describe and explain motion due to a uniform velocity in one direction and a uniform acceleration in a perpendicular direction
Motion with a Uniform Velocity in One Direction and a Uniform Acceleration Perpendicular to It
1. Objective
Describe and explain the motion of a particle that moves with a constant velocity in one direction while simultaneously experiencing a constant acceleration in a direction perpendicular to that velocity. The presentation follows the Cambridge International AS & A Level Physics (9702) syllabus, linking each point to the relevant syllabus statements (AO1–AO3).
2. Syllabus Alignment (Quick Reference)
| Syllabus Requirement | Notes Coverage |
|---|---|
| 1.1 – Physical quantities consist of a numerical magnitude and a unit (AO1) | Section 3 includes an SI‑units reminder and a complete symbol table. |
| 1.4 – Vectors and component‑wise treatment (AO1) | Section 4 introduces vector decomposition using the same notation as the syllabus (\(\vec v,\;\vec a\)). |
| 2.1 – Kinematic equations of motion (AO1, AO2) | Section 5 lists the standard kinematic equations and shows how they are obtained from the syllabus table. |
| 2.2 – Use of graphical methods (AO2) | Section 7 describes how to construct and interpret \(x\!-\!t,\;y\!-\!t,\;v\!-\!t\) graphs. |
| 2.3 – Force analysis (AO2, AO3) | Section 8 applies Newton’s II law component‑wise. |
| 2.4 – Work‑energy theorem (AO2, AO3) | Section 9 treats the vertical motion with the work‑energy approach. |
3. Physical Quantities & Units (Topic 1)
All quantities must be expressed in SI units. The table below lists every symbol that appears later; the unit is given next to each symbol so that the requirement “physical quantities consist of a numerical magnitude and a unit” is satisfied.
| Symbol | Quantity | SI Unit |
|---|---|---|
| \(x,\;y\) | Position coordinates | m |
| \(x0,\;y0\) | Initial position | m |
| \(v{x0},\;v{y0}\) | Initial velocity components | m s⁻¹ |
| \(vx,\;vy\) | Velocity components at any instant | m s⁻¹ |
| \(a_y\) | Uniform vertical acceleration (≈ \(g=9.81\) m s⁻² downwards) | m s⁻² |
| \(t\) | Time elapsed since launch | s |
| \(m\) | Mass of the particle | kg |
| \(g\) | Acceleration due to gravity (downwards) | m s⁻² |
Common unit‑mistake: Never mix km h⁻¹ with m s⁻¹.
Convert: \(1\;\text{km h}^{-1}= \dfrac{1000}{3600}=0.278\;\text{m s}^{-1}\).
4. Vectors & Component‑wise Treatment (Topic 1 & 2)
The motion is most easily handled by separating it into two independent one‑dimensional motions.
\[
\vec r(t)=x(t)\,\hat{\imath}+y(t)\,\hat{\jmath},\qquad
\vec v(t)=vx(t)\,\hat{\imath}+vy(t)\,\hat{\jmath},\qquad
\vec a =0\,\hat{\imath}+a_y\,\hat{\jmath}.
\]
- Horizontal direction (\(\hat{\imath}\)): \(\displaystyle ax=0\) → \(\displaystyle vx=v_{x0}= \text{constant}\).
- Vertical direction (\(\hat{\jmath}\)): \(\displaystyle ay=\text{constant}\) (normally \(ay=-g\)).
Because \(\vec a\) is perpendicular to the initial velocity \(\vec v_0\), the two components are mathematically independent – exactly the “vector decomposition” required by the syllabus.
5. Coordinate System
- x‑axis: chosen along the direction of the uniform velocity component \(v_{x0}\) (horizontal).
- y‑axis: chosen along the direction of the uniform acceleration \(ay\) (positive upwards; \(ay=-g\) if downwards).
- Initial position at \(t=0\): \(\vec r0=(x0,\;y_0)\).
