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 d₁₂ and the distance between the points of second and the third hit is d₂₃. find the ratio of d₁₂/d₂₃ .

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

Answer 1

Answer: (B)

$\frac{1}{2}$

Answer 2

If we rotate the coordinate system 80 that the ramp is horizontal then the free fall acceleration will have two components one downward a_y = – g cos q and one horizontal a_x = g sinq. In this frame, the initial velocity will have components given by V_y = V₀cosq & V_x = V₀ sinq time for each bounce is given by t_b = $\frac{2V_{y}}{–ay}$= $\frac{2V_{0}}{g}$

The horizontal displacement

Dx = V_xt + 0.5 a_xt²

given these equation

d₁₂ = V₀ sinq $\left( \frac{2v_{0}}{g} \right)$ + 0.5g sinq $\left( \frac{2V_{0}}{g} \right)^{2}$

= $\frac{4V_{0}^{2}su\theta}{g}$

d₁₃ = V₀sinq$\left( \frac{4V_{0}}{g} \right)$ + 0.5 g sinq $\left( \frac{4V_{0}}{g} \right)^{2}$

= $\frac{12{V_{0}}^{2}\sin\theta}{g}$

Since d₂₃ = d₁₃

d₁₂ = $\frac{8{V_{0}}^{2}\sin\theta}{g}$

$\frac{d_{12}}{d_{23}}$ = $\frac{1}{2}$