Ask Question
6 May, 02:28

A space vehicle is traveling at 3730 km/h relative to Earth when the exhausted rocket motor is disengaged and sent backward. The relative speed between the motor and the command module is then 85 km/h. The mass of the motor is four times the mass of the module. What is the speed of the command module relative to Earth just after the separation?

+1
Answers (2)
  1. 6 May, 02:33
    0
    the speed of the command module relative to Earth = 1017.22 m/s

    Explanation:

    We are given;

    Speed space vehicle is traveling at relative to Earth; Vi = 3730 km/h = (3730 x 10) / 36 m/s = 1036.111 m/s

    Rocket motor speed; Vmc = 85km/h = (85x10) / 36 = 23.611 m/s

    From conservation of linear momentum, we know that;

    Initial momentum = final momentum.

    Thus,

    MV_i = 4m (V_me) + mV_ce

    Where;

    M = the mass of the space vehicle which is the sum of the motor's mass and command mass

    V_i = initial speed

    V_me = speed of the motor relative to the earth

    V_ce = speed of command relative to the earth

    m = the command module mass

    4m = the mass of the motor

    Now, M = the sum of the motor's mass and command mass

    Thus; M = 4m + m = 5m

    So, we now have;

    5mV_i = 4m (V_me) + mV_ce

    Now, the velocity of the motor relative to the earth is;

    V_me = V_mc + V_ce

    Thus, we now have;

    5mV_i = 4m (V_mc + V_ce) + mV_ce

    This gives;

    5mV_i = 4mV_mc + 5mV_ce

    Divide through by m to get;

    5V_i = 4V_mc + 5V_ce

    Let's make V_ce the subject;

    5V_i - 4V_mc = 5V_ce

    Divide through by 5;

    V_ce = V_i - (4/5) V_mc

    Plugging in the relevant values;

    V_ce = 1036.111 - (4/5) •23.611 = 1017.22 m/s
  2. 6 May, 02:51
    0
    Given that,

    Speed of space ship

    V = 3730km/hr relative to earth

    Motor speed is v (mc) = 85km/hr relative to command nodule

    Let mass of command module be m

    Motor mass is 4 times module=4m

    Let command module be V (ce)

    Note, the space vehicle contain the command module and the motor

    Applying conservation of linear momentum

    Pi = Pf

    Momentum is given as p=mv

    For command module p = mv (ce)

    For motor module p=4mV (me)

    Initial momentum before Pi = MV

    M is the total mass of both the command module and motor module

    M=m+4m = 5m

    Pi = Pf

    MV = m•v (ce) + 4m•V (me)

    v (ce) velocity of command relative to the earth

    V (me) velocity of motor relative earth

    V (me) = v (mc) + v (ce)

    5mV = mv (ce) + 4m (v (mc) + v (ce))

    Divide through by m

    5V = v (ce) + 4v (mc) + 4v (ce)

    5V - 4v (mc) = 5v (ce)

    5v (ce) = 5V - 4 v (mc)

    5v (ce) = 5 * 3730 - 4 * 85

    5v (ce) = 18310

    v (ce) = 18310/5

    v (ce) = 3662km/hr

    To m/s v (ce) = 3662*1000/3600

    v (ce) = 1017.22 m/s

    command module relative to Earth just after the separation is 3662km/h or 1017.22 m/s
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “A space vehicle is traveling at 3730 km/h relative to Earth when the exhausted rocket motor is disengaged and sent backward. The relative ...” in 📗 Physics if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers