AHL 4.28 Mixed practice: projectiles, circular motion and delayed objects · AI HL · AHL 4: Vectors

AHL 4.28

Mixed practice: projectiles, circular motion and delayed objects

AHL 4.28

Mixed practice: projectiles, circular motion and delayed objects

By the end of this section

Key points

Projectiles

Use \(g = 9.8\) m s\(^{−2}\). A projectile has acceleration \(\begin{pmatrix}0\\−9.8\end{pmatrix}\), so the horizontal velocity is constant and the vertical velocity is \(v_y = u − 9.8t\).
The angle with the horizontal comes from \(\tan\theta = \dfrac{v_y}{v_x}\). A negative \(v_y\) means the object is falling.
The greatest height is when \(v_y = 0\). It lands when \(y = 0\) again, which is twice the time to the top on level ground.
Integrate \(\mathbf{v}\) with the start at the origin to get \(\mathbf{r} = \begin{pmatrix}v_xt\\ut − 4.9t^2\end{pmatrix}\). On the ground the distance from the origin is the value of \(x\).

Circular motion

If \(|\mathbf{r}|\) is constant the path is a circle about the origin. Write \(\sin^2\theta + \cos^2\theta = 1\) in the working.
For \(\mathbf{r} = \begin{pmatrix}r\cos\omega t\\r\sin\omega t\end{pmatrix}\): \(\mathbf{v} = \begin{pmatrix}-r\omega\sin\omega t\\r\omega\cos\omega t\end{pmatrix}\), speed \(r\omega\), and \(\mathbf{a} = −\omega^2\mathbf{r}\).
\(\mathbf{v}\cdot\mathbf{r} = 0\), so the velocity is perpendicular to \(\mathbf{r}\). \(\mathbf{a} = k\mathbf{r}\) with \(k < 0\), so the acceleration points towards the centre and \(|\mathbf{a}| = \dfrac{v^2}{r}\).
One full circle takes \(T = \dfrac{2\pi}{\omega}\), so the speed is \(\dfrac{2\pi r}{T}\). The number \(\omega\) is the angular speed.

Going back from velocity to position

Integrate each component of \(\mathbf{v}\). The constants of integration come from the starting position, so use it.
Then check that \(|\mathbf{r}|\) is constant. The radius is that constant and the starting position is \(\mathbf{r}(0)\).

Objects that start later

An object that starts \(k\) seconds after the first follows the same rule but has moved for \(t − k\) seconds: replace \(t\) by \(t − k\), and only for \(t \geq k\).
Two objects collide only if their position vectors are equal at the same time \(t\). Use one component to find \(t\), then the other to find the unknown.
The distance between them is \(|\mathbf{r}_1 − \mathbf{r}_2|\). To find the least distance, graph it on the calculator and use the minimum tool.

On the calculator

Minimum of a graph

Enter the distance \(d(t)\) as a function of \(t\).
Graph it for the times when both objects are moving, with a window that shows the lowest point.
Use the minimum tool. The x-value is the time and the y-value is the least distance.

Screenshot and video to come

Explore: circular motion

The object has position r = (r cos ωt, r sin ωt). Choose the radius r and ω, then move the time slider round one full circle.

Radius, r (m) ω (radians per second) Time, t = s

Distance from the centre, |r|
Speed, |v| = rω
Time for one circle, 2π/ω

Ink arrow: the position r from the centre. Teal arrow: the velocity v. Gold arrow: the acceleration a. The arrows are drawn to a convenient length, so read the sizes from the numbers. The calculator is in radians.

Move the time slider to move the object.

Move the time slider. What stays the same and what changes: the distance from the centre, the speed, the direction of the velocity?
Make ω twice as big. What happens to the speed, to the time for one circle and to the size of the acceleration?
Make r twice as big and keep ω. What happens to the speed and to the time for one circle? Why does the time not change?
Look at r · v. Why is it always 0? What does that say about the angle between v and r?

Explore: a projectile under gravity

The velocity is (p, q − 9.8t). Drag the tip of the orange launch arrow to set p and q, and drag the launch point up to give a height. Move the time slider to follow the object. Before you press "Show the landing point", type your prediction of the range.

Horizontal velocity, p (m s⁻¹) Vertical velocity at the start, q (m s⁻¹)

Launch height, h (m) Time, t (s)

A wall (drag its top)

i is horizontal and j is upward. The acceleration is (0, −9.8). The black dot is the object. The orange arrow is the launch velocity and the teal arrow the velocity now, both to the same scale; the dashed lines are the components.

