Question

Помогите пожалуйста!!!Шар массой m=4 кг движется со скоростью V1=5 м/с навстречу шару массой m2=1 кг. После центрального неупругого удара модуль скорости шаров оказался U=3 м/с, а направление такое же, как у первого шара до столкновения.Определите начальную скорость второго шара количество теплоты, выделившейся при ударе.

Answer 1

Импульс системы до удара - m₁*v₁ - m₂*v₂
Импульс системы после удара - (m₁ + m₂) * u
Запишем закон сохранения импульса
 m₁*v₁ - m₂*v₂ = (m₁ + m₂) * u
v₂ =( m₁*v₁ - (m₁ + m₂) * u) / m₂
v₂ = (4 кг * 5 м/с - (4 кг + 1 кг) * 3 м/с) / 1 кг = (20 - 15) / 1 = 5 м/с
Запишем закон сохранения энергии
m₁*v₁²/2 + m₂*v₂²/2 - (m₁ + m₂)*u²/2 = Q
Q = 4 кг * (5 м/с)² / 2 + 1 кг * (5 м/с)² / 2 - (4 кг + 1 кг) * (3 м/с)² /2 = 50 + 12,5 - 22,5 = 40 Дж

Answer 2

Problem Statement

We have a ball with a mass of m = 4 kg moving with a velocity of V1 = 5 m/s towards another ball with a mass of m2 = 1 kg. After a perfectly inelastic collision, the magnitude of the velocities of both balls is U = 3 m/s, and the direction of the second ball is the same as the direction of the first ball before the collision. We need to determine the initial velocity of the second ball and the amount of heat released during the collision.

Solution

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

# Conservation of Momentum

The principle of conservation of momentum states that the total momentum before the collision is equal to the total momentum after the collision. Mathematically, this can be expressed as:

m1 * V1 + m2 * V2 = (m1 + m2) * U where: - m1 is the mass of the first ball (4 kg), - V1 is the initial velocity of the first ball (5 m/s), - m2 is the mass of the second ball (1 kg), - V2 is the initial velocity of the second ball (unknown), - U is the final velocity of both balls after the collision (3 m/s).

We can rearrange this equation to solve for V2:

V2 = (m1 * V1 + m2 * U) / (m1 + m2)

Substituting the given values, we have:

V2 = (4 kg * 5 m/s + 1 kg * 3 m/s) / (4 kg + 1 kg) V2 = (20 kg·m/s + 3 kg·m/s) / 5 kg V2 = 23 kg·m/s / 5 kg V2 = 4.6 m/s

Therefore, the initial velocity of the second ball is 4.6 m/s.

# Conservation of Kinetic Energy

The principle of conservation of kinetic energy states that the total kinetic energy before the collision is equal to the total kinetic energy after the collision. Mathematically, this can be expressed as:

0.5 * m1 * V1^2 + 0.5 * m2 * V2^2 = 0.5 * (m1 + m2) * U^2 where: - m1 is the mass of the first ball (4 kg), - V1 is the initial velocity of the first ball (5 m/s), - m2 is the mass of the second ball (1 kg), - V2 is the initial velocity of the second ball (4.6 m/s), - U is the final velocity of both balls after the collision (3 m/s).

We can rearrange this equation to solve for the amount of heat released during the collision:

Heat = 0.5 * (m1 * V1^2 + m2 * V2^2 - (m1 + m2) * U^2)

Substituting the given values, we have:

Heat = 0.5 * (4 kg * (5 m/s)^2 + 1 kg * (4.6 m/s)^2 - (4 kg + 1 kg) * (3 m/s)^2) Heat = 0.5 * (4 kg * 25 m^2/s^2 + 1 kg * 21.16 m^2/s^2 - 5 kg * 9 m^2/s^2) Heat = 0.5 * (100 kg·m^2/s^2 + 21.16 kg·m^2/s^2 - 45 kg·m^2/s^2) Heat = 0.5 * 76.16 kg·m^2/s^2 Heat = 38.08 J

Therefore, the amount of heat released during the collision is 38.08 J.

Summary

To summarize, the initial velocity of the second ball is 4.6 m/s, and the amount of heat released during the collision is 38.08 J. These values are obtained by applying the principles of conservation of momentum and conservation of kinetic energy.