Two manned satellites approaching one another at a relative...

Two manned satellites approaching one another at a relative speed of 0.400 m/s intend to dock. The first has a mass of 3.50 x 10^3 kg, and the second a mass of 7.50 x 10^3 kg. Assume that the positive direction is directed from the second satellite towards the first satellite. (a) Calculate the final velocity after docking, in the frame of reference in which the first satellite was originally at rest. m/s (b) What is the loss of kinetic energy in this inelastic collision? J (c) Repeat both parts, in the frame of reference in which the second satellite was originally at rest. final velocity m/s loss of kinetic energy J

Transcript

00:01 Okay, in this question here, we have two satellites, satellite one and satellite two.

00:03 They have a relative velocity of 0 .4 metres per second, they're moving towards each other.

00:07 And we're told to consider the right to left direction as positive.

00:12 And in the first scenario, we're to consider satellite one at rest, and therefore satellite two moving towards satellite one at 0 .4 meters per second.

00:20 And we're asked to calculate the final velocity after docking, so they're going to join with each other.

00:23 So we're going to use the conservation of momentum.

00:26 So we have m1 v1 plus m2 v2, this is the momentum of satellite 1 plus the momentum satellite 2.

00:33 That's going to be equal to their combined mass because they join together, m1 plus m2, and then finally v3, the final velocity.

00:43 Well, given that in the first scenario, the first satellite is at rest, it has no momentum.

00:48 So we just consider the momentum of the second satellite.

00:51 So we have 7 .5 times 10 to the 3, multiplied by 0 .4.

00:56 Again, it's positive because the positive direction is from right to left.

01:00 And that's going to give us 11 times 10 to 3, which is the combined mass, multiplied by v3.

01:05 Therefore, we have v3 is equal to 7 .5 times 10 to 3, multiplied by 0 .4, divided by 11 times 10 to 3.

01:15 And that's going to get you to 0 .2727 meters per second.

01:20 This is quite a lot of sig figs.

01:22 However, we need it for the precision in our answers later on.

01:26 Then part b, we'll ask to work out the loss of kinetic energy.

01:30 So the initial kinetic energy, ke initial, is just going to be the kinetic energy due to the second satellite, because the first satellite is not moving, or at least we consider it to be at rest.

01:40 So we have 0 .5, which is half, so we'll write it in just for those who need a reminder of the equation.

01:47 So we have ke initial equals half m2 v2 squared, which is equal to 0 .5 multiplied by 7 .5...

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