Part 2) Design a toy rocket so that when placed on rails next to the dragster tr

Question

Part 2) Design a toy rocket so that when placed on rails next to the dragster track it could start at the same time as the dragster and finish at the same time. (They might not be together for the whole race though.) Here are the simplifying assumptions and rules: The total mass of the rocket is 8.00[kg] The exhaust speed is so large that the mass of the fuel used is negligible (maybe not all that realistic but okay to start with) The rocket engine provides a constant thrust but only lasts for 1.00[s]. You can assume that the rails are frictionless and that air resistance is negligible. (That's probably not realistic at the speed that is needed but we can use it as a starting point and improve the design later.) So the free body diagram for the rocket will just be a horizontal rocket force (when the rocket is on) and vertical weight and normal forces. . . . What you need to find out is the thrust of the rocket engine.

Explanation / Answer

2. mass of rocket, m = 8 kg

exhaust speed = v ( very large, so no fuel used)

thrust is constant = T

thrust duration t = 1 s

no friction and air drag

phase 1, thrust on

T = m*a ( a is acceelration in thrust mode)

a = T/8

now, distance travelled in this time

s = 0.5*a*t^2 = 0.5a = T/16 m

speed at the end of this phase vo = a*t = T/8 m/s

phase 2, thrust off

let coasting deceleration be A

and total distance convered by the dragster be S, time taken by dragster be t'

then

S - T/16 = vo(t' - 1) - 0.5*A*(t' - 1)^2

also, final speed is 0 hence

2*A*(S - T/16) = vo^2 = T^2/16

A = T^2/32(S - T/16) = T^2/2(16S - T)

S - T/16 = T(t' - 1)/8 - T^2(t' - 1)^2/4(16S - T)

T^2(t' - 1)^2/4 - T(t' - 1)(16s - T)/8 + (S - T/16)(16S - T) = 0

T^2(t' - 1)^2/4 - 2T(t' - 1)S + T^2(t' - 1)/8 + 16S^2 - 2ST + T^2/16 = 0

T^2(4t'^2 - 6t' + 3) - 32T(t'S) + 256S^2 = 0

so T = (32t'S +- sqroot((32t'S)^2 - 4*(4t'^2 - 6t' + 3)*256S^2))/2*(4t'^2 - 6t' + 3)

where S is total distance covered by the dragster in time t'

3. when the rocket is launched up

T - mg = ma ( a is acceleration of rocket)

a = (T - mg)/m = T/8 - g

so final speed after this thrust, vo = a*1 = T/8 - g m/s

distance covered in this time s1 = 0.5*a*1^2 = T/16 - g/2 m

after this, height reached by rocekt = s2

2*g*s2 = vo^2 = (T/16 - g/2)^2

s2 = (T/16 - g/2)^2/2g

hence total distaqnce covered before the rocket falls = s1 + s2 =T/16 - g/2 +(T/16 - g/2)^2/2g

where T can be found form the previous part

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