Question

a 70.0 - kg astronaut pushes to the left on a spacecraft with a force in \gravity - free\ space. the spacecraft has a total mass of 1.0×10⁴ kg. during the push, the astronaut accelerates to the right with an acceleration of 0.36 m/s². determine the magnitude of the acceleration of the spacecraft. (hint: consider newtons 3rd law and then newtons 2nd law)

a. 0.36 m/s²
b. 3.97×10⁻⁴ m/s²
c. 2.5×10⁻³ m/s²
d. 51.4 m/s²
e. 7.0×10⁻³ m/s²

Explanation:

Step1: Apply Newton's 3rd Law

According to Newton's 3rd law, the force exerted by the astronaut on the spacecraft $F_{as}$ is equal in magnitude and opposite in direction to the force exerted by the spacecraft on the astronaut $F_{sa}$. So $F_{as}=F_{sa}$.

Step2: Calculate the force on the astronaut using Newton's 2nd Law

Newton's 2nd law is $F = ma$. For the astronaut, $m_{a}=70.0$ kg and $a_{a}=0.36$ m/s². So $F_{sa}=m_{a}a_{a}=70.0\times0.36$ N.

Step3: Calculate the acceleration of the spacecraft using Newton's 2nd Law

For the spacecraft, $F_{as}=F_{sa}$ and $m_{s}=1.0\times 10^{4}$ kg. From $F = ma$, we have $a_{s}=\frac{F_{as}}{m_{s}}=\frac{m_{a}a_{a}}{m_{s}}$. Substitute $m_{a}=70.0$ kg, $a_{a}=0.36$ m/s² and $m_{s}=1.0\times 10^{4}$ kg into the formula: $a_{s}=\frac{70.0\times0.36}{1.0\times 10^{4}}=2.52\times 10^{-3}\approx2.5\times 10^{-3}$ m/s².

Answer:

C. $2.5\times 10^{-3}$ m/s²