Answer to Question #125059 in Physics for Adeboye oluwaseyifunmi

Question

A ballot mass 200g released from a height of 2.0m hits a horizontal floor and renounce height of 1.8v. calculate the impulse received by the floor.( G=10m/s)

Answer 1

According to the energy conservation law, the kinetic energy of the ball just before the touching the floor is equal to its potential energy before it was released. Thus:

$$K_1 = U_1 = mgh_1$$

where $m = 0.2kg$, $h = 2m$, $g = 10 m/s^2$.

Similarly, the kinetic energy of the ball just after bounce is equal to its potential energy at the height is renounced. Thus:

$$K_2 =U_2 = mgh_2$$

where $h_2 = 1.8m$.

As far as the kinetic energy and momentum are connected in the following way:

$$K = \dfrac{p^2}{2m}$$

the impulse received by the floor will be equal to the sum of the modules of the initial and final impulses, as far as they point in the different diractions:

$$J = p_1 + p_2 = \sqrt{2m} (\sqrt{K_1}+\sqrt{K_2}) = \sqrt{2g}m(\sqrt{h_1}+\sqrt{h_2})$$

Substituting the numerical values:

$$J = \sqrt{2\cdot 10}\cdot0.2\cdot(\sqrt{2}+\sqrt{1.8})\approx 2.46 kg\cdot m/s$$

Answer. 2.46 kg*m/s.