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

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

$\dfrac{1}{2}m_1u_1^2+\dfrac{1}{2}m_2u_2^2=\dfrac{1}{2}m_1v_1^2+\dfrac{1}{2}m_2v_2^2$

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

$e=\dfrac{v_1-v_2}{u_2-u_1}$

$1=\dfrac{v_1-v_2}{u_2-u_1}$

$v_1-v_2=u_2-u_1$

$v_1-v_2=0.25$ (given )

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