My prediction of the range m

Move the time slider to follow the object.

Set the launch height to 0. How does the time of flight compare with the time to the top? Why?
Move the time slider. Which part of the velocity changes and which part stays the same? What in the acceleration explains this?
Raise the launch point. Is the landing time still twice the time to the top? Is the landing speed still the launch speed?
Choose "Speed and angle" with θ = 45° and add the wall. Change u until the object only just clears the wall. What is the least speed? Check it by solving for u.

Worked examples

Problems like the ones in class. Pick one, try it, then open the working.

Angle, top of the flight and greatest height

A ball is hit from the origin. Its velocity at time \(t\) seconds is \(\mathbf{v} = \begin{pmatrix}16\\14 − 9.8t\end{pmatrix}\) m s⁻¹, with \(\mathbf{i}\) horizontal and \(\mathbf{j}\) vertically up. (a) Find the angle the velocity makes with the horizontal at the start and when \(t = 2\). (b) Find the time at which the ball reaches its greatest height. (c) Find an expression for the displacement from the origin, and the greatest height.

Working

Angles
\(\tan\theta = \dfrac{v_y}{v_x}\): at the start \(41.2^\circ\) above the horizontal; at \(t = 2\), \(v_y = −5.6\), \(19.3^\circ\) below

The angle comes from the two components. A negative vertical part means the ball is falling.
Top of the flight
\(14 − 9.8t = 0 \Rightarrow t = 1.43\) s

At the greatest height the vertical velocity is zero.
Displacement
\(\mathbf{r} = \begin{pmatrix}16t\\14t − 4.9t^2\end{pmatrix}\)

Integrate \(\mathbf{v}\) with \(\mathbf{r}(0) = \mathbf{0}\).
Greatest height
\(y = 14\times1.43 − 4.9\times1.43^2 = 10\) m

Use the unrounded time in the calculator: \(t = \frac{10}{7}\).

How far it lands from the origin

A shot is launched from the origin on level ground. Its velocity at time \(t\) seconds is \(\mathbf{v} = \begin{pmatrix}15\\11.76 − 9.8t\end{pmatrix}\) m s⁻¹. Find the distance from the origin to where it lands.

Working

Displacement
\(\mathbf{r} = \begin{pmatrix}15t\\11.76t − 4.9t^2\end{pmatrix}\)

Integrate with the start at the origin.
Landing time
\(y = 0:\ t(11.76 − 4.9t) = 0 \Rightarrow t = 2.4\)

The root \(t = 0\) is the launch. The other root is the landing.
Horizontal distance
\(x = 15\times 2.4 = 36\)

The ball is on the ground, so \(y = 0\) and the distance from the origin is just \(x\).

Showing motion in a circle

An object has position vector \(\mathbf{r} = \begin{pmatrix}7\cos 2t\\7\sin 2t\end{pmatrix}\) m at time \(t\) seconds. (a) Show that it moves in a circle and state the radius. (b) Find the velocity vector. (c) Show that the velocity is perpendicular to the position vector. (d) Show that the speed is constant and find it.

Working

Radius
\(|\mathbf{r}| = \sqrt{49\cos^2 2t + 49\sin^2 2t} = 7\)

Say that \(\sin^2\theta + \cos^2\theta = 1\). The distance from the origin is constant, so it is a circle of radius 7 m.
Velocity
\(\mathbf{v} = \begin{pmatrix}-14\sin 2t\\14\cos 2t\end{pmatrix}\)

Differentiate each part. The 2 from the chain rule multiplies the 7.
Perpendicular
\(\mathbf{v}\cdot\mathbf{r} = −98\sin 2t\cos 2t + 98\sin 2t\cos 2t = 0\)

A scalar product of zero means the vectors are perpendicular.
Speed
\(|\mathbf{v}| = \sqrt{196\sin^2 2t + 196\cos^2 2t} = 14\) m s⁻¹

No \(t\) is left, so the speed is constant.

Acceleration towards the centre

An object moves with position vector \(\mathbf{r} = \begin{pmatrix}5\cos 0.4t\\5\sin 0.4t\end{pmatrix}\) m. (a) Find the speed from the velocity vector. (b) Find the acceleration and show that \(\mathbf{a} = k\mathbf{r}\), where \(k\) is a constant to find. (c) State the direction of the acceleration. (d) Show that its magnitude is constant and find it.

