GSO001 Problem 5

This is a standard problem combining the definition of the coefficient of restitution with conservation of energy for free fall and upward projection.

1. Velocity Just Before the First Collision

The speed $v_1$v1​ just before the ball falling from height $H$H hits the floor, by conservation of mechanical energy:

$$m g H = \frac{1}{2} m v_1^2 \implies v_1 = \sqrt{2gH}$$mgH=21​mv12​⟹v1​=2gH​

2. Velocity Just After the First Collision and Height Reached

By the definition of the coefficient of restitution $e$e, the speed just after hitting the floor $v'_1$v1′​ (taking upward as positive) is:

$$v'_1 = e v_1 = e \sqrt{2gH}$$v1′​=ev1​=e2gH​

The height $h_1$h1​ reached by the ball with this upward speed:

$$\frac{1}{2} m (v'_1)^2 = m g h_1 \implies h_1 = \frac{(e v_1)^2}{2g} = \frac{e^2 \cdot 2gH}{2g} = e^2 H$$21​m(v1′​)2=mgh1​⟹h1​=2g(ev1​)2​=2ge2⋅2gH​=e2H

3. Velocity Just After the Second Collision and Height Reached

Similarly, letting $v_2$v2​ be the speed just before the second collision and $v'_2$v2′​ be the speed just after:

$$v_2 = \sqrt{2g h_1} = \sqrt{2g e^2 H} = e \sqrt{2gH}$$v2​=2gh1​​=2ge2H​=e2gH​ $$v'_2 = e v_2 = e^2 \sqrt{2gH}$$v2′​=ev2​=e22gH​

The height $h_2$h2​ reached after this collision:

$$h_2 = \frac{(v'_2)^2}{2g} = \frac{(e^2 \sqrt{2gH})^2}{2g} = \frac{e^4 \cdot 2gH}{2g} = e^4 H$$h2​=2g(v2′​)2​=2g(e22gH​)2​=2ge4⋅2gH​=e4H

4. Numerical Calculation

Substituting the given values:

$$h_2 = (0.5)^4 \times 100 = 0.0625 \times 100 = 6.25 \text{ m}$$h2​=(0.5)4×100=0.0625×100=6.25 m

The required answer is this value multiplied by 100:

$$6.25 \times 100 = 625$$6.25×100=625

Answer: 625