Question 18
Determine the force required to launch a rocket with a mass of 2000 kg vertically from rest to a height of 20 km in 20 seconds. Assume that the rocket’s mass and acceleration due to gravity remain constant during the launch and that air resistance is not a factor.
Include a free-body diagram showing the forces at launch. Answer to the nearest whole unit in kN.
The engines have to hold up the weight and also supply the net force
The rocket climbs 20 km in 20 s at a steady acceleration, shown at twice real speed. Watch the velocity-time graph draw itself, then see the launch force split into its two jobs.
The motion gives you the net force, $ma$. The engines produce that much more than the weight, so the launch force is $ma+mg$.
Change the rocket and its climb, then watch the launch force
Set the mass, the height and the time. The question is 2000 kg, 20 km and 20 s. Then switch the weight off to see the answer most people stop at.
This is the question. The motion gives $a=100\text{ m/s}^2$, so the net force is 200 kN. The engines also hold up 19.6 kN of weight, which makes 219.6 kN, or 220 kN to the nearest whole unit.
This is the answer you get when you stop at $F=ma$. It is the net force, the part left over after the weight is beaten. A rocket producing only this much would accelerate at less than you need, so add the weight back.
A gentle climb needs only a small net force, so now the weight is a big share of the launch force. Leaving it out here would be a large error, not a small one.
Halve the time and the acceleration goes up four times, because $t$ is squared in $s=ut+\tfrac12at^2$. The weight only changes with the mass.
Work through the solution one mark at a time
Try each step yourself before you reveal it.
Draw the rocket at launch with two labelled arrows from it: the launch force up and the weight down. Make the launch force the longer arrow, because the rocket accelerates upwards.
No air resistance arrow. The question rules it out.
Marker: provides an appropriate free-body diagram. Note: “An appropriate free-body diagram includes both weight and launch force vectors.”
Find the weight. It is not on the formula sheet, so write it from $F=ma$ with $a=g$.
$$\begin{aligned}mg&=2000\times9.8\\&=19\,600\text{ N}\end{aligned}$$
Marker: correctly determines weight
List the motion. From rest gives $u=0$, the height is $s=20\,000$ m and the time is $t=20$ s. You want $a$ and have no final velocity, so use the equation without $v$. Write it, then write it again with the numbers in.
$$\begin{aligned}s&=ut+\tfrac12at^2\\20\,000&=(0\times20)\\&\quad+\left(\tfrac12\times a\times20^2\right)\end{aligned}$$
Marker: provides correct formula and substituted values for acceleration
Solve for the acceleration.
$$\begin{aligned}20\,000&=\tfrac12\times a\times400\\a&=\frac{40\,000}{400}\\&=100\text{ m/s}^2\end{aligned}$$
The 20 km must go in as 20 000 m. Left as 20, it gives $a=0.1\text{ m/s}^2$, and the launch force comes out at 19.8 kN, barely more than the weight.
Marker: correctly determines acceleration
$F=ma$ gives the net force. The net force is the launch force minus the weight, so add the weight back.
$$\begin{aligned}F_{net}&=F_{\text{launch}}-19\,600\\F_{net}&=ma=2000\times100\\&=200\,000\text{ N}\end{aligned}$$
$$\begin{aligned}F_{\text{launch}}&=200\,000+19\,600\\&=219\,600\text{ N}\end{aligned}$$
The marking guide prints this line as “2019 600N”. It is a typing slip for 219 600 N, and its own final answer is 220 kN. Follow-through is allowed, so a wrong weight or a wrong acceleration from earlier still earns this mark if you combine them correctly.
Marker: correctly determines the answer in kN to the nearest whole unit. Note: “Allow FT error for incorrect: weight when determining the answer, acceleration when determining the answer.”
Each phrase of the question gave you something
The 2022 subject report chose this question as a sample of effective practice. The responses it praised drew a free-body diagram with both the weight and the launch force, and gave the answer in kN to the nearest whole unit, with every step laid out.
Everything you need is in the stem and the chain is a familiar one: weight, acceleration from one motion equation, then the force. The 2022 subject report calls it “This simple familiar question”, so the band is QCAA’s own.
If I had to guess where you might lose marks, it would be stopping at $F=ma=200$ kN. That is the net force, and the engines have to beat the weight before any of it is left over. The other two traps are leaving 20 km as 20 and drawing the diagram with only one arrow.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording, any figures and marking-guide steps are from the 2022 QCAA Engineering external assessment, © State of Queensland (QCAA) 2022, licensed under CC BY 4.0, and have been adapted, except where a third-party source is named. Tangent Tuition is not affiliated with the QCAA.
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