Question

An elastic ball is dropped on a long inclined plane. If bounces, hits the plane again, bounces and so on. let us Label the distance between the point of the first and second hit $d_{12}$ and the distance between the points of second and the third hit is $d_{23}$ . find the ratio of $d_{12}/d_{23}$ .

• A. $\frac{2}{1}$
• B. $\frac{1}{2}$
• C. $\frac{4}{1}$
• D. $\frac{1}{4}$

Answer 1

Answer: (B)

If we rotate the coordinate system so that the ramp is horizontal, then the free fall acceleration will have two components one downward $a_y=-gcos\theta$ and one horizontal $a_x=gsin\theta .$ In this frame, the initial velocity will have components given by $V_y=V_{0}cos\theta$ & $V_x=V_{0}sin\theta$ time for each bounce is given by $t_b=\frac{2V_y}{-ay}=\frac{2V_0}{g}$
The horizontal displacement $\Delta x=V_{x}t+0.5a_x{t}^2$
given these equation
$d_{12}=V_{0}sin\theta \left(\frac{2V_0}{g}\right)+0.5gsin\theta {\left(\frac{2V_0}{g}\right)}^2$
$=\frac{4V_0^{2}sin\theta }{g}$
$d_{13}=V_{0}sin\theta \left(\frac{4V_0}{g}\right)+0.5gsin\theta {\left(\frac{4V_0}{g}\right)}^2$
$=\frac{12V_0^{2}sin\theta }{g}$
Since $d_{23}=d_{13}-d_{12}=\frac{8V_0^{2}sin\theta }{g}$
$\frac{d_{12}}{d_{23}}=\frac{1}{2}$