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 10 % of its mass in 340 s (assume the burn rate is constant).

(a) What is the speed v of the rocket after a burn time of 170.0 s? (suppose that the rocket starts at rest; and enter your answer in m/s)

v=

unanswered
(b) What is the instantaneous acceleration a of the rocket at time 170.0 s after the start of the engines?(in m/s2)

a=

Relevant equation: v_f = -v_ex*ln(m_final/m_initial) - g*burn time

What I tried to do: I plugged in the data, getting -2000*ln(0.95) - g*170, but got the velocity wrong. I have no idea where I went wrong, and I am stuck here, along with finding the acceleration. Can someone please help me get unstuck here. Thanks and sorry to be a bother.

notice that the rocket described in the problem is *not* under any gravitational interaction. So really the only problem with your equation is the "-g*burn time" part.

the equation you want to solve is just

v_f = -v_ex*ln(m_final/m_initial)

How to calculate M initial and M final?

M_final is 0.95*M_initial

0.90Minitial?

To solve this problem, we can use the principle of conservation of momentum.

The rocket ejects burned fuel, so the momentum of the rocket and the fuel combined remains constant throughout the burn time.

Let's break down the steps to solve the problem:

(a) To find the speed v of the rocket after a burn time of 170.0 s:

1. Consider the initial mass of the rocket as m_initial.
2. We know that the rocket burns 10% of its mass. So, the final mass of the rocket (after burning) is 0.9 * m_initial.
3. Use the concept of conservation of momentum:
Initially, the rocket is at rest, so its momentum is zero (since momentum = mass * velocity).
After the burn, the momentum of the rocket and the ejected fuel remains constant.
So, the momentum of the fuel ejected is equal in magnitude but opposite in direction to the momentum of the rocket.
The momentum of the ejected fuel is mass_fuel * velocity_fuel, where mass_fuel is 0.1 * m_initial (since 10% of the mass is burned) and velocity_fuel is 2.0 km/s.
Therefore, the momentum of the rocket is -(0.1 * m_initial * 2.0 km/s) after the burn.
4. Let v be the speed of the rocket after the burn. Use the equation for momentum to find the speed of the rocket:
(0.9 * m_initial * v) = -(0.1 * m_initial * 2.0 km/s).
Solve for v by dividing both sides by (0.9 * m_initial):
v = -(0.1 * m_initial * 2.0 km/s) / (0.9 * m_initial).
Simplifying, we get:
v = -2.2 km/s.

Now, let's move on to part (b):

(b) To find the instantaneous acceleration a of the rocket at time 170.0 s:

1. We know that the rocket burns for 340 s (given in the problem statement).
2. The acceleration of the rocket is the derivative of its velocity with respect to time (a = dv/dt).
3. Use the relevant equation given in the problem statement: v_f = -v_ex * ln(m_final/m_initial) - g * burn time.
Rearrange the equation to solve for acceleration:
a = dv/dt = -v_ex * (-m_initial/m_initial) / (m_final/m_initial) - g.
Simplifying, we get:
a = v_ex * (-1) / (m_final/m_initial) - g,
where v_ex is the ejection speed of the fuel (2.0 km/s), m_initial is the initial mass of the rocket, m_final is the final mass of the rocket (0.9 * m_initial), and g is the acceleration due to gravity.

Now, substitute the values into the equation:

a = (2.0 km/s) * (-1) / (0.9) - g.

Be sure to convert the units consistently to get the correct answer.

Note: The equation mentioned in your question, v_f = -v_ex * ln(m_final/m_initial) - g * burn time, can also be used to find the speed, but you need to consider the negative sign on the right-hand side to handle the direction change.