Question

Брусок массой m1 = 500 г соскальзывает по наклонной плоскости с высоты h = 0,8 м и, двигаясь по горизонтальной поверхности, сталкивается с бруском массой m2 = 300 г, лежащем на горизонтальной плоскости. Считая столкновение абсолютно неупругим, определите скорость второго бруска в результате столкновения. Трением при движении пренебречь.

Answer 1

H = 0.8 м - высота горки, с которой без трения соскальзывает брусок с массой
m₁ = 0.5 кг
m₂ = 0.3 кг - масса покоящегося бруска

v₁ - скорость первого бруска можно определить из закона сохранения механической энергии:
m₁gh = m₁v₁²/2 откуда
v₁ = √2gh = √2·0.8·10 = 4 м/с
Конечные скорости u₁  и u₂ брусков после того, как первый брусок испытал упругое лобовое столкновение с покоящимся бруском можно получить из законов сохранения импульса и сохранения энергии:
m₁v₁ = m₁u₁ + m₂v₂                      (*)
m₁v₁²/2 = m₁u₁²/2 + m₂u₂²/2          (**)
Выразив скорость первого бруска из первого уравнения
u₁ = (m₁v₁ - m₂u₂)/m₁ (***)
cледует подставить это выражение во второе. Решая его относительно u₂, получим:
u₂ = 2m₁v₁/(m₁ + m₂) = 2*0.5*4/0.8 = 5 м/с
Ответ: скорость второго бруска равна 5 м/с

PS
Вдруг да понадобится для однотипных задач, чтоб заново не выводить.
Получить конечную скорость первого бруска можно, подставив  u₂ в выражение для u₁ (***) после чего получится:
u₁ = v₁(m₁ - m₂)/(m₁ + m₂) = 4*0.2/0.8 = 1 м/с
В том, что вроде бы ни в чём не проврались можно убедиться, подставив значения для m₁ m₂ v₁ u₁ u₂ в исходные уравнения (*) и (**).

Answer 2

Problem Analysis

We are given a scenario where a block of mass m1 = 500 g slides down an inclined plane from a height h = 0.8 m and collides with another block of mass m2 = 300 g that is initially at rest on a horizontal surface. The collision is assumed to be perfectly inelastic, and we need to determine the velocity of the second block after the collision. We are also told to neglect friction during the motion.

Solution

To solve this problem, we can use the principle of conservation of mechanical energy and the principle of conservation of linear momentum.

1. Conservation of Mechanical Energy: - The initial potential energy of the first block is converted into kinetic energy as it slides down the inclined plane. - The kinetic energy of the first block is then transferred to the second block during the collision. - After the collision, the system will have a new total kinetic energy, which is equal to the kinetic energy of the second block. - Since there is no friction, we can assume that the total mechanical energy of the system is conserved.

2. Conservation of Linear Momentum: - The principle of conservation of linear momentum states that the total momentum of an isolated system remains constant before and after a collision. - In this case, the system consists of the two blocks. - Before the collision, the first block has momentum in the downward direction due to its velocity, and the second block has zero momentum since it is at rest. - After the collision, the two blocks will move together with a common velocity.

Let's calculate the velocity of the second block after the collision using these principles.

Calculation

1. Conservation of Mechanical Energy: - The initial potential energy of the first block is given by m1 * g * h, where g is the acceleration due to gravity. - The final kinetic energy of the second block is given by (m1 + m2) * v^2 / 2, where v is the common velocity of the two blocks after the collision. - Equating these two energies, we have: m1 * g * h = (m1 + m2) * v^2 / 2

2. Conservation of Linear Momentum: - The initial momentum of the first block is given by m1 * v1, where v1 is the velocity of the first block before the collision. - The final momentum of the system is given by (m1 + m2) * v, where v is the common velocity of the two blocks after the collision. - Equating these two momenta, we have: m1 * v1 = (m1 + m2) * v

We have two equations with two unknowns (v and v1). We can solve these equations simultaneously to find the values of v and v1.

Substitution and Calculation

Let's substitute the given values into the equations and solve for v.

From the equation of conservation of mechanical energy: m1 * g * h = (m1 + m2) * v^2 / 2

Substituting the values: m1 = 500 g = 0.5 kg m2 = 300 g = 0.3 kg h = 0.8 m g = 9.8 m/s^2

0.5 kg * 9.8 m/s^2 * 0.8 m = (0.5 kg + 0.3 kg) * v^2 / 2

Simplifying the equation: 3.92 = 0.8 v^2

Dividing both sides by 0.8: v^2 = 4.9

Taking the square root of both sides: v = 2.2 m/s

Therefore, the velocity of the second block after the collision is 2.2 m/s.

Answer

The velocity of the second block after the collision is 2.2 m/s.