A rocket with initial mass of 7.50×103 kg is fired in the vertical direction. Its exhaust gases are ejected at the rate of 4.50×101 kg/s with a relative velocity of 2.200×103 m/s. What is the initial acceleration of the rocket?
What is the acceleration after 20 s have elapsed?
F=ma
mass*velocity= impulse
in the first instance
massrate*velocitygas= force
45.0kg/s*2200m/s=ForceinNEwtons
in twenty seconds, the mass of the rocket is reduced to 7.50E3-45.0*20
a= force/mass= force above/newmass
To find the initial acceleration of the rocket, we can use Newton's Second Law of Motion, which states that the force acting on an object is equal to its mass multiplied by its acceleration.
1. First, we need to calculate the force exerted by the rocket's exhaust gases. This force is equal to the rate of change of momentum of the expelled gas. The rate of change of momentum is given by the mass flow rate of the gas (4.50×10^1 kg/s) multiplied by the velocity of the expelled gas (2.200×10^3 m/s).
F = (mass flow rate) x (exhaust gas velocity)
F = (4.50×10^1 kg/s) x (2.200×10^3 m/s)
F = 9.90×10^4 N
2. Since the rocket and the expelled gas are in an action-reaction pair, the force experienced by the rocket (due to the expelled gas) is equal in magnitude but opposite in direction. So, the force acting on the rocket is -9.90×10^4 N.
3. The net force acting on the rocket is the difference between the upward force exerted by the rocket engines (F) and the downward force due to gravity (mg), where m is the mass of the rocket and g is the acceleration due to gravity (9.8 m/s^2).
Net force = F - mg
Net force = -9.90×10^4 N - (7.50×10^3 kg)(9.8 m/s^2)
Net force = -9.90×10^4 N - 7.35×10^4 N
Net force = -1.63×10^5 N
4. Now we can determine the initial acceleration of the rocket using Newton's Second Law.
Net force = mass x acceleration
-1.63×10^5 N = (7.50×10^3 kg) x acceleration
Solving for acceleration:
acceleration = (-1.63×10^5 N) / (7.50×10^3 kg)
acceleration ≈ -21.73 m/s^2
Therefore, the initial acceleration of the rocket is approximately -21.73 m/s^2.
To find the acceleration after 20 seconds have elapsed, we can use the principle of conservation of momentum.
1. The mass of the rocket decreases as it ejects gas, so the mass at any time t can be calculated using the initial mass (7.50×10^3 kg) and the mass flow rate of the gas (4.50×10^1 kg/s).
mass at time t = initial mass - (mass flow rate x t)
mass at time t = 7.50×10^3 kg - (4.50×10^1 kg/s x t)
For t = 20 seconds:
mass at time 20s = 7.50×10^3 kg - (4.50×10^1 kg/s x 20s)
mass at time 20s = 7.50×10^3 kg - 9.00×10^2 kg
mass at time 20s = 6.60×10^3 kg
2. The net force acting on the rocket at any time t is the difference between the force due to the expelled gas and the force due to gravity.
Net force at time t = (mass flow rate x gas velocity) - (mass at time t x gravity)
Net force at time t = (4.50×10^1 kg/s x 2.200×10^3 m/s) - (6.60×10^3 kg x 9.8 m/s^2)
Net force at time t = 9.90×10^4 N - 6.47×10^5 N
Net force at time t = -5.48×10^5 N
3. Now we can find the acceleration at time t using Newton's Second Law.
Net force at time t = mass at time t x acceleration at time t
-5.48×10^5 N = (6.60×10^3 kg) x acceleration at time t
Solving for acceleration at time t:
acceleration at time t = (-5.48×10^5 N) / (6.60×10^3 kg)
acceleration at time t ≈ -83.03 m/s^2
Therefore, the acceleration after 20 seconds have elapsed is approximately -83.03 m/s^2.