Question

Два человека на роликовых коньках стоят друг против друга. Масса первого человека m1 = 70 кг, а второго m2 = 80 кг. Первый бросает второму груз массой m = 10 кг. со скоростью, горизонтальная составляющая которой V = 5 м/с относительно земли. Определить скорость первого человека после броска и второго после того, как он поймает груз. