Help with this long response physics problem

Question

Students conduct an experiment to study the motion of two toy rockets. In the first experiment, rocket  X of mass  mR is launched vertically upward with an initial speed  v0 at time t=0. The rocket continues upward until it reaches its maximum height at time  t1. As the rocket travels upward, frictional forces are considered to be negligible. The rocket then descends vertically downward until it reaches the ground at time  t2. The figure above shows the toy rocket at different times of its flight. In a second experiment, which has not yet been conducted by the students, rocket  Y of mass  MR, where  MR>mR, will be launched vertically upward with an initial speed  v0 at time  t=0 until it reaches its maximum height. Rocket  Y will then descend vertically downward until it reaches the ground.Two students in the group make predictions about the motion of rocket  Y compared to that of rocket  X. Their arguments are as follows.

Student 1: “Rocket  Y will have a smaller maximum vertical displacement than rocket  X, although it is launched upward with the same speed as rocket  X and has more kinetic energy than rocket  X. Because rocket  Y will have a smaller maximum vertical displacement than rocket  X, I predict that it will take less time for rocket  Y to reach the ground compared with rocket  X.”

Student 2: “Rocket  Y will have the same maximum vertical displacement as rocket  X because both rockets have the same kinetic energy. Since both rockets will have the same maximum vertical displacement, I predict that it will take both rockets the same amount of time to reach the ground.”

(b) Use quantitative reasoning, including equations as needed, to derive expressions for the maximum heights achieved by rocket  X and rocket  Y. Express your answer in terms of  v0,  mR,  MR,  g, and/or other fundamental constants as appropriate.

Use quantitative reasoning, including equations as needed, to derive expressions for the time it takes rocket  Xand rocket  Y to reach the ground after reaching their respective maximum heights,  HX and  HY. Express your answer in terms of  v0,  mR,  MR,  HX,  HY,  g, and/or other fundamental constants as appropriate.

Answer 1

This answer is predicated on two assumptions - 1) we can neglect air resistance (stated in the question) on both the upward and downward paths, and 2) the rockets are in free-fall (not propelled) after launch.

The coordinate system is defined with up positive and down negative.

If these assumptions are valid, then Student 2 has the correct answer for the wrong reason.

Max height: The maximum height obtained by the rockets is independent of rocket mass under the assumptions above. This can be demonstrated through conservation of energy:

At the launch, all the energy is kinetic energy, E₀ = K = (1/2) m v₀².

At the maximum height, all the energy is potential energy, E_max = P = mgh_max. Because we are neglecting drag (friction, air resistance),

E₀ = E_max ==> K = P ==> (1/2) m v₀² = mgh_max ==> mass can be canceled from both sides of the equation, so the maximum height is independent of the mass of the rockets and depends only on the initial velocity and acceleration due to gravity, h_max = v₀² / (2g).

This is true regardless of which rocket is considered.

If the rocket is propelled upward with a force due to the rocket engine, then the problem gets a little bit more difficult and mass of the rocket cannot be neglected. If it is assumed that the rocket has an engine that exerts a force, F, for some time tE, then the height calculation needs to take into consideration this force. Newton's 2nd Law says that the net force acting is F - mg = ma; the net acceleration experienced by the rocket during the time the force is applied is a = (F - mg)/m. This acceleration, a, may be positive or negative depending on the magnitude of the force F - it might be positive for Rocket X and negative for Rocket Y. For Rocket X the acceleration is aX = (F - mR g)/mR and for Rocket Y it is aY = (F - MR g)/MR. Since MR > mR, aY < aX

The rocket will reach some height, hE while powered by the rocket engine. This height is given by hE = v0(tE) + (1/2) a (tE)², where a is the either aX or aY, depending on the rocket. The maximum height would be the sum of the height to this point and the height in free-fall acting only under gravity. The velocity when the engine stops power is vE = v0 + a (tE). Each rocket will have a different vE and hE because a is different, depending on the mass.

The maximum height reached, hmax = hE + vE(t1 - tE) - (1/2) g (t1 - tE)²;

Fall time: Obviously the fall time will depend on the final height. Making the assumption that the rockets fall is not slowed by a parachute (assuming free-fall), the motion equation is hf = hmax - (1/2) g t² where t is the time to fall from the maximum height to the ground. Since we are assuming the rockets fall to the ground (not down a well or on top of a house), hf = 0. Therefore the equation above can be solved for the fall time:

t = √(2(hmax)/g). Under the projectile motion assumption, the fall times for the two rockets will be equal and independent of mass. If hmax is substituted in for the time, the time to fall from max height to the ground is t = v0/g

If it is not projectile motion, then the equations for max height with appropriate substitutions for vE, hE, and a will need to be made in the original time equation.