I'm trying to solve this problem from Halliday & Resnick in a creative way:
A block is projected up a frictionless inclined plane with initial speed $v_0 = 3.50$ m/s. The angle of incline is $\theta = 32.0°$. (a) How far up the plane does the block go? (b) How long does it take to get there? (c) What is its speed when it gets back to the bottom?
I could solve the problem the standard way ($v_f^2=v_i^2+2ad$) but I want to try some other methods.
I assumed the block will travel the same distance vertically up the ramp as a ball thrown straight up but with $a = g \sin\theta$ instead of $a = g$. Then we get
$$d=\frac{(3.5)^2}{2(g \sin 32)} = 1.18 \text{ m} \qquad \text{and} \qquad t = \frac{3.5}{g \sin 32} = 0.67 \text{ s}$$
and the return speed is equivalent to the launch speed. I was surprised to see all my answers are correct. My conclusion is that motion on an inclined plane can be simply modelled as a projectile motion with $a = g \sin\theta$ instead of $a=g$.
Questions:
- Why does this work? Is it just a fluke?
- Have I reframed this problem in a trivial way?
- Is there something deeper underlying this?
Best Answer
This is not correct conclusion - the two motions you describe are not equivalent. There are some similarities in equations for purely vertical motion and motion on an inclined plane, but this is far from saying the two motions are equivalent.
In projectile motion the trajectory is a parabola and there are two independent axes: (i) in horizontal axis there is no acceleration, and (ii) in vertical axis the acceleration equals $a = g$ (pointing downwards). In motion on an inclined plane the trajectory is a straight line and there is only one independent axis (along the inclined plane) with acceleration $a = g \sin\theta$ (pointing downwards). You could define a coordinate system such that you have two axes (horizontal and vertical) for motion on the inclined plane, but using single axis along the inclined plane is sufficient and much simpler.
If by "the same distance vertically up" you mean the distance along the ramp and not the height difference, this is true. Below I prove this observation in two different ways: (i) via energy approach, and (ii) via equations of motion. We will use the following coordinate system:
Energy approach
The work-energy theorem states that change in object's kinetic energy (final value minus initial value) equals total (net) work done on the object
$$\Delta K = \vec{F}_\text{net} \cdot \Delta \vec{s}$$
where $\vec{F}_\text{net}$ is the net force acting on the object, and $\Delta\vec{s}$ is the displacement vector.
In case of projectile motion there is only one force acting on the object (gravitational force) and the work is defined as
$$W = (-mg \hat{\jmath}) \cdot (\Delta x \hat{\imath} + \Delta y \hat{\jmath}) = -m g \Delta x (\hat{\imath} \cdot \hat{\jmath}) -m g \Delta y (\hat{\jmath} \cdot \hat{\jmath}) = -m g \Delta y$$
where $\Delta x$ and $\Delta y$ are horizontal and vertical displacement. For a projectile going up $\Delta y$ is positive which means $W$ is negative which means $\Delta K$ is negative and the kinetic energy decreases.
In case of motion on an inclined plane, there are two forces acting on the body (gravitational and normal force), but only gravitational force does work since normal force is perpendicular to the displacement. The work is defined as
$$W = (-mg \hat{\jmath}) \cdot (\Delta x \hat{\imath} + \Delta y \hat{\jmath}) = (-mg \hat{\jmath}) \cdot (\Delta s \cos\theta \hat{\imath} + \Delta s \sin\theta \hat{\jmath}) = -m (g \sin\theta) \Delta s$$
where $\Delta \vec{s} = \Delta x \hat{\imath} + \Delta y \hat{\jmath} = \Delta s \angle \theta$ is the displacement along the ramp.
At the highest point, the change in kinetic energy $\Delta K = 0-\frac{1}{2} m v_0^2$ is the same for (purely) vertical motion and motion on inclined plane. We can then compare two expressions for work
$$W = -mg \Delta y \qquad \text{and} \qquad W = -m (g \sin\theta) \Delta s$$
and conclude that $\Delta s$ equals vertical distance a projectile would have travelled if thrown vertically with $a = g \sin\theta$. It must be noted that this conclusion is valid only for purely vertical motion, because in projectile motion the final kinetic energy is not zero at the highest point!
Equations of motion
Projectile motion
If we neglect air resistance and some other minor effects, there is only one force acting on the body (gravitational force), and the net force is
$$\vec{F}_\text{net} = m \vec{a} = -m g \hat{\jmath} \quad \rightarrow \quad a_x = 0, \quad a_y = -g$$
The equations of motion for the two axes are
$$ \begin{aligned} x(t) &= \frac{1}{2} a_x t^2 + v_{0,x} t = (v_0 \cos\theta) t \\ y(t) &= \frac{1}{2} a_y t^2 + v_{0,y} t = -\frac{1}{2} g t^2 + (v_0 \sin\theta) t \end{aligned} $$
where starting point is assumed to be at the origin: $x_0 = 0$ and $y_0 = 0$.
The object reaches maximum altitude when vertical velocity drops to zero
$$\left. \frac{dy(t)}{dt} \right|_{t=t_m} = -g t_m + v_0 \sin\theta = 0 \qquad \rightarrow \qquad t_m = \frac{v_0 \sin\theta}{g}$$
The object position at time $t_m$ is
$$x(t_m) = \frac{v_0^2}{2g} \sin 2\theta \qquad \text{and} \qquad y(t_m) = \frac{v_0^2}{2g} \sin^2\theta$$
In a special case when $\theta = 90^\circ$ (purely vertical motion), the above equations become
$$t_m = \frac{v_0}{g} \qquad \text{and} \qquad x(t_m) = 0 \qquad \text{and} \qquad y(t_m) = \frac{v_0^2}{2g} \tag 1$$
Motion on an inclined plane
We usually model this motion with only one axis which points along the inclined plane. Here I use the same two axes as for projectile motion in order to show differences between the two motions.
If we neglect friction and some other minor effects, there are two forces acting on the body
$$\vec{F}_\text{net} = m \vec{a} = \vec{w} + \vec{n}$$
where $\vec{w}$ is the gravitational force and $\vec{n}$ is normal force that is perpendicular to the inclined plane. The two forces are defined as
$$\vec{w} = -mg \hat{\jmath}, \qquad \vec{n} = mg \cos\theta \angle (\theta + 90^\circ) = -mg \sin\theta \cos\theta \hat{\imath} + mg \cos^2\theta \hat{\jmath}$$
The net acceleration is then
$$a_x = -g \sin\theta \cos\theta, \qquad a_y = -g + g \cos^2 \theta = -g \sin^2 \theta$$
The equations of motion for the two axes are
$$ \begin{aligned} x(t) &= \frac{1}{2} a_x t^2 + v_{0,x} t = (-\frac{1}{2} g \sin\theta \cos\theta) t^2 + (v_0 \cos\theta) t \\ y(t) &= \frac{1}{2} a_y t^2 + v_{0,y} t = (-\frac{1}{2} g \sin^2\theta) t^2 + (v_0 \sin\theta) t \end{aligned} $$
The object reaches maximum altitude when vertical velocity drops to zero
$$\left. \frac{d y(t)}{dt} \right|_{t=t_m} = (-g \sin^2\theta) t_m + (v_0 \sin\theta) = 0 \qquad \rightarrow \qquad t_m = \frac{v_0}{g \sin\theta}$$
The object position at time $t_m$ is
$$x(t_m) = \frac{v_0^2}{2g} \frac{1}{\tan\theta} \qquad \text{and} \qquad y(t_m) = \frac{v_0^2}{2g} \qquad \text{and} \qquad \Delta s = \frac{v_0^2}{2g} \frac{1}{\sin\theta}$$
where $\Delta s$ is the distance travelled up the ramp.
Comparison between two types of motions
If you equations for (purely) vertical motion (see Eq. (1)) and replace $g$ with $g \sin\theta$ you get
$$t_m = \frac{v_0}{g \sin\theta} \qquad \text{and} \qquad y(t_m) = \frac{v_0^2}{2g} \frac{1}{\sin\theta}$$
These two equations are equal to $t_m$ and $\Delta s$ in the motion along inclined plane.