6. Kinematic Equations – Standard Form (Topic 2)
The syllabus provides the following generic equations for constant acceleration:
\[
s = ut + \tfrac12 a t^{2},\qquad
v = u + a t,\qquad
v^{2}=u^{2}+2as,
\]
where \(s\) is displacement, \(u\) the initial speed in the considered direction and \(a\) the constant acceleration.
Applying these to each independent direction gives:
| Direction | Equation of motion (derived from the generic form) |
|---|---|
| Horizontal (\(x\)) | \(x = x0 + v{x0}\,t\) (since \(a_x=0\)) |
| Vertical (\(y\)) | \(y = y0 + v{y0}\,t + \tfrac12 a_y t^{2}\) |
Eliminating the time variable – Parabolic trajectory
From the horizontal equation, \(t = \dfrac{x-x0}{v{x0}}\). Substituting into the vertical equation yields
\[
y = y0 + \frac{v{y0}}{v{x0}}\,(x-x0) + \frac{ay}{2v{x0}^{2}}\,(x-x_0)^{2},
\]
which is a quadratic (parabolic) relationship between \(x\) and \(y\). This is the hallmark of projectile motion.
A‑Level extensions – useful derived formulas
When the particle is launched from the origin \((x0=y0=0)\) with speed \(v\) at an angle \(\theta\) above the horizontal (\(v{x0}=v\cos\theta,\;v{y0}=v\sin\theta\)), the following results are obtained by simple algebraic manipulation of the equations above:
- Time of flight (until the particle returns to the launch height):
\[
t_{\text{flight}} = \frac{2v\sin\theta}{g}.
\]
- Horizontal range:
\[
R = v\cos\theta \; t_{\text{flight}} = \frac{v^{2}\sin2\theta}{g}.
\]
- Maximum height:
\[
H_{\max}= \frac{(v\sin\theta)^{2}}{2g}.
\]
These formulas are frequently required in Cambridge A‑Level exam questions and are therefore listed explicitly.
7. Graphical Methods (Topic 2 – “use graphical methods to represent motion”)
Graphs provide a visual check of the algebraic results and are useful when analysing experimental data.
- \(x\) vs \(t\) graph – straight line; slope = \(v_{x0}\). The area under the line is the horizontal displacement.
- \(y\) vs \(t\) graph – parabola opening downwards (if \(ay=-g\)). The gradient at any instant equals the instantaneous vertical velocity \(vy(t)=v{y0}+ay t\). The curvature (second derivative) gives the constant vertical acceleration \(a_y\).
- \(vy\) vs \(t\) graph – straight line; gradient = \(ay\), intercept = \(v_{y0}\). This graph is a quick way to verify the value of the acceleration from experimental data.
When plotting real data, students should:
- Label axes with both quantity and unit (e.g. “\(x\) (m)”).
- Draw a best‑fit straight line (for \(x\!-\!t\) or \(vy\!-\!t\)) and use the slope‑intercept method to obtain numerical values of \(v{x0}\) or \(a_y\).
- Check that the area under the \(v_y\!-\!t\) graph equals the measured vertical displacement, confirming the work‑energy relationship.
8. Force Analysis (Dynamics – Topic 2.3)
Newton’s II law applied to each component:
- \(\displaystyle \sum Fx = m ax = 0 \;\Rightarrow\; \text{no horizontal net force, so } v{x}=v{x0}\text{ (constant).}\)
- \(\displaystyle \sum Fy = m ay = mg \;\Rightarrow\; a_y = -g\) (downwards). The only external force (ignoring air resistance) is the weight.
Because the horizontal net force is zero, horizontal momentum is conserved throughout the motion – a point often examined in AO2 questions.
9. Energy Approach (Work‑Energy Theorem – Topic 2.4)
- Horizontal motion: No net work is done because the horizontal force component is zero. Hence the horizontal kinetic energy \(\tfrac12 m v_{x}^{2}\) remains constant.
