A 160 kg astronaut (including space suit) acquires a speed of 2.20 m/s by pushing off with his legs from an 1950 kg space capsule.
What is the change in speed of the space capsule? As the reference frame, use the position of the space capsule before the push.
If the push lasts 0.47 s, what is the average force exerted on the astronaut by the space capsule?
To find the change in speed of the space capsule, we can use the principle of conservation of momentum. The total momentum before the push is equal to the total momentum after the push.
The momentum of an object is given by the product of its mass and its velocity. We need to calculate the initial momentum of the astronaut and the space capsule, and then the final momentum of both after the push.
Initial momentum of astronaut: p₁ = m₁ * v₁
where m₁ = mass of astronaut = 160 kg
v₁ = velocity of astronaut before the push = 0 m/s (since the astronaut is at rest initially)
Initial momentum of space capsule: p₂ = m₂ * v₂
where m₂ = mass of space capsule = 1950 kg
v₂ = velocity of space capsule before the push = 0 m/s (since the space capsule is also at rest initially)
Total initial momentum: p₁ + p₂ = (m₁ * v₁) + (m₂ * v₂) = (160 kg * 0 m/s) + (1950 kg * 0 m/s) = 0 kg·m/s
Final momentum of astronaut: p₁' = m₁ * v₁'
where v₁' = velocity of astronaut after the push = 2.20 m/s
Final momentum of space capsule: p₂' = m₂ * v₂'
where v₂' = velocity of space capsule after the push
Total final momentum: p₁' + p₂' = (m₁ * v₁') + (m₂ * v₂') = (160 kg * 2.20 m/s) + (1950 kg * v₂')
Since the reference frame is the position of the space capsule before the push, the final velocity of the space capsule is the negative of the final velocity of the astronaut, i.e., v₂' = -v₁'.
Therefore, the equation becomes:
(160 kg * 2.20 m/s) + (1950 kg * (-v₁')) = 0 kg·m/s
Now, we can solve for v₁':
(160 kg * 2.20 m/s) - (1950 kg * v₁') = 0 kg·m/s
352 kg·m/s - 1950 kg·v₁' = 0 kg·m/s
1950 kg·v₁' = 352 kg·m/s
v₁' = 352 kg·m/s / 1950 kg
v₁' ≈ 0.18 m/s
Therefore, the change in speed of the space capsule (Δv) is equal to the magnitude of the final velocity of the space capsule:
Δv = |v₂' - v₂| = |0.18 m/s - 0 m/s| = 0.18 m/s
So, the change in speed of the space capsule is approximately 0.18 m/s.
To calculate the average force exerted on the astronaut by the space capsule, we can use Newton's second law which states that force is equal to the rate of change of momentum:
Force = Δp / Δt
Here, Δp is the change in momentum of the astronaut and Δt is the duration of the push, which is given as 0.47 s.
The change in momentum of the astronaut is given by:
Δp = m₁ * Δv
= 160 kg * 0.18 m/s
≈ 28.8 kg·m/s
Plugging in the values, the average force exerted on the astronaut is:
Force = Δp / Δt
= 28.8 kg·m/s / 0.47 s
≈ 61.28 N
Therefore, the average force exerted on the astronaut by the space capsule is approximately 61.28 Newtons.