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A 600 kg rocket motor is released from a 280 kg capsule. The combined shuttle was traveling at 6800 m/s. The capsule is ejected in the opposite direction at 7300 m/s. What is the speed of the rocket m

Physics

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Answer

6.2

13380 m/s\text{13380 m/s}

Step-by-step solution

6 steps

Part 6.1

The principle of the conservation of momentum states that the total linear momentum of an isolated system remains constant in both magnitude and direction, provided no external forces act on the system.

Part 6.2

13380 m/s\text{13380 m/s}

  1. Identify the known values and define a positive direction.

    Let the initial direction of the space shuttle be positive.
    Mass of rocket motor, mR=600 kgm_R = 600 \, \text{kg}
    Mass of capsule, mC=280 kgm_C = 280 \, \text{kg}
    Initial velocity of the combined shuttle, vi=+6800 m/sv_i = +6800 \, \text{m/s}
    Final velocity of the capsule, vC=−7300 m/sv_C = -7300 \, \text{m/s} (opposite direction)
    We need to find the final velocity of the rocket motor, vRv_R.

  2. Apply the principle of conservation of linear momentum.

    The total momentum before the release is equal to the total momentum after the release.

    (mR+mC)vi=mRvR+mCvC(m_R + m_C) v_i = m_R v_R + m_C v_C

  3. Substitute the values and solve for vRv_R.

    (600 kg+280 kg)(6800 m/s)=(600 kg)vR+(280 kg)(−7300 m/s)(600 \, \text{kg} + 280 \, \text{kg}) (6800 \, \text{m/s}) = (600 \, \text{kg}) v_R + (280 \, \text{kg}) (-7300 \, \text{m/s}) (880 kg)(6800 m/s)=600vR−2044000 kg⋅m/s(880 \, \text{kg}) (6800 \, \text{m/s}) = 600 v_R - 2044000 \, \text{kg}\cdot\text{m/s} 5984000 kg⋅m/s=600vR−2044000 kg⋅m/s5984000 \, \text{kg}\cdot\text{m/s} = 600 v_R - 2044000 \, \text{kg}\cdot\text{m/s} 5984000+2044000=600vR5984000 + 2044000 = 600 v_R 8028000=600vR8028000 = 600 v_R vR=8028000600v_R = \frac{8028000}{600} vR=13380 m/sv_R = 13380 \, \text{m/s}

    The speed of the rocket motor immediately after it is released from the capsule is 13380 m/s\boxed{\text{13380 m/s}}.

Part 6.3· 3 steps

The "collision" (separation) is inelastic.

  1. Calculate the total kinetic energy before the separation.

    KEi=12(mR+mC)vi2KE_i = \frac{1}{2} (m_R + m_C) v_i^2 KEi=12(600 kg+280 kg)(6800 m/s)2KE_i = \frac{1}{2} (600 \, \text{kg} + 280 \, \text{kg}) (6800 \, \text{m/s})^2 KEi=12(880 kg)(46240000 m2/s2)KE_i = \frac{1}{2} (880 \, \text{kg}) (46240000 \, \text{m}^2/\text{s}^2) KEi=440×46240000 JKE_i = 440 \times 46240000 \, \text{J} KEi=20345600000 JKE_i = 20345600000 \, \text{J}

  2. Calculate the total kinetic energy after the separation.

    KEf=12mRvR2+12mCvC2KE_f = \frac{1}{2} m_R v_R^2 + \frac{1}{2} m_C v_C^2 KEf=12(600 kg)(13380 m/s)2+12(280 kg)(−7300 m/s)2KE_f = \frac{1}{2} (600 \, \text{kg}) (13380 \, \text{m/s})^2 + \frac{1}{2} (280 \, \text{kg}) (-7300 \, \text{m/s})^2 KEf=300×179024400 J+140×53290000 JKE_f = 300 \times 179024400 \, \text{J} + 140 \times 53290000 \, \text{J} KEf=53707320000 J+7460600000 JKE_f = 53707320000 \, \text{J} + 7460600000 \, \text{J} KEf=61167920000 JKE_f = 61167920000 \, \text{J}

  3. Compare the initial and final kinetic energies.

    Since KEf=61167920000 JKE_f = 61167920000 \, \text{J} and KEi=20345600000 JKE_i = 20345600000 \, \text{J}, the kinetic energy is not conserved (KEf≠KEiKE_f \neq KE_i). In fact, the kinetic energy has increased. This increase comes from the chemical energy stored in the rocket motor's fuel, which is converted into kinetic energy during the separation process. Therefore, the event is inelastic.

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