Solved Problem on Two-dimensional Motion and Relative Motion

Solved Problem on Two-dimensional Motion

A ball rolls on a horizontal table of height H, with constant speed v₀, frictionless, until it falls over the edge. Calculate:
a) The time for the ball to hit the ground;
b) The horizontal distance, from the edge of the table, where the ball hits the floor;
c) The equation motion;
d) The speed with which the ball hits the ground.

Problem data:

Initial velocity of the ball on the table: v₀;
Table height: H.

Problem diagram:

We choose a frame of reference on the table with the axis Ox pointing to the right and Oy pointing down, the acceleration due to gravity pointing down, and the point where the ball falls from the table at (x₀, y₀) = (0, 0), (Figure 1).

Figure 1

Solution

The motion can be decomposed in the x and y directions. The initial velocity v₀, with which the ball rolls on the table in the x direction, will be the only speed until the instant when the ball falls over the edge of the table, in the y direction the initial speed will be zero, in magnitude

\[ \begin{gather} v_{0 x} = v_0 \tag{I}\\[10pt] v_{0 y} = 0 \tag{II} \end{gather} \]

Decomposing the motion in the x direction there is no acceleration acting on the ball, it is in Uniform Rectilinear Motion and its motion is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {S_x=S_{0x}+v_xt} \end{gather} \]

since in uniform motion v_x = v_0x is constant, we can substitute v_x by the value of (I) and S_0x = 0

\[ \begin{gather} S_x=0+v_0t\\[5pt] S_x=v_0t \tag{III} \end{gather} \]

In the y direction, the ball is under the action of the acceleration due to gravity, so it is in free fall given by

\[ \begin{gather} \bbox[#99CCFF,10px] {S_y=S_{0y}+v_{0y}t+\frac{g}{2}t^2} \end{gather} \]

\[ \begin{gather} \bbox[#99CCFF,10px] {v_y=v_{0y}+gt} \end{gather} \]

substituting v_0y with the value given in (II) and S_0y = 0

\[ \begin{gather} S_y=0+0\times t+\frac{g}{2}t^2\\[5pt] S_y=\frac{g}{2}t^2 \tag{IV} \end{gather} \]

\[ \begin{gather} v_y=gt \tag{V} \end{gather} \]

with constant g (the positive sign indicates that the acceleration due to gravity is in the same direction of the reference frame).

In Figure 2, we see that in the movement along the x direction we have that for equal time intervals, we have equal displacement (Δx₁ = Δx₂ = Δx₃ = Δx₄). In the y direction, we have, at the instant the ball falls from the table, the velocity v_y starts to increase from zero under the action of gravity, for equal time intervals we have increasingly larger displacements (Δ₁ < Δ₂ < Δ₃ < Δ₄).

Figure 2

a) The time interval for the ball to hit the ground will be obtained from equation (IV) with the condition that on the ground, the height of the fall is the height of the table S_y = H

\[ \begin{gather} H=\frac{g}{2}t^2\\[5pt] t^2=\frac{2H}{g} \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {t=\sqrt{\frac{2H}{g}\;}} \end{gather} \]

b) The time interval calculated above, for the ball to fall to the ground, is also the time interval it will take to go from the origin to point D along the x-axis, substituting the answer to the previous item in the expression ( III)

\[ \begin{gather} \bbox[#FFCCCC,10px] {D=v_0\sqrt{\frac{2H}{g}\;}} \end{gather} \]

c) To obtain the equation of motion (Figure 1), we have to have y as a function of x, or y = f(x), using equations (III) and (IV) for the motions in x and y. Solving the expression (III) for time

\[ \begin{gather} t=\frac{S_x}{v_0} \end{gather} \]

substituting this value in equation (IV)

\[ \begin{gather} S_y=\frac{g}{2}\left(\frac{S_x}{v_0}\right)^2 \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {S_y={\frac{g}{2v_0^2}S_x^2}} \end{gather} \]

Making the association with a Quadratic Equation of the type y = ax²+bx+c

\[ \begin{array}{c} S_y & = & {\dfrac{g}{2v_0^2}} & S_x^2 & + & 0 & S_x & + & 0\\ \downarrow & & \downarrow & \downarrow & & \downarrow & \downarrow & & \downarrow \\ y & = & a & x^2 & + & b & x & + & c \end{array} \]

we see that we obtained a function of the type S_y = f(S_x) with the coefficient a > 0, which indicates that the trajectory is a parabola with the concave in the same direction of y-axis (in this case in the downwards direction contrary to what is usually done ).

d) When the ball hits the ground its velocity has components in the x and y directions (Figure 3). The velocity in the x direction is given by expression (I), and the velocity in the y direction is obtained from expression (V) where the time is substituted by the value found in item (a)

\[ \begin{gather} v_y=g\sqrt{\frac{2H}{g}\;}\\[5pt] v_y=\sqrt{\frac{2Hg^{\cancel 2}}{\cancel g}\;}\\[5pt] v_y=\sqrt{2gH\;} \end{gather} \]

Figure 3

The speed of the ball will be given by the vector sum

\[ \begin{gather} \vec v={\vec v}_x+{\vec v}_y \end{gather} \]

The magnitude is obtained by applying the Pythagorean Theorem

\[ \begin{gather} v^2=v_x^2+v_y^2\\[5pt] v^2=v_0^2+\left(\sqrt{2gH}\right)^2\\[5pt] v^2=v_0^2+2gH \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {v=\sqrt{v_0^2+2gH\;}} \end{gather} \]

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