Question

Вагон массой m1=1кг с автоматической сцепкой, движущийся со скоростью v1=3м/с, догоняет такой же скоростью v2, и сцепляется с ним. Двигаясь дальше вместе, оба вагона сталкиваются со стоящим на рельсах третьим вагоном массы m3=4кг. Найти скорости движения вагонов на разных участках пути. Трением пренебречь решение с пояснением

Answer 1

Ответ:  2,5м/с; 2м/с

Объяснение:

По закону сохранения импульса импульс до сцепки после равны

m1*V1+m2*V2 = (m1+m2)*V'

Тогда скорость после сцепки равна

V'=(m1*V1+m2*V2)/(m1+m2)= (2*5+3*6)/(5+6)=2,5[м/с]

По закону сохранения импульса импульс до сцепки после равны

((m1+m2)V'+m3*V3) = (m1+m2+m3)*V''

Тогда скорость после сцепки равна

V''=((m1*+m2*)*V'+V*m3)/(m1+m2+m3)= ((2,5(5+6)+0*3)/(5+6+3)=2 [м/с]

Answer 2

Problem Analysis

We have three wagons: wagon 1 with a mass of m1 = 1 kg, wagon 2 with an unknown mass, and wagon 3 with a mass of m3 = 4 kg. Wagon 1 is moving with a velocity of v1 = 3 m/s and catches up with wagon 2, which is also moving with a velocity of v2. After the collision, both wagons collide with wagon 3, which is stationary. We need to find the velocities of the wagons on different sections of the track.

Solution

Let's assume that after the collision, the two wagons move together with a common velocity v. We can use the principle of conservation of momentum to solve this problem. According to this principle, the total momentum before the collision is equal to the total momentum after the collision.

The momentum of an object is given by the product of its mass and velocity: momentum = mass * velocity.

Before the collision, the momentum of wagon 1 is m1 * v1, and the momentum of wagon 2 is m2 * v2. After the collision, the momentum of the combined wagons is (m1 + m2) * v.

Using the conservation of momentum, we can write the equation: m1 * v1 + m2 * v2 = (m1 + m2) * v

Since we know the values of m1, v1, and v2, we can substitute them into the equation and solve for v.

m1 * v1 + m2 * v2 = (m1 + m2) * v 1 kg * 3 m/s + m2 * v2 = (1 kg + m2) * v

Now, let's consider the collision between the combined wagons and wagon 3. Since wagon 3 is stationary, its initial momentum is zero. After the collision, the momentum of the combined wagons is (m1 + m2 + m3) * v'.

Using the conservation of momentum again, we can write the equation: (m1 + m2) * v = (m1 + m2 + m3) * v'

Substituting the known values, we have: (1 kg + m2) * v = (1 kg + m2 + 4 kg) * v'

Now we have two equations with two unknowns (m2 and v'). We can solve these equations simultaneously to find the values.

Let's solve the equations step by step.

From the first equation: 1 kg * 3 m/s + m2 * v2 = (1 kg + m2) * v

Simplifying: 3 m/s + m2 * v2 = v + m2 * v

Rearranging the terms: m2 * v2 - m2 * v = v - 3 m/s

Factoring out m2: m2 * (v2 - v) = v - 3 m/s

Dividing both sides by (v2 - v): m2 = (v - 3 m/s) / (v2 - v)

Now, let's substitute this value of m2 into the second equation: (1 kg + m2) * v = (1 kg + m2 + 4 kg) * v'

Substituting the value of m2: (1 kg + (v - 3 m/s) / (v2 - v)) * v = (1 kg + (v - 3 m/s) / (v2 - v) + 4 kg) * v'

Expanding and simplifying: (v2 - v + v - 3 m/s) * v = (v2 - v + v - 3 m/s + 5(v2 - v)) * v'

Simplifying further: (v2 - 3 m/s) * v = (6 v2 - 2 v - 3 m/s) * v'

Dividing both sides by v: v2 - 3 m/s = (6 v2 - 2 v - 3 m/s) * v'

Expanding and rearranging the terms: v2 - 3 m/s = 6 v2 * v' - 2 v * v' - 3 m/s * v'

Rearranging the terms: 6 v2 * v' - v2 + 2 v * v' = 3 m/s * v' - 3 m/s

Factoring out v': v' * (6 v2 + 2 v - 3 m/s) = 3 m/s - v2

Dividing both sides by (6 v2 + 2 v - 3 m/s): v' = (3 m/s - v2) / (6 v2 + 2 v - 3 m/s)

Now we have the values of m2 and v'. We can substitute them back into the equations to find the velocities of the wagons on different sections of the track.

Results

Using the equations derived above, we can calculate the values of m2 and v'.

From equation : m2 = (v - 3 m/s) / (v2 - v)

From equation : v' = (3 m/s - v2) / (6 v2 + 2 v - 3 m/s)

Substituting the given values: m2 = (v - 3 m/s) / (v2 - v) v' = (3 m/s - v2) / (6 v2 + 2 v - 3 m/s)

Please provide the values of v and v2 to calculate the velocities of the wagons on different sections of the track.