Question

Шарик массой 250 г падает без начальной скорости с высоты H = 7,2 м на горизонтальный пол. После отскока от пола шарик поднимается на высоту H/9. Найдите модуль изменения импульса в процессе отскока шарика от пола. Пожалуйста с решением!!!

Answer 1

Ответ:

Объяснение:

Дано:

m = 250 г =0,250 кг

H = 7,2 м

h = H / 9 = 0,8 м

________________

Δp - ?

Скорость шарика в момент удара о пол:

V₁ = √ (2·g·H) = √ (2·10·7,2) = 12 м/с      (вектор скорости направлен вниз)

Скорость шарика в момент отскока от пола:

V₂ = √ (2·g·h) = √ (2·10·0,8) = 4  м/с   (вектор скорости направлен вверх)

Считаем положительным направлением направление вертикально вверх. Тогда изменение импульса шарика:

Δp = m·(V₂ - (-V₁)) = m·(V₂ + V₁) = 0,250·(4+12) = 4 кг·м/с

Answer 2

Problem Analysis

We are given the following information: - The mass of the ball, m = 250 g = 0.25 kg. - The initial height from which the ball falls, H = 7.2 m. - The ball rebounds from the floor and reaches a height of H/9.

We need to find the magnitude of the change in momentum during the rebound process.

Solution

To solve this problem, we can use the principle of conservation of mechanical energy. According to this principle, the total mechanical energy of a system remains constant if no external forces are acting on it.

The mechanical energy of the ball at the initial height is given by the sum of its potential energy and kinetic energy:

E_initial = mgh

where: - m is the mass of the ball, - g is the acceleration due to gravity, and - h is the initial height from which the ball falls.

The mechanical energy of the ball at the maximum height it reaches during the rebound is given by:

E_max = mgh_max/9

where: - h_max is the maximum height the ball reaches during the rebound.

Since no external forces are acting on the ball during the rebound, the total mechanical energy before and after the rebound should be the same:

E_initial = E_max

Substituting the expressions for E_initial and E_max, we get:

mgh = mgh_max/9

We can cancel out the mass, m, from both sides of the equation:

gh = gh_max/9

Now, we can solve for h_max, which is the maximum height the ball reaches during the rebound:

h_max = 9h

The change in momentum during the rebound can be calculated using the equation:

Change in momentum = 2mv

where: - m is the mass of the ball, and - v is the velocity of the ball during the rebound.

To find the velocity, v, we can use the equation for conservation of mechanical energy:

E_initial = E_max = 1/2 mv^2

Substituting the expressions for E_initial and E_max, we get:

mgh = 1/2 mv^2

We can cancel out the mass, m, from both sides of the equation:

gh = 1/2 v^2

Solving for v, we get:

v = sqrt(2gh)

Substituting the value of h_max into the equation, we get:

v = sqrt(2g(9h)) = 3sqrt(2gh)

Now, we can calculate the change in momentum:

Change in momentum = 2mv = 2(0.25)(3sqrt(2gh)) = 0.15sqrt(2gh)

Therefore, the magnitude of the change in momentum during the rebound process is 0.15sqrt(2gh).

Answer

The magnitude of the change in momentum during the rebound process is 0.15sqrt(2gh).