Question 19
An object is swinging at the end of a 0.5 m length of string in a vertical circular path with a constant angular speed, completing each revolution in 0.24 seconds.
The object is projected from a height of 0.3 m above the ground in a vertical plane and just passes over a narrow pole as shown in the diagram. The pole is 2.05 m high and its base is 14 m horizontally from where the object was projected.
A flat-topped vehicle of length 0.6 m and height 0.4 m is initially at rest against the pole as shown in the diagram. At the instant that the object is projected, the vehicle moves in a horizontal direction away from the pole in the same vertical plane with an acceleration of magnitude of $a\ \text{m s}^{-2}$. The object strikes the middle of the top of the vehicle.
Assuming that air resistance is negligible, use vector calculus to model the motion of the projectile in order to determine the value of $a$.
Circular motion sets the speed, then it becomes a projectile
Watch the object swing, get released, clear the pole and land on the moving vehicle. This one is drawn to scale, with the flight slowed down about four times.
The object and the vehicle start moving at the same instant, so they share the same flight time.
Find the release angle that just clears the pole
The speed is fixed at 13.09 m s⁻¹, so the only thing you can change is the angle. Move the slider until the path just touches the top of the pole. There are two angles that do it.
Work through the solution one mark at a time
Try each step yourself before you reveal it. This follows the QCAA’s first method, with the origin on the ground directly below the point of release.
The object leaves the circle along the tangent, at the speed it was moving round. Find the angular speed from the period, then use $v=r\omega$.
$$\omega=\frac{2\pi}{0.24}\approx26.18\text{ rad s}^{-1}\qquad v=r\omega=0.5\times26.18\approx13.09\text{ m s}^{-1}$$
The exact value is $v=\frac{25\pi}{6}$.
Marker: correctly determines the tangential velocity of the object
Put the origin on the ground below the release point, with $\hat{\mathbf{i}}$ horizontal and $\hat{\mathbf{j}}$ up. Let $\theta$ be the release angle. Integrate the acceleration twice.
$$\mathbf{a}=-9.8\hat{\mathbf{j}}\qquad\mathbf{v}=13.09\cos(\theta)\,\hat{\mathbf{i}}+\left(13.09\sin(\theta)-9.8t\right)\hat{\mathbf{j}}$$
$$\mathbf{r}=13.09\cos(\theta)\,t\,\hat{\mathbf{i}}+\left(0.3+13.09\sin(\theta)\,t-4.9t^{2}\right)\hat{\mathbf{j}}$$
So $x=13.09\cos(\theta)\,t$ … (1) and $y=0.3+13.09\sin(\theta)\,t-4.9t^{2}$ … (2).
Marker: identifies the parametric form of the projectile path
From (1), $t=\frac{x}{13.09\cos(\theta)}$ … (3). Substitute into (2) to remove $t$.
$$y=0.3+x\tan(\theta)-\frac{0.0286x^{2}}{\cos^{2}(\theta)}$$
You keep follow-through marks from here if you slipped earlier.
Marker: determines the Cartesian form of the projectile path
“Just passes over” means the path goes through the top of the pole, $(14,\ 2.05)$. Substitute and solve for $\theta$ on your calculator.
$$2.05=0.3+14\tan(\theta)-\frac{0.0286(14)^{2}}{\cos^{2}(\theta)}\quad\Rightarrow\quad\theta\approx0.644$$
That is about 36.9°. The steeper angle $\theta\approx1.05$ also passes over the top, and the QCAA accept answers that follow from it.
Marker: determines the angle of release of the object
The object lands on the top of the vehicle, which is 0.4 m high. Put $y=0.4$ into the path.
$$0.4=0.3+x\tan(0.644)-\frac{0.0286x^{2}}{\cos^{2}(0.644)}\quad\Rightarrow\quad x\approx0.134\ \text{or}\ 16.66$$
The first value is when the object passes 0.4 m on the way up, before it reaches the pole. So the impact is at $x\approx16.66$ m.
Marker: determines the horizontal displacement at impact
Use (3) to find the flight time.
$$t=\frac{16.66}{13.09\cos(0.644)}\approx1.592\text{ s}$$
Marker: determines the time of flight
The vehicle starts from rest against the pole. Its middle is half its length past the pole, at $14.3$ m. Model it with vectors too, then match the $\hat{\mathbf{i}}$ components at $t=1.592$.
$$\mathbf{r}_{V}=\left(\frac{at^{2}}{2}+14.3\right)\hat{\mathbf{i}}+0.4\hat{\mathbf{j}}\qquad\frac{a(1.592)^{2}}{2}+14.3=16.66$$
$a\approx1.86\ \text{m s}^{-2}$
Using $s=ut+\frac{1}{2}at^{2}$ with $s=2.36$ m is also accepted. The steeper angle gives $a\approx0.27\ \text{m s}^{-2}$, which is accepted too.
Marker: determines the acceleration of the trolley
Each phrase of the question gave you something
You have circular motion, projectile motion and a second moving object all in one question, and you are not given the launch speed or the angle. You have to get the speed from the swing and the angle from the pole before you can think about the vehicle.
The detail I would check first is the 14.3. The QCAA often put a small length like the vehicle’s 0.6 m into the stem, and it is easy to use 14 or 14.6 instead of the middle.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording and marking-guide steps are from the 2020 QCAA Specialist Mathematics external assessment, © State of Queensland (QCAA) 2020, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.