Question

A 70 kg astronaut floating in space in a 110 kg MMU (manned maneuvering unit) experiences an acceleration of 0.029m/s² when he fires one of the MMU

Answer 1

a)
F=ma = mass astronaut + mmu * a

F = v * dm/dt, v = 490

dm/dt is mass use per second

put the value and get answer

b)
53g/5 * 490

Answers:(a) In 5.0 s, 53 g of fuel will be ejected; (b) the thrust is 5.19 N..

Answer 2

a)
F=ma = mass astronaut + mmu * a

F = v * dm/dt, v = 490
In 5.0 s, 53 g of fuel will be ejected
dm/dt is mass use per second

go figure


b)
53g/5 * 490

the thrust is 5.19 N

Answer 3

a)

writing the balancing equation of force on the rocket,

$(m+M)(g-a)= M_{g}\frac{dv}{dt}$

$(70+110)(9.8-0.029)= M_{g}\frac{490}{5}$

$\Rightarrow M_{g}=\frac{1758.78}{98}=17.9467Kg$

b) Thrust with the final velocity

$F = M_{g}\frac{dv}{dt}=17.9467 *490/5=1758.708N$

Answer 4

a)
F=ma = mass astronaut + mmu * a

F = v * dm/dt, v = 490

dm/dt is mass use per second

b)
53g/5 * 490

Answers:(a) In 5.0 s, 53 g of fuel will be ejected; (b) the thrust is 5.19 N..

The thrust is F = dm/dt * v where dm/dt is the amount of mass being ejected in kg/second. v is the velocity of the ejected material.

The acceleration on the astronaut is a = F/M where F is the thrust and m is the total mass 70kg + 110kg = 180kg

so a = dm/dt * v / M