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Sagot :
Answer:
Approximately [tex]9.64\; {\rm m\cdot s^{-1}}[/tex] (assuming that the velocity of the exhaust is [tex](-100)\; {\rm m\cdot s^{-1}}[/tex].
Explanation:
When an object of mass [tex]m[/tex] travels at a velocity of [tex]v[/tex], the momentum [tex]p[/tex] of this object will be [tex]p = m\, v[/tex].
Assume that there is no external force on this spaceship. The total momentum of the ship and the exhaust will be conserved. In other words,
[tex]\begin{aligned}& (\text{Momenum of Spaceship, before}) \\ &+ (\text{Momentum of Exhaust, before}) \\ =\; & (\text{Momenum of Spaceship, after}) \\ &+ (\text{Momentum of Exhaust, after})\end{aligned}[/tex].
Rearrange to find the momentum of the spaceship after releasing the exhaust:
[tex]\begin{aligned} & (\text{Momenum of Spaceship, after}) \\ =\; & (\text{Momenum of Spaceship, before}) \\ &+ (\text{Momentum of Exhaust, before}) \\ &- (\text{Momentum of Exhaust, after})\end{aligned}[/tex].
It is given that the spaceship and the exhaust were initial stationary. Hence, initial momentum will be [tex]0\; {\rm kg \cdot m\cdot s^{-1}}[/tex] for both the ship and the exhaust.
[tex]\begin{aligned} & (\text{Momenum of Spaceship, after}) \\ =\; & (0\; {\rm kg \cdot m\cdot s^{-1}}) \\ &+ (0\; {\rm kg \cdot m\cdot s^{-1}}) \\ &- (\text{Momentum of Exhaust, after})\end{aligned}[/tex].
Since the exhaust is of mass [tex]10600\; {\rm kg}[/tex] and velocity [tex](-100)\; {\rm m\cdot s^{-1}}[/tex], the momentum of the exhaust after release will be:
[tex]\begin{aligned} & (\text{Momenum of Exhaust, after}) \\ =\; & (\text{mass of Exhaust})\, (\text{Velocity of Exhaust, after}) \\ =\; & (10600.0\; {\rm kg})\, ((-100)\;{\rm m \cdot s^{-1}}) \\ =\; & (-1.06000\times 10^{6})\; {\rm kg \cdot m\cdot s^{-1}}\end{aligned}[/tex].
Divide the momentum of the spaceship by mass to find velocity:
[tex]\begin{aligned} & (\text{Velocity of Spaceship, after}) \\ =\; & \frac{(\text{Momentum of Spaceship})}{(\text{mass of Spaceship})} \\ =\; & \frac{((-1.06000\times 10^{6})\; {\rm kg \cdot m\cdot s^{-1}})}{(110000.0\; {\rm kg})} \\ \approx\; & 9.64\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].
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