Question

Two manned satellites approach one another at a relative speed of 0.300 m/s, intending to dock. The first has a mass of 2.00 ✕ 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 in meters per second? (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.)

_____m/s

Answer 1

Given m₁ = 2x10³ = 2000 kg and m₂ = 7.50x10³ =7500 kg,

By adopting 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. In the specified reference frame, the first satellite has negative momentum.

v₁ = - 0.3 m/s

For this problem, note that v₂ = 0 and use conservation of momentum. Thus,

p₁ = p ′ ₁ + p ′ ₂

or m₁v₁ = m₁v ′ ₁ + m₂v ′ ₂

Using conservation of internal kinetic energy and that v2 = 0

(1/2)m₁v₁² = (1/2)m₁v ′₁² + (1/2)m₂v ′₂².

Solving the first equation (momentum equation) for v ′₂, we obtain

Substituting this expression into the second equation (internal kinetic energy equation) eliminates the variable v′₂, leaving only v ′₁ as an unknown (the algebra is left as an exercise for the reader). There are two solutions to any quadratic equation; in this example, they are

substituting known values v1, m1 and m2 we will get the quadratic equation as below

v'₁= 0.173 m/s or v'₁ = -0.3 m/s

In this case, the second solution is the same as the initial condition. The second solution thus represents the situation before the collision and is discarded. The first solution (v ′₁ = 0.173 m/s) is positive, meaning that the first object bounces backward. When this positive value of v ′₁ is used to find the velocity of the second object after the collision, we get