A space vehicle is traveling at 5425 km/h relative to the Earth when the exhausted rocket motor (mass 4m) is disengaged and sent backward with a speed of 81 km/h relative to the command module (mass m). What is the speed of the command module relative to Earth just after the separation

Explanation:

Given data

Space vehicle speed=5425 km/h relative to earth

The rocket motor speed=81 km/h  and mass 4m

The command has mass m

From the conservation of momentum as the system isolated

Since the motion in on direction we can drop the unit vector direction

Where M is the mass of space vehicle which equals to sum of the motors mass and command mass.

The velocity of the motor relative to the earth equals the velocity of the motor relative to command plus the velocity of the command relative to earth

Where Vmc is the velocity of motor relative to command

This yields

Substitute the given values

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft