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}=- \, g \, cos \theta $ and one horizontal $a_{x}=g \, sin \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 $t_{b}=\frac{2 V_{y}}{- a y}=\frac{2 V_{0}}{g}$
The horizontal displacement $\Delta x=V_{x}t \, + \, 0.5 \, a_{x}t^{2}$
given these equation
$d_{12}=V_{0} \, sin \theta \left(\frac{2 V_{0}}{g}\right) \, + \, 0.5g \, sin ⁡ \theta \, \left(\frac{2 V_{0}}{g}\right)^{2}$
$= \, \frac{4 V_{0}^{2} sin \theta }{g}$
$d_{13}=V_{0}sin \theta \left(\frac{4 V_{0}}{g}\right) \, + \, 0.5 \, g \, sin ⁡ \theta \left(\frac{4 V_{0}}{g}\right)^{2}$
$= \, \frac{12 V_{0}^{2} sin \theta }{g}$
Since $d_{23}=d_{13}- \, d_{12}= \, \frac{8 V_{0}^{2} sin \theta }{g}$
$\frac{d_{12}}{d_{23}}=\frac{1}{2}$