Two manned satellites approach one another at a relative velocity of v = 0.150 m/s, intending to dock. The first has a mass of m 1 = 4.00 × 10 3 kg and the second a mass of m 2 = 7.50 × 10 3 kg. If the two satellites collide elastically rather than dock, what is their final relative velocity?

their final relative velocity, v₂ - v₁ = u₁ -u₂ = 0.15 m/s

Explanation:

Given;

mass of the first satellite, m₁ = 4000 kg

mass of the second satellite, m₂ = 7500 kg

their relative velocity, u₁ -u₂ = 0.15 m/s

Since, the two satellites collide elastically, then total momentum before and after collision are conserved, also the total kinetic energy is also conserved.

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

m₁u₁ -m₁v₁ = m₂v₂ - m₂u₂

m₁(u₁-v₁) = m₂(v₂-u₂ ) ------------equation (i)

where;

u is the initial velocity

v is the final velocity

Also, kinetic energy is conserved

Total initial kinetic energy = Total final kinetic energy

¹/₂m₁u₁² + ¹/₂m₂u₂² = ¹/₂m₁v₁² +  ¹/₂m₂v₂²

multiple through by 2

m₁u₁² + m₂u₂² = m₁v₁² + m₂v₂²

m₁u₁² - m₁v₁²  = m₂v₂² - m₂u₂²

m₁(u₁² - v₁²) = m₂(v₂²- u₂²) --------------equation (ii)

divide equation (ii) by equation (i)

0.15 m/s = v₂ - v₁

Thus, the relative velocity before collision is equal to the relative velocity of separation of collision.

answer; the correct answer is technician a; ///100%good luck

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