Consider a rocket in space that ejects burned fuel at a speed of vex= 2.0 km/s with respect to the rocket. The rocket burns 8 % of its mass in 300 s (assume the burn rate is constant).
(a) What is the speed v of the rocket after a burn time of 150.0 s? (suppose that the rocket starts at rest; and enter your answer in m/s)
v=
(b) What is the instantaneous acceleration a of the rocket at time 150.0 s after the start of the engines?(in m/s2)
a=
To solve this problem, we need to apply the principle of conservation of momentum. The momentum of the rocket and the ejected fuel before the burn is equal to the momentum of the rocket after the burn.
(a) To find the speed of the rocket after 150.0 s, we need to determine the mass of the rocket before the burn. Assuming the initial mass of the rocket is M, the mass of the fuel burned would be 0.08M (since 8% of the mass is burned). Therefore, the remaining mass of the rocket after the burn is (1 - 0.08)M = 0.92M.
Applying the conservation of momentum, the initial momentum of the system (rocket + fuel) is zero since the rocket starts at rest. The final momentum after the burn is the product of the mass of the rocket after the burn and its velocity:
0 = (0.92M)(v) - (0.08M)(vex)
Simplifying the equation:
0 = 0.92Mv - 0.08M(2.0 km/s)
To convert km/s to m/s, we need to multiply by 1000:
0 = 0.92Mv - 0.08M(2.0 km/s)(1000 m/km)
0 = 0.92Mv - 0.08M(2000 m/s)
0 = 0.92Mv - 160M
Now, we can solve for v:
0.92Mv = 160M
v = 160M / 0.92M
v = 173.91 m/s
Therefore, the speed of the rocket after 150.0 s of burn time is 173.91 m/s.
(b) To find the instantaneous acceleration of the rocket at time 150.0 s, we can calculate the change in velocity over a small time interval. The instantaneous acceleration (a) can be approximated as the change in velocity (Δv) divided by the time interval (Δt). In this case, we'll take a very small time interval, Δt, around 150.0 s.
To find Δv, we can subtract the velocity at 150.0 s from the velocity at 150.0 + Δt s:
Δv = v(150.0 + Δt) - v(150.0)
Since the rocket is in space, there are no external forces acting on it, except for the ejected fuel. Therefore, the average force acting on the rocket during the burn is given by the rate of change of momentum:
Force = (Change in momentum) / (Change in time)
Force = [(0.92Mv) - (0.92Mvex)] / Δt
The rocket experiences a constant force during the burn, so the average force can be approximated as the instantaneous force at any time during the burn.
The mass flow rate of the fuel (dm/dt) is given by 0.08M / 300 s since 8% of the mass is burned in 300 s. This can also be written as (0.08M / 300) kg/s.
The instantaneous force at any time during the burn is given by the product of the mass flow rate of the fuel and the ejection velocity:
Force = (dm/dt) * vex
Substituting (0.08M / 300) kg/s for dm/dt and 2.0 km/s for vex:
Force = (0.08M / 300) kg/s * (2.0 km/s)(1000 m/km)
Force = (0.08M / 300) kg/s * (2000 m/s)
Force = (0.08M * 2000) kg⋅m/s² / 300
Force = 0.53M kg⋅m/s² / 300
Since the force acting on the rocket is equal to its mass multiplied by its acceleration:
0.53M kg⋅m/s² / 300 = M * a
Simplifying the equation:
0.53 / 300 = a
Now, let's calculate the value of a:
a = 0.53 / 300
a ≈ 0.00177 m/s²
Therefore, the instantaneous acceleration of the rocket at time 150.0 s after the start of the engines is approximately 0.00177 m/s².