Question

brainbuster question! 12. an 80.0 kg novice astronaut named andrew panics during a spacewalk, when he realizes he appears to have no safety cable * connecting him to the space shuttle! despite his panic, he does remember the law of action and reaction (newtons third law) from his high school physics course, which he passed with a c - grade. so, he throws a 16.0 kg tv camera toward the east at 2.5 m/s, and his 15.0 kg tool kit toward the north at 2.0 m/s. with what speed, and in what direction, will andrew travel as a result of throwing these objects away from himself? (assume that, relative to the space shuttle, he is initially at rest.)

Explanation:

Step1: Apply the law of conservation of momentum in two - dimensions

The law of conservation of momentum states that the initial momentum of the astronaut - object system is zero (since initially at rest), and the sum of the final momenta of the astronaut and the objects must also be zero. Let the mass of the astronaut be $m_{a}=80.0$ kg, the mass of the TV camera be $m_{1}=16.0$ kg with velocity $v_{1}=2.0$ m/s east, and the mass of the tool kit be $m_{2}=15.0$ kg with velocity $v_{2}=2.5$ m/s north.

Step2: Calculate the momentum of the TV camera

The momentum of the TV camera in the x - direction (east - west) is $p_{1x}=m_{1}v_{1}$. Substituting the values, we have $p_{1x}=16.0\times2.0 = 32$ kg·m/s east.

Step3: Calculate the momentum of the tool kit

The momentum of the tool kit in the y - direction (north - south) is $p_{2y}=m_{2}v_{2}$. Substituting the values, we get $p_{2y}=15.0\times2.5=37.5$ kg·m/s north.

Step4: Calculate the components of the astronaut's velocity

Since the total initial momentum is zero, the momentum of the astronaut $p_{a}$ must be equal in magnitude and opposite in direction to the sum of the momenta of the two objects. Let the velocity of the astronaut be $\vec{v}_{a}=(v_{ax},v_{ay})$. Then $m_{a}v_{ax}=-p_{1x}$ and $m_{a}v_{ay}=-p_{2y}$.
$v_{ax}=-\frac{p_{1x}}{m_{a}}=-\frac{32}{80.0}=- 0.4$ m/s (west)
$v_{ay}=-\frac{p_{2y}}{m_{a}}=-\frac{37.5}{80.0}=-0.46875$ m/s (south)

Step5: Calculate the magnitude of the astronaut's velocity

The magnitude of the velocity of the astronaut is $v_{a}=\sqrt{v_{ax}^{2}+v_{ay}^{2}}$.
$v_{a}=\sqrt{(-0.4)^{2}+(-0.46875)^{2}}=\sqrt{0.16 + 0.2197265625}=\sqrt{0.3797265625}\approx0.616$ m/s

Step6: Calculate the direction of the astronaut's velocity

The direction $\theta$ of the astronaut's velocity is given by $\tan\theta=\frac{v_{ay}}{v_{ax}}$.
$\tan\theta=\frac{- 0.46875}{-0.4}=1.171875$. So $\theta=\arctan(1.171875)\approx49.5^{\circ}$ south of west.

Answer:

The speed of Andrew is approximately $0.62$ m/s and the direction is approximately $49.5^{\circ}$ south of west.