Question

Heavy elastic ball falls freely from point A at a height H₀ onto the smooth horizontal surface of an elastic plate. As the ball strikes the plate another such ball is dropped from the same point A. At what time t, after the second ball is dropped, and at what height will the balls meet ?

• A. $\sqrt{\frac{H_{0}}{2g}}$; $\frac{3H_{0}}{4}$
• B. $\sqrt{\frac{2H_{0}}{g}}$; $\frac{3H_{0}}{4}$
• C. $\sqrt{\frac{H_{0}}{2g}}$; $\frac{H_{0}}{4}$
• D. $\sqrt{\frac{H_{0}}{g}}$, $\frac{H_{0}}{4}$

Answer 1

Answer: (A)

$\sqrt{\frac{H_{0}}{2g}}$; $\frac{3H_{0}}{4}$

Answer 2

t = $\sqrt{\frac{\mathbf{H}_{\mathbf{0}}}{\mathbf{2g}}}$ ; h₁ = $\frac{\mathbf{3}}{\mathbf{4}}$H₀

The velocity of the first ball at the moment it strikes the plate will be v₀ = $\sqrt{2gH_{0}}$. Since the impact is elastic, the ball will begin to rise after the impact with a velocity of the same magnitude v₀. During the time t the first ball will rise to a height

h₁ = v₀t – $\frac{gt^{2}}{2}$

During this time the second ball will move down from a point A a distance

h₂ = $\frac{gt^{2}}{2}$

At the moment the balls meet, h₁ + h₂ = H₀. Hence,

t = $\frac{H_{0}}{v_{0}}$ = $\sqrt{\frac{H_{0}}{2g}}$.