Two manned satellites are approaching one another, at a relative speed of 0.26 m

Question

Two manned satellites are approaching one another, at a relative speed of 0.26 m/s, intending to dock. The first has a mass of 3.8 x 103 kg, and the second a mass of 7.75 x 103 kg.

A.) Calculate the final velocity, in meters per second, of the two satellites after docking, in the frame of reference in which the first satellite was originally at rest. Take the velocity of the second satellite to be in the postitive direction.

B.) What is the change in kinetic energy, in joules, in this inelastic collision?

C.) Calculate the final velocities of the satellites, in meters per second, in the frame of reference in which the second satellite was originally at rest. Take the velocity of the first satellite to be in the negative direction.

D.) What is the change of kinetic energy, in joules, using the frame of reference in which the second satellite was originally at rest?

Explanation / Answer

m1 = 3.8x10^3 kg , m2 =7.75x10^3 kg

(A) u1 =0, u2 =0.26 m/s

From conservation of momentum

m1u1+m2u2 =(m1+m2)v

0 + (7.75x103 x 0.26)=(3.8+7.75)x103 v

v = 2.015/11.55 =0.174 m/s

(B) Ki =(1/2)m1u1^2 +(1/2)m2u2^2

Ki =0+(1/2)(7.75x103)(0.26x0.26)

Ki = 262.95 J

Kf =(1/2)(m1+m2)v^2

Kf =0+(1/2)(3.8x103+7.75x103)(0.174x0.174)

Kf =174.8 J

Lost in kinetic energy = KI -Kf = 262.95 -174.8 = 88.18 J

(C) u1 = -0.26m/s, u2 =0

From conservation of momentum

m1u1+m2u2 =(m1+m2)v

- (3.8x103 x 0.26)+0=(3.8+7.75)x103 v

v = - 0.988/11.55 = -0.086 m/s

(D) Ki =(1/2)m1u1^2 +(1/2)m2u2^2

Ki =(1/2)(3.8x103)(-0.26)2

Ki = 128.44 J

Kf =(1/2)(m1+m2)v2

Kf =0+(1/2)(3.38x103+7.75x103)(-0.086)2

Kf =42.7 J

Lost in kinetic energy = KI -Kf = 128.44 -42.7 = 85.74 J

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