emplar
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iv) State and explain how you would expect the acceleration of the spacecraft to change, if at all, while the engine is running.

As the mass is decreasing, the resultant force will be getting smaller due to F = rate of change of momentum/time and the mass decreasing. So then F = ma and as F decreases and m decreases, a stays the same.

However the answer says the Force is constant so as the mass decreases, the acceleration increases.

Why does the force stay constant?
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Graphittte
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The mass is being ejected at a constant rate.
Force = rate of change of momentum but since the mass is decreasing at a constant rate then the force must be constant as the rate of change of momentum stays the same.
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emplar
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(Original post by Graphittte)
The mass is being ejected at a constant rate.
Force = rate of change of momentum but since the mass is decreasing at a constant rate then the force must be constant as the rate of change of momentum stays the same.
F = m(v-u)/t if m decreases constantly, wouldn't, both v and u have to decrease at a constant rate for the change in momentum to be the same? As if only m decreased at a constant rate than surely the change in momentum would decrease as well?
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Eimmanuel
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(Original post by emplar)
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iv) State and explain how you would expect the acceleration of the spacecraft to change, if at all, while the engine is running.

As the mass is decreasing, the resultant force will be getting smaller due to F = rate of change of momentum/time and the mass decreasing. So then F = ma and as F decreases and m decreases, a stays the same.

However the answer says the Force is constant so as the mass decreases, the acceleration increases.

Why does the force stay constant?
(Original post by emplar)
F = m(v-u)/t if m decreases constantly, wouldn't, both v and u have to decrease at a constant rate for the change in momentum to be the same? As if only m decreased at a constant rate than surely the change in momentum would decrease as well?
The general equation of motion that describe this system is

 {\mathbf  {F}}_{{{\mathrm  {ext}}}}+{\mathbf  {v}}_{{{\mathrm  {rel}}}}{\dfrac  {{\mathrm  {d}}m}{{\mathrm  {d}}t}}=m \dfrac{{\mathrm{d}}{\mathbf  v}}{\mathrm{d}t }  -----eqn(1)

where Fext is the net external force on the body, vrel is the relative velocity of the escaping or incoming mass with respect to the center of mass of the body, and v is the velocity of the body.

As the external force on the system is zero, we can reduce the equation to

 {\mathbf {v}}_{{{\mathrm  {rel}}}}{\dfrac  {{\mathrm  {d}}m_\text{xenon}}{{\mathrm  {d}}t}} = m_\text{spacecraft} \dfrac{{\mathrm{d}}{\mathbf  v}}{\mathrm{d}t } -----eqn(2)


 {\mathbf {v}}_{{{\mathrm {rel}}}}{\dfrac {{\mathrm {d}}m_\text{xenon}}{{\mathrm {d}}t}}  is constant and this is the force that the MS is referring to as constant.

From eqn(2), on the right hand side, the mass of the spacecraft is decreasing and the acceleration will have to increase to ensure that product is equal to the constant force on the left hand side.
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