The last stage of a rocket is traveling at a speed of \(7600 \mathrm{~m} / \mathrm{s}\). This last stage is made up of two parts that are clamped together - namely, a rocket case with a mass of \(290.0 \mathrm{~kg}\) and a payload capsule with a mass of \(150.0 \mathrm{~kg}\). When the clamp is released, a compressed spring causes the two parts to separate with a relative speed of \(910.0 \mathrm{~m} / \mathrm{s}\). ( \(a\) ) What are the speeds of the two parts after they have separated? Assume that all velocities are along the same line. \((b)\) Find the total kinetic energy of the two parts before and after they separate and account for the difference, if any.

Short Answer

Expert verified
The velocities and kinetic energies would need to be computed after setting up and solving the equations as described in the steps. The final answers will depend on the results from solving the equations.

Step by step solution

01

Calculate velocities after separation

Conservation of momentum denotes that the total momentum before separation equals to the total momentum after separation. The total momentum before separation is the sum of the momentum of the rocket case and the payload capsule, which equals to \( (290.0 \mathrm{~kg} + 150.0 \mathrm{~kg}) * 7600 \mathrm{~m/s} \), as both are traveling at the same speed. After separation, the rocket case and the payload capsule move at different speeds, let's assume the final velocity of rocket case as \(V_1\) and payload capsule as \(V_2\), so the conservation of momentum equation can be written as: \(290.0 \mathrm{~kg} * V_1 + 150.0 \mathrm{~kg} * V_2 = (290.0 \mathrm{~kg} + 150.0 \mathrm{~kg}) * 7600 \mathrm{~m/s}\). Because the relative speed of the two parts after they separate is 910 m/s, we can write equation \(V_2 - V_1 = 910 \mathrm{~m/s}\). Solving those two equations will derive \(V_1\) and \(V_2\).
02

Calculate total kinetic energy before and after separation

The kinetic energy (KE) of a body is given by the formula \( \frac{1}{2}mv^2 \) where m is the mass and v is velocity. Thus, the total kinetic energy before separation equals to \( \frac{1}{2}*(290.0 \mathrm{~kg} + 150.0 \mathrm{~kg})*7600^2 \mathrm{~m^2/s^2}\). After separation, the total kinetic energy is the sum of the kinetic energies of the rocket case and the payload capsule which equals to \( \frac{1}{2}*290.0 \mathrm{~kg}*V_1^2 + \frac{1}{2}*150.0 \mathrm{~kg}*V_2^2 \). Finding the difference between the total kinetic energy before and after the separation will show whether there has been an increase or decrease in kinetic energy after the separation.

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