Working

Velocity and speed
\(\mathbf{v} = \begin{pmatrix}-2\sin 0.4t\\2\cos 0.4t\end{pmatrix}\), \(|\mathbf{v}| = 2\) m s⁻¹

The speed is the magnitude of \(\mathbf{v}\).
Acceleration
\(\mathbf{a} = \begin{pmatrix}-0.8\cos 0.4t\\−0.8\sin 0.4t\end{pmatrix}\)

Differentiate \(\mathbf{v}\) once more.
Compare with r
\(\mathbf{a} = −0.16\begin{pmatrix}5\cos 0.4t\\5\sin 0.4t\end{pmatrix} = −0.16\mathbf{r}\), so \(k = −0.16\)

Take out the factor that turns \(\mathbf{r}\) into \(\mathbf{a}\).
Direction
k < 0

The acceleration points opposite to \(\mathbf{r}\): towards the centre of the circle.
Magnitude
\(|\mathbf{a}| = 0.16\times 5 = 0.8\) m s⁻²

It is constant. Check: \(\dfrac{v^2}{r} = \dfrac{2^2}{5} = 0.8\).

From velocity back to the circle

An object has velocity \(\mathbf{v} = \begin{pmatrix}-1.6\sin 2t\\1.6\cos 2t\end{pmatrix}\) m s⁻¹. At \(t = 0\) its position vector is \(\begin{pmatrix}0.8\\0\end{pmatrix}\) m. (a) Find the position vector at time \(t\), show that the object moves in a circle about the origin and state the radius. (b) Write down the starting position. (c) Find the time \(T\) when it first returns to its starting position. (d) Verify that the speed equals \(\dfrac{2\pi r}{T}\).

Working

Integrate
\(\mathbf{r} = \begin{pmatrix}0.8\cos 2t + c_1\\0.8\sin 2t + c_2\end{pmatrix}\), and \(\mathbf{r}(0) = \begin{pmatrix}0.8\\0\end{pmatrix}\) gives \(c_1 = c_2 = 0\)

The constants come from the starting position.
Circle and radius
\(|\mathbf{r}| = 0.8\sqrt{\cos^2 2t + \sin^2 2t} = 0.8\)

A circle about the origin of radius 0.8 m. The start is \(\begin{pmatrix}0.8\\0\end{pmatrix}\).
Time to return
\(2T = 2\pi \Rightarrow T = \pi = 3.14\) s

The angle \(2t\) must reach \(2\pi\).
Check the speed
\(\dfrac{2\pi\times0.8}{\pi} = 1.6\) m s⁻¹

This equals \(|\mathbf{v}| = 1.6\).

An object that starts later

A buggy starts at the origin when \(t = 0\). Its velocity is \(\mathbf{v} = \begin{pmatrix}2 + 0.5t\\6 − 0.2t^2\end{pmatrix}\) m s⁻¹. A second buggy follows the same path with the same velocity, but leaves the origin 3 seconds later. (a) Find the displacement of the first buggy at time \(t\). (b) Find the distance between the buggies 2 seconds after the second one leaves.

Working

Integrate
\(\mathbf{r} = \begin{pmatrix}2t + 0.25t^2\\6t − \dfrac{t^3}{15}\end{pmatrix}\)

Integrate each part, with the start at the origin.
Times
First: \(t = 5\). Second: \(t = 2\) (it has travelled 2 s).

The second buggy follows the same path 3 s behind: its position is \(\mathbf{r}(t − 3)\).
Positions
\(\mathbf{r}(5) = \begin{pmatrix}16.25\\21.7\end{pmatrix}\), \(\mathbf{r}(2) = \begin{pmatrix}5\\11.5\end{pmatrix}\)

Substitute each time into the same expression.
Distance
\(d = \sqrt{(11.25)^2 + (10.2)^2} = 15.2\) m

Subtract the position vectors, then find the magnitude.

Two projectiles that collide

A particle is projected from the origin with initial velocity \(\begin{pmatrix}9\\25\end{pmatrix}\) m s⁻¹ and acceleration \(\begin{pmatrix}0\\−9.8\end{pmatrix}\) m s⁻². A second particle is projected from the origin 1 second later with initial velocity \(\begin{pmatrix}12\\b\end{pmatrix}\) m s⁻¹ and the same acceleration. (a) Find the displacement of the first particle at time \(t\). (b) Find the displacement of the second particle at time \(t\). (c) The particles collide. Find \(b\). (d) Find how far from the origin they collide.

