Conservation of Momentum Space Ship Problem

[AnhTran]

Homework Statement

The payload of a spaceship accounts for 20% of its total mass. The ship is traveling in a straight line at 2100km/hr relative to some inertial observer O. When the time is right, the spaceship ejects the payload, which is moving away from the ship at 500km/hr immediately after the ejection. How fast is the spaceship now moving, as observed by O

Homework Equations

total momentum before launch equal total momentum after launch
p=mv
v(p)=v(p/s)+v(s) (relative velocity of the payload after launch)
m(s): mass of the body of the ship
m(p): mass of the payload
v(si): velocity of the body of the ship before launch
v(pi): velocity of the payload before launch
v(sf): velocity of the body of the ship after launch
v(pf): velocity of the body of the ship after launch

The Attempt at a Solution

since the total mass is not given, I let the mass of the body of the ship equal 1 and the mass of the payload equal 0.25
my equation: m(s)*v(si)+m(p)*v(pi)=m(s)*v(sf)+m(p)*v(pf)
substituting for the variable: 1(2100)+0.25(2100)=1(2100-v(s))+0.25(500+v(s))
v(s) equal -533.33km/hr, which is greater than 500, which shows that the actual velocity of the payload is opposite of velocity of the entire ship before launch. I don't know how I did it wrong so if someone can help me on this I will be very grateful.

 

[kuruman]

The problem does not specify the direction of motion of the payload relative to the direction of the initial motion. Let's assume that the direction is opposite to the velocity of the ship. In the momentum conservation equation all velocities must be relative to the inertial frame. We know what they are initially. If the payload is moving away from the ship backward at 500 km/h as seen from a passenger on the ship and the ship is moving forward at 2100 km/h, what is the velocity of the payload as seen by someone at rest on the inertial frame?

 

[AnhTran]

the velocity of the payload as seen by someone at rest at the inertial frame would be 2100km/hr-500km/hr, or 1600km/hr. So if we do not assume the direction, then how do we know which velocity of the payload is the correct one? The velocity of the payload could be 2100+500=2600.

 

[kuruman]

So if we do not assume the direction, then how do we know which velocity of the payload is the correct one?

Your guess is as good as mine. Perhaps there is a figure that accompanies the problem. If this has been assigned to you by someone, you can ask them.

 

[haruspex]

the velocity of the payload as seen by someone at rest at the inertial frame would be 2100km/hr-500km/hr, or 1600km/hr. So if we do not assume the direction, then how do we know which velocity of the payload is the correct one? The velocity of the payload could be 2100+500=2600.

The payload could have been ejected in any direction, including at right angles to the motion of the ship. But whichever the direction, neither of the two expressions you have written can be correct. The question states:

away from the ship at 500km/hr immediately after the ejection

You have taken it to be at 500km/hr relative to the motion of the ship before ejection.