Question

You lay a rectangular board on the horizontal floor and then tilt the board about one edge until it slopes at angle \(\theta\) with the horizontal. Choose your origin at one of the two corners that touch the floor, the \(x\) axis pointing along the bottom edge of the board, the \(y\) axis pointing up the slope, and the \(z\) axis normal to the board. You now kick a frictionless puck that is resting at \(O\) so that it slides across the board with initial velocity \(\left(v_{\mathrm{ox}}, v_{\mathrm{oy}}, 0\right) .\) Write down Newton's second law using the given coordinates and then find how long the puck takes to return to the floor level and how far it is from \(O\) when it does so.

Short answer

The puck returns after \(\frac{2v_{oy}}{g\cos(\theta)}\) seconds and is \(\frac{2v_{oy}}{g\cos(\theta)}(v_{ox} - v_{oy}\tan(\theta))\) units from O.

Step-by-step solution

Understanding the System and Forces

The puck is sliding on an inclined plane. The forces acting on the puck include the gravitational force. Since the puck is frictionless, we consider only the component of gravity acting along the plane. The gravitational force has a component along the x-axis \(-mg\sin(\theta)\) and a component along the y-axis \(-mg\cos(\theta)\). Newton's second law states \(F=ma\), so we will apply this along the axes.

Application of Newton's Second Law

For the x-axis, the only force is \(-mg\sin(\theta)\), giving: \[ ma_x = -mg\sin(\theta) \Rightarrow a_x = -g\sin(\theta) \]For the y-axis, the force is \(-mg\cos(\theta)\), giving: \[ ma_y = -mg\cos(\theta) \Rightarrow a_y = -g\cos(\theta) \]The z-axis has no forces as it's normal to the board, hence \(a_z = 0\).

Solving for Velocity Components

Starting with the initial velocity \((v_{ox}, v_{oy}, 0)\), find the velocity as functions of time:For the x-component:\[ v_x(t) = v_{ox} - g\sin(\theta)t \]For the y-component:\[ v_y(t) = v_{oy} - g\cos(\theta)t \]For the z-component:\[ v_z(t) = 0 \]

Position Functions in Terms of Time

Integrate the velocity functions to get position.For the x-component:\[ x(t) = v_{ox}t - \frac{1}{2}g\sin(\theta)t^2 \]For the y-component:\[ y(t) = v_{oy}t - \frac{1}{2}g\cos(\theta)t^2 \]The z-component remains 0 because there's no motion perpendicular to the board.

Time to Return to the Floor Level

The puck returns to floor level when \(y(t) = 0\). Set the equation for \(y(t)\) to 0:\[ v_{oy}t - \frac{1}{2}g\cos(\theta)t^2 = 0 \]Factor out \(t\):\[ t(v_{oy} - \frac{1}{2}g\cos(\theta)t) = 0 \]This gives two solutions, \(t = 0\) and \(t = \frac{2v_{oy}}{g\cos(\theta)}\). The non-zero solution is our answer.

Distance from Origin When Returning

Find the position along the x-axis at \(t = \frac{2v_{oy}}{g\cos(\theta)}\):\[ x(\frac{2v_{oy}}{g\cos(\theta)}) = v_{ox}\frac{2v_{oy}}{g\cos(\theta)} - \frac{1}{2}g\sin(\theta)(\frac{2v_{oy}}{g\cos(\theta)})^2 \]Simplify the expression to find:\[ x(t) = \frac{2v_{oy}v_{ox}}{g\cos(\theta)} - \frac{2v_{oy}^2\tan(\theta)}{g\cos(\theta)} \]

Key concepts

Inclined Plane Dynamics

Inclined plane dynamics are essential when analyzing the motion of objects sliding down a slope. The motion across an inclined surface involves the interplay of gravitational forces, which act at an angle to the surface. In this specific problem, a board is inclined at an angle \( \theta \) with the horizontal plane, forming a slope. The dynamics here are crucial because they dictate how forces are broken down into components parallel and perpendicular to the surface. This breakdown is vital for applying Newton's second law accurately. For a frictionless object on such a plane, we only consider gravitational force components parallel and perpendicular to the incline, simplifying our analysis by eliminating frictional forces. This simplification makes it easier to predict and calculate the motion of objects like our puck.

Frictionless Puck Motion

In this scenario, a frictionless puck is kicked across the sloped board. The absence of friction ensures that the only forces acting on the puck are due to gravity. This concept is central in learning how objects move on inclined planes. The lack of friction means no energy is lost to heat, and the puck's motion is purely the result of gravitational influences. This results in a straightforward application of Newton's second law, as forces opposing motion do not need to be accounted for. Consequently, we can graphically visualize how the puck maintains its trajectory while its speed and direction are influenced by the slope's gravitational component. The simplicity of frictionless motion enables us to clearly see how theoretical dynamics directly drive an object's path.

Gravitational Force Components

Gravitational force on an inclined plane can be split into two components: one parallel and one perpendicular to the plane. Understanding these components helps in accurately applying Newton's Second Law. The parallel component acts along the plane and is responsible for accelerating the puck down the slope. Mathematically, it is given by \(-mg\sin(\theta)\). The perpendicular component acts towards the surface, calculated as \(-mg\cos(\theta)\). While this component affects the contact force, it does not influence the motion along the plane in a frictionless scenario. With these components, students can predict how an object's speed changes as it travels on the inclined plane. Grasping how to resolve gravitational force into these components simplifies many physics problems involving slopes and inclines.

Velocity and Position Equations

The velocity and position equations describe an object's movement in terms of time. For our puck, the velocity equations derive from the integration of acceleration functions. Initially, the puck has a velocity of \((v_{ox}, v_{oy}, 0)\). The velocity change over time has both an x and y component, yielding equations: \(v_x(t) = v_{ox} - g\sin(\theta)t\) and \(v_y(t) = v_{oy} - g\cos(\theta)t\). These provide insights into how the puck's motion evolves due to the slope's gravitational pull. By integrating these, we obtain position equations: \(x(t) = v_{ox}t - \frac{1}{2}g\sin(\theta)t^2\) for the x-direction, and \(y(t) = v_{oy}t - \frac{1}{2}g\cos(\theta)t^2\) in the y-direction. These functions enable us to determine where the puck is at any point in time. They reflect how initial velocities and gravitational forces interplay to define the path taken by the puck.