Question

The SpaceExx delivery craft is cruising at 3550 km/h relative to the Earth when it ejects the empty rocket motor backward with a speed of 490 km/h relative to the payload module. The mass of the motor is five times the mass of the payload module. What is the speed (km/h) of the payload relative to Earth after the separation?

Answer #1

we can solve this problem using the concept of conservation of momentum

which state that with reference to one point momentum of a system is conserved.

Initial momentum = final momentum

first we will convert all velocity related to earth

Speed of rocket before seperation related to earth = 3550 Km/h

Let the speed of payload after seperation related to earth = v Km/h

and Mass of the payload after seperation = m Kg

So the speed of the motor after seperation relaed to earth = (v-490) Km/h

and Mass of the motor after seperation = 5m Kg

So the total mass of the rocket before seperation = m+5m = 6m Kg

Initial momentum = final momentum

**So the speed of the payload relative to Eart after
seperation is 3958.33 Km/h**

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