(50 points) A rocket of mass m is launched straight up with thrust Frust a) (3 points) Model: How will you model this problem? b) (6 points) Visualize: Draw a free body diagram of the forces acting on the rocket. (6 points) Visualize: Draw a pictorial representation of the rocket and unknown kinematic quantities. d) (15 points) Solve: Find the acceleration of the rocket in terms of known quantities m, thrust, and q e) (15 points) Solve: Find an expression for the rockets speed at height h if air resistance is neglected. (5 points)Assess: Verify that your expressions for parts d) and e) give you the correct dimensions of L/T2 and L/T, respectively. f)

a) use the law of conservation of the moment where the system is formed by the rocket plus the ejected fuel

c)  Δm v ’= m Δv,   v = v’ ln (m₀ / m)

d)  a = - v’/m  dm/dt

e)   y = v’/C  [(m- m ln ( m)]

d)

Explanation:

a) To model this problem we can use the law of conservation of the moment where the system is formed by the rocket plus the ejected fuel that creates the propulsion

b) in the attached we can see a diagram of the free body the cohetees the aja grade and the fuel the small box

  The ejected mass is m ’= Δm

c) we write the initial moment

                 p₀ = (m + Δm) v

     Final moment

                pf = m (v + Δv) + Δm (v- v ’)

Where v ’is the gas velocity relative to the rocket

              p₀ = pf

             (m + Δm) v = m (v + Δv) + Δm (v- v ’)

             Δm v ’= m Δv

The amount Δm is negative because it is a decrease in rocket masses

                  Δv = -v’ Δm / m

                  dv = -v' dm/m

We integrate

                 v = v’ ln m

We evaluate between the lower limit v = 0 m = m0 and the upper limit v = v for m = m

              v = v’ ln (m₀ / m)

d) acceleration is defined

             a = dv / dt

             a = dv / dm dm / dt

             a = - v’/m  dm/dt

The amount dm/dt in the amount of fuel burned per unit of time, in general is given for a specific engine

e) the rocket speed is found in part c

The position, we can find it from the define of the speed

              v = dy / dt = dy /dm   dm/dt

Dm / dt = C

               dy = v dm / C

               dy = v’/C   ln (mo / m) dm

We integrate

               y = v’/C  [(m- m ln ( m)]

f)  dimensional analysis

   a = - v’/m  dm/dt

  [ m/s²] =[m/s]/[km] [km/s] = [ m/s²]

    y = v’/C  [(m- m ln ( m)]

  [m] = [m/s]/[kg/s] [kg] = [m]

note: the given question is incomplete.

complete question is given below.

which is an example of kinetic energy?

a. the energy stored in ethanol  

b. a hockey puck sliding across the ice  

c. a compressed spring  

d. a ball sitting at the top of a ramp

solution: the kinetic energy is the energy posses by the virtue of the motion of the object.

in the given options, a, c and d are the example of potential energy because it is the energy posses by the virtue of the rest of the object. but option b has the kinetic energy as the object is in motion.

hence, a hockey puck sliding across the ice   is the apex example of kinetic energy.

so, the option (b) a hockey puck sliding across the ice   is the correct option.