An open train car moves with speed 11.3 m/s on a flat frictionless railroad track, with no engine pulling the car. It begins to rain. The rain falls straight down and begins to fill the train car.
The speed of the car 1. increases.
2. decreases.
3. stays the same.
A(n) 71.6 kg astronaut becomes separated from the shuttle, while on a space walk. She
finds herself 73.6 m away from the shuttle and moving with zero speed relative to the shuttle. She has a(n) 0.679 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 minutes.
Answer in units of min
22.1
1. The speed of the train car will stay the same. The rain falling vertically down will not exert any horizontal force on the train car. Therefore, the train car will continue moving with a speed of 11.3 m/s.
2. To find out how long it will take for the astronaut to reach the shuttle, we can use the principle of conservation of momentum. The initial momentum of the astronaut and the camera combined is zero, and this momentum should be conserved throughout the motion.
The momentum of an object is given by the product of its mass and velocity: momentum = mass × velocity.
Let's assume the astronaut's mass is represented by "m", the camera's mass is represented by "M", and the velocity at which the camera is thrown away from the shuttle is represented by "v".
Initially, the momentum of the astronaut and camera is zero, so we have:
(m + M) × 0 = m × 0 + M × v
Simplifying this equation, we get:
0 = M × v
From this equation, we can conclude that the mass of the camera does not affect the time it takes for the astronaut to reach the shuttle.
Now, let's consider the motion of the astronaut. From the information given, we know that the astronaut is initially at rest relative to the shuttle. After throwing the camera, the astronaut will experience a reaction force that propels her in the opposite direction. This force is equal to the momentum change of the camera divided by the time it takes for the throw to happen.
Since the initial momentum of the astronaut and the camera combined is zero, the momentum change of the camera will be equal in magnitude but opposite in direction to the momentum of the astronaut after the throw.
The final momentum of the astronaut is given by:
momentum_f = m × velocity_astronaut
Now, the momentum change of the camera can be calculated using the initial velocity of the camera and its mass:
momentum_change = M × (-velocity_camera)
Since momentum is conserved, the momentum change of the camera is equal in magnitude but opposite in direction to the momentum of the astronaut after the throw:
momentum_change = momentum_f
We can equate these two equations and solve for the velocity of the astronaut:
m × velocity_astronaut = M × (-velocity_camera)
Simplifying the equation, we get:
velocity_astronaut = (-M/M) × velocity_camera
Since the mass of the camera (M) cancels out, we get:
velocity_astronaut = -velocity_camera
Substituting the given values:
velocity_astronaut = -12 m/s
The negative sign indicates that the astronaut is propelled in the opposite direction of the throw.
Now, we can use the equation of motion to find the time it takes for the astronaut to reach the shuttle. The equation is:
distance = velocity × time
Given that the distance to the shuttle is 73.6 m and the velocity of the astronaut is -12 m/s, we can solve for time:
73.6 m = -12 m/s × time
Dividing both sides of the equation by -12 m/s:
time = 73.6 m / -12 m/s
Simplifying the equation, we get:
time = -6.1333 s
However, we are asked to provide the answer in minutes. To convert seconds to minutes, we divide the time by 60 (1 minute = 60 seconds):
time = (-6.1333 s) / 60
Calculating the result, we find:
time ≈ -0.1022 min
The negative sign indicates that the motion is in the opposite direction of the throw. However, in the context of the problem, time cannot be negative. Therefore, we can disregard the negative sign and write:
time ≈ 0.1022 min
So, it will take approximately 0.1022 minutes (or 6.1333 seconds) for the astronaut to reach the shuttle.