I am definitely lost on this problem. Can someone help me out?

A(n) 51.3 kg astronaut becomes separated
from the shuttle, while on a spacewalk. She
finds herself 50.3 m away from the shuttle
and moving with zero speed relative to the
shuttle. She has a(n) 0.537 kg camera in her
hand and decides to get back to the shuttle
by throwing the camera at a speed of 12 m/s
in the direction away from the shuttle.
How long will it take for her to reach the
shuttle? Answer in units of min.

Think conservation of momentum here. Relative to the shuttle, the momentum of the astronaut and camera is 0 (zero velocity realtive to the shuttle). After she throws the camera, her momentum plus the momentum of the camera is still 0.

Consider a coordinate sytaem moving with the shuttle. Apply the law of conservation of momentumin that coordinate system.

Since total momentum will be conserved when she throws away the camera, Mastronaut*Vrecoil= = Mcamera*Vcamera
Vrecoil = (.537)/51.3)*12 m/s
= 0.1256 m/s
Time to reach shuttle = 50.3 m/0.1256 m/s = 400 seconds = 6 2/3 minutes

To solve this problem, we can use the principle of conservation of momentum. The momentum before she throws the camera is zero since she has zero speed. Therefore, the momentum after she throws the camera should also be zero, since no external forces act on her.

Let's denote the velocity of the astronaut after throwing the camera as V_a, and the velocity of the camera as V_c. The mass of the astronaut is m_a = 51.3 kg, and the mass of the camera is m_c = 0.537 kg. The initial distance between the astronaut and the shuttle is d = 50.3 m.

According to the law of conservation of momentum, the momentum before throwing the camera equals the momentum after throwing the camera. The momentum is defined as the product of mass and velocity: momentum = mass * velocity.

Initially, the momentum of the astronaut is zero, and the momentum of the camera is m_c * V_c. After she throws the camera, the momentum of the astronaut is m_a * V_a, and the momentum of the camera becomes zero.

So, we can write the conservation of momentum equation as:

0 + m_c * V_c = m_a * V_a + 0

Next, let's consider the motion of the astronaut before and after she throws the camera. Before throwing the camera, she had zero initial velocity. After throwing the camera, she has a final velocity V_a, and the distance she traveled is d.

We can use the equation of motion to relate the distance, velocity, and time. The equation is:

d = V * t

where d is the distance, V is the velocity, and t is the time.

Now, we can solve the conservation of momentum equation for V_a:

V_a = (m_c * V_c) / m_a

Then we substitute the equation of motion for the distance:

d = V_a * t

Substituting the value of V_a from the conservation of momentum equation:

d = ((m_c * V_c) / m_a) * t

Finally, solve for t:

t = (m_a * d) / (m_c * V_c)

t = (51.3 kg * 50.3 m) / (0.537 kg * 12 m/s)

t = 4829.26 s

To convert the time to minutes, divide by 60:

t = 4829.26 s / 60 s/min

t ≈ 80.49 min

Therefore, it will take her approximately 80.49 minutes to reach the shuttle.