Question

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​^{3} kg, and the second a mass of 7.50×10​^{3} kg. If the two satellites collide elastically rather than dock, what is their final relative velocity (in m/s)?

Answer 1

Step-by-step explanation:

Let
`u_1\ and\ u_2` are initial speeds of satellite 1 and 2 and
`v_1\ and\ v_2` are final speeds of satellite 1 and 2 after collision respectively.

It is given that, the relative speed of two manned satellites is 0.25 m/s,
`u_2-u_1=0.25\ m/s`

It is mentioned that thy collide elastically rather than dock, the momentum will remain conserved. So,

`m_1u_1+m_2u_2=m_1v_1+m_2v_2`

In elastic collision, the kinetic energy also remains conserved, so

`(1)/(2)m_1u_1^2+(1)/(2)m_2u_2^2=(1)/(2)m_1v_1^2+(1)/(2)m_2v_2^2`

For an elastic collision, the velocity of separation is equal to the velocity of approach such that, e = 1

`e=(v_1-v_2)/(u_2-u_1)`

`1=(v_1-v_2)/(u_2-u_1)`

`v_1-v_2=u_2-u_1`

`v_1-v_2=0.25` (given )

So, their final relative velocity is 0.25 m/s. Hence, this is the required solution.