- Vertical motion: The weight does work \(W = mg\,(y-y_0)\). Using the work‑energy theorem,
\[
\frac12 m vy^{2} - \frac12 m v{y0}^{2}= mg\,(y-y_0),
\]
which rearranges to the same vertical‑motion equation obtained from kinematics.
- Total mechanical energy: Not conserved because gravity does work, but the separation of horizontal and vertical contributions makes the analysis transparent.
10. Worked Example – General Launch Angle
Problem: A projectile is launched from ground level with speed \(v=20\;\text{m s}^{-1}\) at an angle \(\theta =30^{\circ}\) above the horizontal. Determine (a) the time of flight, (b) the horizontal range, and (c) the speed just before impact. Air resistance is ignored.
- Resolve the initial velocity:
\[
v_{x0}=v\cos\theta =20\cos30^{\circ}=17.3\;\text{m s}^{-1},
\qquad
v_{y0}=v\sin\theta =20\sin30^{\circ}=10.0\;\text{m s}^{-1}.
\]
- Time of flight: Set \(y=0\) in the vertical equation
\[
0 = v_{y0}t + \tfrac12(-g)t^{2}
\;\Rightarrow\;
t\bigl(v_{y0}-\tfrac12 g t\bigr)=0.
\]
The non‑zero solution gives
\[
t{\text{flight}} = \frac{2v{y0}}{g}= \frac{2\times10.0}{9.81}=2.04\;\text{s}.
\]
- Horizontal range:
\[
R = v{x0}\,t{\text{flight}} = 17.3 \times 2.04 \approx 35.3\;\text{m}.
\]
- Velocity just before impact:
\[
vy = v{y0} - g t_{\text{flight}} = 10.0 - 9.81\times2.04 = -10.0\;\text{m s}^{-1}\;(downwards),
\]
\[
vx = v{x0}=17.3\;\text{m s}^{-1},
\]
\[
v = \sqrt{vx^{2}+vy^{2}} = \sqrt{17.3^{2}+10.0^{2}} \approx 20.0\;\text{m s}^{-1}.
\]
The speed equals the launch speed because gravity does no work in the horizontal direction and the loss of vertical kinetic energy is exactly compensated by the gain in gravitational potential energy during ascent.
11. Common Misconceptions
- Horizontal speed changes because of vertical acceleration: It does not. Perpendicular forces affect only the component in their own direction.
- Neglecting the initial vertical component \(v_{y0}\) when the launch is not perfectly horizontal: This leads to incorrect range and time‑of‑flight values.
- Assuming the trajectory is a straight line: The quadratic term in the vertical equation makes the path a parabola.
- Inconsistent units: Always convert to SI before using the equations (e.g. km h⁻¹ → m s⁻¹).
12. Suggested Diagram
13. Summary
When a particle experiences a constant velocity in one direction and a constant acceleration perpendicular to that direction, the motion can be treated as two independent one‑dimensional motions. Using the syllabus notation:
\[
x = x0 + v{x0}\,t,\qquad
y = y0 + v{y0}\,t + \tfrac12 a_y t^{2}.
\]
Eliminating \(t\) gives the parabolic relation
\[
y = y0 + \frac{v{y0}}{v{x0}}(x-x0) + \frac{ay}{2v{x0}^{2}}(x-x_0)^{2}.
\]
Key points required by the Cambridge AS & A Level Physics syllabus:
- All symbols are accompanied by SI units (AO1).
- Vector decomposition and component‑wise kinematic equations are explicitly shown (AO1).
- Standard kinematic equations are identified, and the derived projectile formulas (range, maximum height, time of flight) are presented (AO2).
- Graphical representation of \(x\!-\!t,\;y\!-\!t,\;v_y\!-\!t\) is explained (AO2).
- Force analysis demonstrates why the horizontal velocity remains constant (AO2, AO3).
- Work‑energy theorem is applied to the vertical motion (AO2, AO3).
This framework underpins all projectile‑motion problems encountered in the 9702 syllabus and provides a clear, exam‑ready structure for students.