A space vehicle is traveling at 4500 km/h relative to the Earth when the exhausted rocket motor is disengaged and sent backward with a speed of 70 km/h relative to the command module. The mass of the motor is four times the mass of the module. What is the speed of the command module relative to Earth after the separation? km/h
Question
A space vehicle is traveling at 4500 km/h relative to the Earth when the exhausted rocket motor is disengaged and sent backward with a speed of 70 km/h relative to the command module. The mass of the motor is four times the mass of the module. What is the speed of the command module relative to Earth after the separation? km/h
Solution
To solve this problem, we can use the principle of conservation of momentum. The total momentum before the separation is equal to the total momentum after the separation.
Step 1: Define the variables Let's denote:
- M as the mass of the module
- m as the mass of the motor, which is 4M
- V as the speed of the module after the separation
- v as the speed of the motor after the separation, which is 70 km/h relative to the module, so it's V - 70 km/h relative to the Earth
- V0 as the initial speed of the vehicle, which is 4500 km/h
Step 2: Write down the conservation of momentum equation Before the separation, the total momentum is (M + m) * V0 = (M + 4M) * 4500 = 5M * 4500. After the separation, the total momentum is M * V + m * v = M * V + 4M * (V - 70). So we have the equation 5M * 4500 = M * V + 4M * (V - 70).
Step 3: Solve the equation We can simplify the equation by dividing all terms by M, which gives us 5 * 4500 = V + 4 * (V - 70). Solving this equation gives us V = 4540 km/h.
So, the speed of the command module relative to Earth after the separation is 4540 km/h.
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