Further Mathematics (9665) | Projectiles launched onto inclined planes

Welcome to Projectiles on Inclined Planes!

This chapter takes the familiar world of projectiles and gives it a tilt! If you felt comfortable with standard 2D motion (where the ground is flat), don't worry—the underlying principles haven't changed. We are still using SUVAT, but we need a smarter coordinate system to make the calculations manageable.

Mastering inclined plane problems is crucial because they test your understanding of vector resolution and how to adapt your modeling to challenging geometry.

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1. The Need for a New Coordinate System (Why Rotate?)

In standard projectile problems, we use a horizontal $x$-axis and a vertical $y$-axis. This works perfectly because gravity acts straight down (along the $y$-axis), and the ground is $y=0$.

When the projectile lands on an inclined plane, the landing condition is no longer $y=0$. The plane itself is a sloping line. Trying to use horizontal/vertical axes makes the landing condition extremely complicated (involving trigonometry and simultaneous equations).

The clever trick is to rotate the coordinate axes so they align perfectly with the slope.

Key Definitions for the Inclined Plane

• The inclined plane makes an angle \(\beta\) (beta) with the horizontal.
• We define the new axes:
  • \(x'\) (x-prime): Parallel to the slope.
  • \(y'\) (y-prime): Perpendicular (normal) to the slope.

Think of it this way: If you were sliding down a ramp, the important directions are "down the ramp" and "into the ramp." These are your new \(x'\) and \(y'\) axes.

2. Resolving the Acceleration (Gravity)

This is the single most important step. Because we rotated our axes, the acceleration due to gravity (\(g\)), which still acts vertically downwards, must be resolved into components along the \(x'\) and \(y'\) directions.

Acceleration Components (a)

Imagine the gravity vector, \(g\), making an angle \(\beta\) with the \(y'\) axis (the perpendicular axis).

1. Parallel to the Plane (\(x'\) direction):
\(\mathbf{a}_{x'} = g \sin \beta\)
Note: If the projectile is launched up the slope, this acceleration component acts to slow it down (down the slope). We often define 'positive' \(x'\) as up the slope, so \(a_{x'}\) would be \(-g \sin \beta\). Be careful with your sign conventions!

2. Perpendicular to the Plane (\(y'\) direction):
\(\mathbf{a}_{y'} = -g \cos \beta\)
This component always acts towards the plane. Since we usually define positive \(y'\) as away from the plane, this component must be negative.

Key Takeaway: The acceleration is not simply \(-g\) or \(0\). It is split into components: \((-g \sin \beta, -g \cos \beta)\) if positive \(x'\) is *up* the slope and positive \(y'\) is *away* from the slope.

3. Setting Up the Initial Conditions (u)

The initial velocity \(U\) is launched at an angle \(\alpha\) relative to the inclined plane. Since we defined our axes based on the inclined plane, resolving \(U\) is straightforward, exactly like standard 2D motion.

Initial Velocity Components (U)

Assuming the launch angle \(\alpha\) is measured upwards from the plane:

• Parallel to the Plane (\(x'\) direction):
  \(\mathbf{U}_{x'} = U \cos \alpha\)
• Perpendicular to the Plane (\(y'\) direction):
  \(\mathbf{U}_{y'} = U \sin \alpha\)

Did you know? In many inclined plane problems, the launch point is taken as the origin \((0, 0)\). This simplifies the position vectors: \(\mathbf{r}_0 = (0, 0)\).

4. The Equations of Motion (SUVAT)

Since the acceleration components are constant (no air resistance assumed), we can use the familiar SUVAT equations in the separated \(x'\) and \(y'\) directions.

Position (\(s\)) after time \(t\)

Using the SUVAT formula \(\mathbf{s} = \mathbf{u}t + \frac{1}{2}\mathbf{a}t^2\):

Position Parallel to the Plane (\(x'\)):
\(\mathbf{x}'(t) = (U \cos \alpha) t + \frac{1}{2} (-g \sin \beta) t^2\)

Position Perpendicular to the Plane (\(y'\)):
\(\mathbf{y}'(t) = (U \sin \alpha) t + \frac{1}{2} (-g \cos \beta) t^2\)

Step-by-Step Tip: Always write out the four components first (U_x', U_y', a_x', a_y') before substituting into the SUVAT equations. It prevents sign errors!

5. Finding the Time of Flight and Range

A. Time of Flight (\(T\))

The projectile stops flying when it hits the inclined plane. In our rotated coordinate system, this moment occurs when the perpendicular distance from the plane returns to zero.

The critical landing condition is: \(\mathbf{y}'(T) = 0\) (where \(T\) is the total time of flight, and \(T \neq 0\)).

