Question

Two manned satellites approach one another at a relative velocity of ????=0.210v=0.210 m/s, intending to dock. The first has a mass of ????1=4.00×103m1=4.00×103 kg and the second a mass of ????2=7.50×103m2=7.50×103 kg. If the two satellites collide elastically rather than dock, what is their final relative velocity?

Answer 1

This is not correct question the correct one is

Two manned satellites approach one another at a relative speed of 0.250 m/s, intending to dock. The first has a mass of 4.00×10^3kg, and the second a mass of 7.50×10^3kg. If the two satellites collide elastically rather than dock, what is their final relative velocity?

Answer:

v=2.50 m/s

Explanation:

v is velocity after the collision, u before  

v₁ = (u₁(m₁–m₂) + 2m₂u₂) / (m₁ + m₂)  

v₂ = (u₂(m₂–m₁) + 2m₁u₁) / (m₁ + m₂)  

take their initial speeds at 0.125 m/s each, first +, second

v₁ = (0.125(4-7.5) – 2•7.5•0.125) / (4+7.5)  

v₂ = (–0.125(7.5-4) + 2•4•0.125) / (4+7.5)  

v₁ = (–0.125•3.5 – 15•0.125) / (11.5)  

v₂ = (–0.125•3.5 + 8•0.125) / (11.5)  

v₁ = (–0.4375 – 1.875) / (11.5)  

v₂ = (–0.4375 + 1) / (11.5)  

v₁ = –0.2010 m/s

v₂ = 0.0489 m/s

and relative velocity is the sum 0.250 m/s

Answer 2

Explanation:

It is given that,

Relative velocity of two manned satellites, v = 0.210 m/s

Mass of first satellite, $m_1=4\times 10^3\ kg$

Mass of other satellite, $m_2=7.5\times 10^3\ kg$

The two satellites collide elastically rather than dock, we need to find their final velocities. Let v' is the final velocity. We know that in an elastic collision, both momentum and kinetic energy remains constant.

Also, the relative velocity of approach between two satellites before collision is same as the relative velocity of separation after collision. So, their final relative velocity is 0.210 m/s but in opposite direction.