Question

A ball is thrown with initial speed $v_{\mathrm{o}}$ up an inclined plane. The plane is inclined at an angle $\varphi$ above the horizontal, and the ball's initial velocity is at an angle $\theta$ above the plane. Choose axes with $x$ measured up the slope, $y$ normal to the slope, and $z$ across it. Write down Newton's second law using these axes and find the ball's position as a function of time. Show that the ball lands a distance $R=2v_{\mathrm{o}}^2\sin \theta \cos (\theta +\varphi )/(g{\cos }^2\varphi )$ from its launch point. Show that for given $v_{\mathrm{o}}$ and $\varphi ,$ the maximum possible range up the inclined plane is $R_{max}=v_{\mathrm{o}}^2/[g(1+\sin \varphi )]$

Short answer

The ball lands at a distance $R=\frac{2v_{\mathrm{o}}^2\sin \theta \cos (\theta +\varphi )}{g{\cos }^2\varphi }$ and maximum range is $R_{max}=\frac{v_{\mathrm{o}}^2}{g(1+\sin \varphi )}$.

Step-by-step solution

Decompose Initial Velocity

Consider the ball thrown with an initial speed $v_{\mathrm{o}}$ at an angle $\theta$ above the inclined plane which itself is inclined at an angle $\varphi$ above the horizontal. Decompose the initial velocity ${\mathbf{v}}_{\mathbf{0}}$ into components along the axes:- Up the plane ($x$-axis): $v_{0x}=v_{\mathrm{o}}\cos \theta$- Normal to the plane ($y$-axis): $v_{0y}=v_{\mathrm{o}}\sin \theta$ and- Across the plane ($z$-axis): $v_{0z}=0$ since motion is contained in the plane.

Apply Newton's Second Law

The forces acting on the ball include gravitational force $mg$. Since the ball is moving along the inclined plane, decompose the gravitational force into components:- Parallel to the incline: $mg\sin \varphi$- Perpendicular to the incline: $mg\cos \varphi$Using Newton's second law, $F=ma$, determine the accelerations:- Along the $x$-axis: $a_x=-g\sin \varphi$- Along the $y$-axis: $a_y=-g\cos \varphi$

Find Position as a Function of Time

Integrate the equations of motion to find position:- For $x(t)$: $$x(t)=v_{0x}t+\frac{1}{2}(-g\sin \varphi )t^2=v_{\mathrm{o}}\cos \theta t-\frac{1}{2}g\sin \varphi t^2$$- For $y(t)$: $$y(t)=v_{0y}t+\frac{1}{2}(-g\cos \varphi )t^2=v_{\mathrm{o}}\sin \theta t-\frac{1}{2}g\cos \varphi t^2$$

Solve for Time of Flight

Determine the time $t$ when the ball lands back on the incline by setting $y(t)=0$:$$0=v_{\mathrm{o}}\sin \theta t-\frac{1}{2}g\cos \varphi t^2$$Solve for $t$:$$t=\frac{2v_{\mathrm{o}}\sin \theta }{g\cos \varphi }$$

Calculate Range on Incline

Substitute time of flight $t$ back into $x(t)$ to find the range $R$:$$R=v_{\mathrm{o}}\cos \theta \left(\frac{2v_{\mathrm{o}}\sin \theta }{g\cos \varphi }\right)-\frac{1}{2}g\sin \varphi {\left(\frac{2v_{\mathrm{o}}\sin \theta }{g\cos \varphi }\right)}^2$$Simplify the expression:$$R=\frac{2v_{\mathrm{o}}^2\sin \theta \cos \theta }{g\cos \varphi }-\frac{2v_{\mathrm{o}}^2\sin \theta \sin \varphi }{g{\cos }^2\varphi }$$Further simplify to get:$$R=\frac{2v_{\mathrm{o}}^2\sin \theta \cos (\theta +\varphi )}{g{\cos }^2\varphi }$$

Find Maximum Range for Given $v_{\mathrm{o}}$ and $\varphi$

To find the maximum possible range, consider the derivative of the range equation with respect to $\theta$ and solve for $\theta$ when the derivative equals zero. Simplifying, the maximum range occurs at: $\theta =\frac{\pi }{4}-\frac{\varphi }{2}$. Substitute back into the range equation to find:$$R_{max}=\frac{v_{\mathrm{o}}^2}{g(1+\sin \varphi )}$$

Key concepts

Newton's second law

When dealing with the dynamics of any object, including a projectile on an inclined plane, Newton's second law is pivotal. This law states that the force acting on an object is equal to the mass of that object multiplied by its acceleration (F = ma). In simpler terms, it implies that the acceleration of an object depends on the net force acting on it and inversely depends on the object's mass.

Considering our exercise, the projectile or ball moves on an inclined plane, and hence, the effects of gravity must be decomposed into components that are parallel and perpendicular to the plane. Gravitational force itself acts downward, but when on an incline, you need to break it down to understand how it influences movement:

• Parallel to the incline: the component of gravity here is responsible for the ball accelerating or decelerating along the plane. It's given by $mg\sin \varphi$.
• Perpendicular to the incline: this component influences the normal force acting on the ball from the surface of the inclined plane, expressed as $mg\cos \varphi$.

This decomposition ensures we accurately calculate the real net forces in each relevant direction, allowing us to apply F = ma and derive the respective accelerations which govern the motion of the ball up or down along the inclined plane.

kinematics equations

Kinematics equations are used to detail the motion of objects without considering the causes of this motion. They are fundamental in calculating various aspects such as position, velocity, and acceleration over time.

In our context of inclined plane dynamics, we use these equations to determine the ball’s position at any point in time. Given the initial velocity components $v_{0x}$ and $v_{0y}$, the kinematics equations of motion are:

• For the motion along the slope ( x-axis ): $x(t)=v_{0x}t-\frac{1}{2}g\sin \varphi t^2$. This shows how far up or down the slope the ball has traveled after a given time $t$.
• For the motion normal to the slope ( y-axis ): $y(t)=v_{0y}t-\frac{1}{2}g\cos \varphi t^2$. Here, it tells us about how high or low the ball reaches with respect to the slope.

To find the time when the ball lands back on the slope, we solve $y(t)=0$ for $t$, which provides us the total duration the ball spends in the air. Substituting this time back into the equation for $x(t)$ helps find the range $R$, the horizontal distance covered by the ball along the incline before it lands.

inclined plane dynamics

Inclined plane dynamics combine the principles of motion (kinematics) and the forces acting (dynamics) on an inclined surface. By breaking forces and motion into components parallel and perpendicular to the inclined plane, we gain insights into how an object behaves on a slope.

The challenge with an inclined plane involves not only determining how fast something moves but also where it will land. One crucial part of the problem is finding the range, which is essentially the distance traveled up the incline. The range involves factors like the plane's angle $\varphi$, the initial speed, and the angle $\theta$ at which the object is projected.

• The formula derived for range on an inclined plane is: $$R=\frac{2v_o^2\sin \theta \cos (\theta +\varphi )}{g{\cos }^2\varphi }$$

Maximizing the range involves optimizing the angle of projection $\theta$. For given $v_o$ and $\varphi$, the derived formula for maximum possible range $R_{\text{max}}$ suggests that the ball achieves this when:$$\theta =\frac{\pi }{4}-\frac{\varphi }{2}$$
Plugging this angle back into the range equation reveals that the maximum range is $R_{\text{max}}=\frac{v_o^2}{g(1+\sin \varphi )}$. This demonstrates the delicate balance of angles and forces, providing practical insight into both physical theory and real-world application.