Working

First particle
\(\mathbf{r} = \begin{pmatrix}9t\\25t − 4.9t^2\end{pmatrix}\)

Integrate twice, with the start at the origin.
Second particle
\(\mathbf{r} = \begin{pmatrix}12(t − 1)\\b(t − 1) − 4.9(t − 1)^2\end{pmatrix}\)

Replace \(t\) by \(t − 1\): it has been moving for one second less.
Same x at the same t
\(9t = 12(t − 1) \Rightarrow t = 4\)

Collision means the same position at the same time.
Same y
\(25\times4 − 4.9\times16 = 3b − 4.9\times9\), so \(21.6 = 3b − 44.1\) and \(b = 21.9\)

Put \(t = 4\) into both vertical parts.
Distance from the origin
\(|\mathbf{r}| = \sqrt{36^2 + 21.6^2} = 42.0\) m

The collision point is \(\begin{pmatrix}36\\21.6\end{pmatrix}\).

Least distance between two launches

A projectile is launched from the origin with initial velocity \(\begin{pmatrix}40\\60\end{pmatrix}\) m s⁻¹ and acceleration \(\begin{pmatrix}0\\−9.8\end{pmatrix}\) m s⁻². A second projectile is launched from the origin with the same initial velocity 4 seconds later. (a) Find an expression for the distance between them \(t\) seconds after the first launch, for \(t \geq 4\). (b) Find the time of least distance while both are in flight.

Working

Both displacements
\(\mathbf{r}_1 = \begin{pmatrix}40t\\60t − 4.9t^2\end{pmatrix}\), \(\mathbf{r}_2 = \begin{pmatrix}40(t − 4)\\60(t − 4) − 4.9(t − 4)^2\end{pmatrix}\)

The second launch replaces \(t\) by \(t − 4\).
The difference
\(\mathbf{r}_1 − \mathbf{r}_2 = \begin{pmatrix}160\\318.4 − 39.2t\end{pmatrix}\)

Subtract and simplify each component.
The distance
\(d = \sqrt{160^2 + (318.4 − 39.2t)^2}\)

The distance is the magnitude of the difference.
Minimum
From the calculator: \(t = 8.12\) s

Graph \(d\) for \(4 \leq t \leq 12.2\). The first lands at 12.2 s, so both are in flight until then.

The general circle

A particle has position vector \(\mathbf{r} = \begin{pmatrix}r\cos\omega t\\r\sin\omega t\end{pmatrix}\). (a) Show that the speed is \(r\omega\), the time for one circle is \(\dfrac{2\pi}{\omega}\) and the magnitude of the acceleration is \(\dfrac{v^2}{r}\). (b) Show that the velocity is perpendicular to the position vector and that the acceleration points towards the centre. (c) For \(r = 2.5\) m and \(\omega = 1.6\) rad s⁻¹ find the magnitude of the acceleration.

Working

Velocity
\(\mathbf{v} = \begin{pmatrix}-r\omega\sin\omega t\\r\omega\cos\omega t\end{pmatrix}\), \(|\mathbf{v}| = r\omega\)

The speed is \(r\omega\), and \(\sin^2 + \cos^2 = 1\) removes \(t\).
Period
\(\omega T = 2\pi \Rightarrow T = \dfrac{2\pi}{\omega}\)

One full circle is an angle of \(2\pi\).
Perpendicular
\(\mathbf{v}\cdot\mathbf{r} = -r^2\omega\sin\omega t\cos\omega t + r^2\omega\sin\omega t\cos\omega t = 0\)

Zero scalar product.
Acceleration
\(\mathbf{a} = −\omega^2\mathbf{r}\), \(|\mathbf{a}| = r\omega^2 = \dfrac{v^2}{r}\)

A negative multiple of \(\mathbf{r}\): towards the centre. This is constant if \(\omega\) is.
Numbers
\(v = 2.5\times1.6 = 4,\quad |\mathbf{a}| = \dfrac{4^2}{2.5} = 6.4\) m s⁻²

Use the exact values.

Menu

Vectors

Mixed practice: projectiles, circular motion and delayed objects

By the end of this section

Key points

Projectiles

Circular motion

Going back from velocity to position

Objects that start later

On the calculator

Minimum of a graph

Explore: circular motion

Explore: a projectile under gravity

Worked examples

Angle, top of the flight and greatest height

Working

How far it lands from the origin

Working

Showing motion in a circle

Working

Acceleration towards the centre

Working

From velocity back to the circle

Working

An object that starts later

Working

Two projectiles that collide

Working

Least distance between two launches

Working

The general circle

Working

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