Question

Two manned satellites approaching one another at a relative speed of 0.150 m/s intend to dock. The first has a mass of 2.50 ✕ 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? 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.

Answer 1

When two satellites collide elastically, the total momentum before the collision is equal to the total momentum after the collision. However, the final velocities and the change in kinetic energy cannot be determined without additional information.

Step-by-step explanation:

When two satellites collide elastically, the total momentum before the collision is equal to the total momentum after the collision. We can start by calculating the initial momentum:

Initial momentum = mass of the first satellite × velocity of the first satellite + mass of the second satellite × velocity of the second satellite

Plugging in the values:

Initial momentum = (2.50 × 10³ kg)(0.150 m/s) + (7.50 × 10³ kg)(0)

Since the second satellite is initially at rest (velocity = 0), the initial momentum simplifies to:

Initial momentum = (2.50 × 10^3 kg)(0.150 m/s) = 375 kg·m/s

Since momentum is conserved, the total momentum after the collision is also 375 kg·m/s. Let's assume the final velocities of the satellites are v1 and v2:

Final momentum = (2.50 × 10³ kg)(v1) + (7.50 × 10³ kg)(v2)

Setting the initial and final momenta equal to each other:

Initial momentum = Final momentum

375 kg·m/s = (2.50 × 10³ kg)(v1) + (7.50 × 10³ kg)(v2)

We can't solve for the individual velocities with only this equation, but we can use the fact that the satellites collide elastically. In an elastic collision, the total kinetic energy before the collision is equal to the total kinetic energy after the collision.

Using the formula for kinetic energy:

Kinetic energy = 1/2 × mass × velocity²

The initial kinetic energy is:

Initial kinetic energy = (1/2)(2.50 × 10³ kg)(0.150 m/s)² + (1/2)(7.50 × 10³ kg)(0)²

Simplifying:

Initial kinetic energy = (1/2)(2.50 × 10³ kg)(0.150 m/s)² = 56.25 J

The final kinetic energy is:

Final kinetic energy = (1/2)(2.50 × 10³ kg)(v1)² + (1/2)(7.50 × 10³ kg)(v2)²

Setting the initial and final kinetic energies equal to each other:

Initial kinetic energy = Final kinetic energy

56.25 J = (1/2)(2.50 × 10³ kg)(v1)² + (1/2)(7.50 × 10³ kg)(v2)²

Unfortunately, this equation cannot be solved with the given information and assumptions. We would need additional information, such as the angle of collision, to solve for the individual velocities and kinetic energies.