Can you help with parts b, c, and d? (7%) Problem 10: Two manned satellites are

Question

Can you help with parts b, c, and d?

(7%) Problem 10: Two manned satellites are approaching one another at a relative speed of 0.235 ms, intending to dock. The first has a mass of 3.8 x 103 kg, and the second a mass of 7.5 x 103 kg 25% Part (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 is initially at rest. Take the initial velocity of the second satellite to be in the positive direction. 25% Part (b) What is the change in kinetic energy in joules, in this inelastic collision? Grade Summary Deductions Potential 3% 97% sinO cotanasin) atan()acotanO coshotanh) cotanho 7 8 9 tan() acosO inh0 cos | Submissions Attempts remaining: 5 (1 % per attempt) 1 23 0 detailed view 1% 1% 1% Degrees Radians DEL CLEAR Submit Hint Feedback Feedback: 2 for a 0% deduction You may have lost the mass of the second satellite before docking I give up! Hints: 190 deduction per hint. Hints remaining: 1 2596 Part (c) Calculate the final velocity of the satellites, in meters per second. in the frame of reference in which the second satellite is initially at rest. Take the initial velocity of the first satellite to be in the negative direction 25% Part (d) what is the change in kinetic energy, in joules, in this frame of reference

Explanation / Answer

m1 = 3800 kg

m2 = 7500 kg

v = 0.235 m/s

(a) m2*v2 + m1*0 = (m1+m2)*v
7500kg * 0.235m/s + 0 = 11300kg * v
v = 0.156 m/s

(b) KEi = ½ * 7500kg * (0.235m/s)² = 207.1 J
KEf = ½ * 11300kg * (0.156m/s)² = 137.5 J
so KE loss = 69.6 J

(c) 0 - 3800kg * 0.235m/s = 11300kg * v
v = -0.079 m/s

KEi = ½ * 3800kg * (-0.235m/s)² = 104.9 J
KEf = ½ * 11300kg * (-0.079m/s)² = 35.3 J
So the KE loss is 69.6 J

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