Two manned satellites approaching one another, at a relative speed of 0.250 m/s, intending to dock. The first has a mass of, and the second a mass of (a) Calculate the final velocity (after docking) by using the frame of reference in which the first satellite was originally at rest. (b) What is the loss of kinetic energy in this inelastic collision

Complete Question:

Two manned satellites approaching one another, at a relative speed of 0.250 m/s, intending to dock. The first has a mass of 4.00x10^3 kg, and the second a mass of 7.50x10^3 kg (a) Calculate the final velocity (after docking) by using the frame of reference in which the first satellite was originally at rest. (b) What is the loss of kinetic energy in this inelastic collision?

Answer:

a) - 0.163 m/s

b) - 81.60 J

Explanation:

When a collision happens the momentum (Q) is conserved (Q before = Q after). The collision can be elastic, partially elastic or inelastic. When the collision is elastic, the objects move away from one another, and the kinetic energy is conserved. In the partially elastic collision, they still move away, but the kinetic energy after the collision is now less than the initial.

When the collision is inelastic, the objects move together with the same velocity, and the kinetic energy is smaller than the initial one.

a) By the momentum conservation:

Qbefore = Qafter

m1*V1 + m2*V2 = (m1 + m2) * V (m is the mass and V the velocity)

Because the first satelity is at rest, V1 = 0. The relative speed is V1 - V2 = 0.250, thus V2 = - 0.250 m/s

4.00x10³ * 0 + 7.50x10³ * (-0.250) = (4.00x10³ + 7.50x10³) * V

-1875 = 11.5x10³V

V = - 0.163 m/s

b) The loss of kinetic energy is the initial kinetic energy (E1) less than the final kinetic energy (E2). The kinetic energy is the mass multiplied by the squared speed, divided by two.

E1 - E2

(m1*V1²/2 + m2*V2²/2) - (m1 + m2) * V²/2

0 + [7.50x10³ * (-0.250) ²]/2 - [11.5x10³ * (-0.163) ²]/2

234.375 - 152.77175

81.60 J

Because it was lost, ΔE = - 81.60 J