We set the \(y'\) position equation to zero and solve for \(T\):
\(0 = (U \sin \alpha) T - \frac{1}{2} (g \cos \beta) T^2\)

Factoring out \(T\) (since \(T \neq 0\)):
\(T \left[ U \sin \alpha - \frac{1}{2} (g \cos \beta) T \right] = 0\)

Solving the bracket for \(T\):
\(\mathbf{T} = \frac{2 U \sin \alpha}{g \cos \beta}\)

Don't worry if this seems complicated; notice how similar this is to the standard time of flight formula \(\frac{2U \sin \alpha}{g}\), except here \(U \sin \alpha\) is divided by the effective gravitational component perpendicular to the slope, \(g \cos \beta\).

B. The Range (\(R\))

The range \(R\) is the distance traveled along the inclined plane. We find this by substituting the total time of flight \(T\) back into the \(x'\) position equation.

\(\mathbf{R} = \mathbf{x}'(T)\)
\(\mathbf{R} = (U \cos \alpha) T - \frac{1}{2} (g \sin \beta) T^2\)

Substituting the expression for \(T\) gives a lengthy, but solvable, equation for the range in terms of \(U, \alpha, g,\) and \(\beta\).

Key Takeaway: The time of flight is governed entirely by the perpendicular motion (\(y'\)) and the range is then calculated using that time in the parallel motion (\(x'\)) equation.

6. Finding the Maximum Range

A common exam requirement is to find the angle of projection, \(\alpha\), that maximizes the range \(R\) up (or down) the slope for a fixed initial speed \(U\) and fixed slope angle \(\beta\).

To find this maximum, you must differentiate the range equation \(R\) with respect to the variable angle \(\alpha\):
\(\frac{dR}{d\alpha} = 0\)

The Crucial Maximum Range Result

When the complex differentiation is performed (often simplified using trigonometric identities like \(\sin(A+B)\)), it leads to a beautiful geometrical result:

For maximum range up or down the inclined plane, the direction of projection bisects the angle between the vertical and the inclined plane.

Mathematically, if \(\theta_{max}\) is the optimum launch angle measured from the horizontal, the result is:
\(\mathbf{\theta}_{max} = \frac{\pi}{4} + \frac{\beta}{2}\) (launching down the slope, where \(\beta\) is positive)
\(\mathbf{\theta}_{max} = \frac{\pi}{4} - \frac{\beta}{2}\) (launching up the slope)

Note: You may be required to derive the optimum angle or simply use the result. If asked to find the actual maximum range, you substitute \(\theta_{max}\) back into the full range equation.

7. Determining the Landing Position (Bounce Problems)

The syllabus mentions determining whether a projectile lands at a higher or lower point on the plane after a bounce. This links the projectile motion with the topic of collisions (specifically, Newton's Experimental Law and the coefficient of restitution, \(e\)).

The Bounce Mechanism

When the projectile hits the plane, the velocity parallel to the plane (\(v_{x'}\)) remains unchanged (assuming a smooth plane).
The velocity perpendicular to the plane (\(v_{y'}\)) changes direction and magnitude based on the coefficient of restitution, \(e\).

If the velocity just before impact is \(\mathbf{v}_{y, \text{before}}'\) and just after is \(\mathbf{v}_{y, \text{after}}'\):
\(\mathbf{v}_{y, \text{after}}' = -e \times \mathbf{v}_{y, \text{before}}'\)

The subsequent flight path (after the bounce) is then calculated using the new initial velocity components: \(U_{\text{new}, x'} = V_{\text{before}, x'}\) and \(U_{\text{new}, y'} = V_{\text{after}, y'}\).

To find out if it lands higher or lower, you calculate the new time of flight \(T_2\) and the resulting displacement \(R_2\). You compare the net change in \(x'\) displacement after the second flight segment to the total change needed to cross the peak of the slope (if the geometry is complex).

Tip: If \(e=1\), the bounce is elastic, and the flight path would simply repeat itself, landing at the exact same location relative to the slope (but further down/up the overall incline). Since \(e<1\) for real collisions, the second range \(R_2\) will be shorter than \(R_1\).

8. Summary and Common Mistakes

Key Takeaways

• Resolution is King: The difficulty lies solely in correctly resolving \(g\) and \(U\) relative to the slope angle \(\beta\).
• Landing Condition: \(\mathbf{y}' = 0\) dictates the time of flight \(T\).
• Maximum Range: The optimal launch angle bisects the angle between the vertical and the plane.

Common Mistakes to Avoid

1. Angle Confusion: Accidentally using the angle to the horizontal (\(\theta\)) where the formula requires the angle to the plane (\(\alpha\)), or vice versa. Always check how \(\alpha\) is defined in the question.

2. Incorrect Signs: Forgetting that \(a_{y'}\) acts back towards the plane (\(-g \cos \beta\)) and \(a_{x'}\) acts down the slope (\(g \sin \beta\)). Ensure your positive directions are clearly defined at the start of the problem.

3. Forgetting the Half: Missing the \(\frac{1}{2}\) in the \(\frac{1}{2}\mathbf{a}t^2\) term when calculating position or range.

4. Mixing Axes: Solving for \(T\) using the \(x'\) equation or calculating \(R\) using the \(y'\) equation. Keep them separate!