3 and 4 the final velocities of satellite 1 and 2 respectively. We know, based on our chosen frame, that v₂ = 0, so that falls out to leave m₁v₁ = m₁v₃+m₂v₄, but we don't know v₃ or v₄, so we need another equation. Let's set up conversation of energy (elastic collisions conserve energy), where we only have to worry about kinetic energy (K = 1/2mv²) for each satellite before and after the collision. So we get 1/2m₁v₁²+1/2m₂v₂² = 1/2m₁v₃²+1/2m₂v₄². Now we have 2 equations and two unknown variables so let's solve with substitution. Let's solve the momentum equation for v₃, v₃ = (m₁v₁ - m₂v₄)/m₁, sub that into the energy equation, cancel the 1/2's and let's drop the v₂ terms since it's zero and we get: m₁v₁² = m₁((m₁v₁ - m₂v₄)/m₁)²+m₂v₄², then after some algebra we get v₄ = sqrt(m₁v₁/((v₁ - m₂/m₁)²+m₂)), then we plug in numbers v₄ = sqrt((4.5*10³*0.9/((0.9-(7.5/4.5))²+7.5*10³) = 0.73 m/s for the 2nd satellite after the collision. Then go back to v₃ = (m₁v₁ - m₂v₄)/m₁ and plug in numbers now that we know v₄ and we get v₃ = (4.5*10³*0.9 - 7.5*10³*0.73)/(4.5*10³) = -0.3167 m/s for the 1st satellite.