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Valentina, the Russian Cosmonaut, goes outside her ship for a spacewalk, butwhen she is floating 15 m from the ship, her tether catches on a sharp piece ofmetal and is severed. Valentina tosses her 2.0-kg camera away from thespaceship with a speed of 12 m/s. a) How fast will Valentina, whose mass isnow 68 kg, travel toward the spaceship? b) Assuming the spaceship remainsat rest with respect to Valentina, how long will it take her to reach the ship?

Respuesta :

Answer:

a) 0.35 m/s

b) 42.5 s

Explanation:

Part a)

By the conservation of momentum, we can write the following equation

[tex]\begin{gathered} p_i=p_f \\ 0=m_cv_c+m_vv_v \end{gathered}[/tex]

Where mc and vc are the mass and velocity of the camera and mv and Vv are the mass and velocity of Valentina. Solving for Vv, we get:

[tex]\begin{gathered} -m_cv_c=m_vv_v \\ v_v=\frac{-m_cv_c}{m_v} \end{gathered}[/tex]

If we take the direction away from the spaceship as negative, we can replace mc = 2.0 kg, vc = -12 m/s, and mv = 68 kg to get

[tex]v_v=\frac{-2.0\text{ kg \lparen-12 m/s\rparen}}{68\text{ kg}}=0.35\text{ m/s}[/tex]

Therefore, Valentina has a velocity of 0.35 m/s towards the spaceship.

Part b)

We know that the speed is 0.35 m/s and the distance that she needs to travel is 15 m. Then, the time that she takes to reach the ship is calculated as

[tex]\begin{gathered} v=\frac{d}{t}\Rightarrow t=\frac{d}{v} \\ t=\frac{15\text{ m}}{0.35\text{ m/s}}=42.5\text{ s} \end{gathered}[/tex]

Therefore, Valentina will take 42.5 s to reach the ship.

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