Question

Two manned satellites approaching one another at a relative speed of 0.300 m/s intend to dock. The first has a mass of 4.00 ✕ 10³ kg, and the second a mass of 7.50 ✕ 10³ 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

Answer 1

Given that satellites approach one another at a relative speed of 0.300 m/sec

u2 - u1 = 0.300 m/sec

Which means relative to 1st satellite, 2nd satellite is moving at 0.300 m/sec and at this moment speed of 1st satellite is at rest

u1 = 0 and u2 = 0.300 m/sec

for inelastic collision we know that

Pi = Pf

m1*u1 + m2*u2 = (m1 + m2)*V

V = (m1*u1 + m2*u2)/(m1 + m2)

m1 = mass of 1st satellite = 4000 kg, & m2 = 7500 kg

So Using above values:

V = (4000*0 + 7500*0.300)/(4000 + 7500)

V = 0.19565 m/s

Part B.

loss in kinetic energy will be given by:

dKE = KEi - KEf

dKE = (1/2)*m1*u1^2 + (1/2)*m2*u2^2 - (1/2)*(m1 + m2)*V^2

Using given values:

dKE = (1/2)*4000*0^2 + (1/2)*7500*0.300^2 - (1/2)*(4000 + 7500)*0.19565^2

dKE = 117.4 J = KE lost

Part C.

Now in the same reference frame, when 2nd satellite was originally at rest, in other words, u2 - u1 = -0.300 m/sec

where u2 = 0 & u1 = -0.300 m/sec

for inelastic collision we know that

Pi = Pf

m1*u1 + m2*u2 = (m1 + m2)*V

V = (m1*u1 + m2*u2)/(m1 + m2)

m1 = mass of 1st satellite = 4000 kg, & m2 = 7500 kg

So Using above values:

V = (4000*(-0.300) + 7500*0)/(4000 + 7500)

V = -0.10435 m/s = final velocity

loss in kinetic energy will be given by:

dKE = KEi - KEf

dKE = (1/2)*m1*u1^2 + (1/2)*m2*u2^2 - (1/2)*(m1 + m2)*V^2

Using given values:

dKE = (1/2)*4000*(-0.300)^2 + (1/2)*7500*0^2 - (1/2)*(4000 + 7500)*(-0.10435)^2

dKE = 117.4 J = KE lost