Question 9
A 1200 kg roller coaster car starts from rest and rolls down a section of track. It reaches a speed of 12 m/s at the bottom. Assume friction is negligible.
What is the vertical height that the car descends?
Not quite. You used $v$ instead of $v^2$: $\dfrac{12}{2 \times 9.8} = 0.61 \approx 0.6\text{ m}$.
That is it. Step through the hints below to see why.
Not quite. You left out the $\frac{1}{2}$ in KE: $\dfrac{12^2}{9.8} = 14.69 \approx 14.7\text{ m}$.
Not quite. You left out $g$: $\dfrac{12^2}{2} = 72.0\text{ m}$.
The track drops by the height the option you are testing gives (dashed orange). Release the car: with no friction, every joule of PE it loses becomes KE.
A 0.6 m drop only gets the car to about 3.4 m/s, not 12 m/s. You used $v$ where the formula needs $v^2$.
A 7.3 m drop (7.347 m before rounding) turns 86.4 kJ of PE into KE, and the car reaches 12 m/s at the bottom. This matches the question.
A 14.7 m drop gets the car to about 17.0 m/s, which is too fast. Leaving out the ½ in KE doubles the height.
A 72 m drop gets the car to about 37.6 m/s, far too fast. Leaving out $g$ makes the height 9.8 times too big.
With no friction, the PE the car loses at the top all becomes KE at the bottom. Set the two formulas equal.
$$\begin{aligned}PE&=mgh\qquad KE=\frac{1}{2}mv^2\\mgh&=\frac{1}{2}mv^2\\h&=\frac{v^2}{2g}\end{aligned}$$
With no friction, all the PE lost becomes KE, so the mass cancels.
Substitute the speed at the bottom and g = 9.8.
$$\begin{aligned}h&=\frac{v^2}{2g}\\&=\frac{12^2}{2\times9.8}\\&=7.347\approx7.3\text{ m}\end{aligned}$$
This is option (B). You never need the 1200 kg.
$mgh=\dfrac{1}{2}mv^2$, so $h=\dfrac{v^2}{2g}=\dfrac{12^2}{2\times9.8}=7.3\text{ m}$.
The past QCAA questions this one was modelled on
Only mark this done when you could do it without help. Reading the solution does not count.
A Tangent Tuition mock exam question. It is not a past QCAA question, and Tangent Tuition is not affiliated with the QCAA.
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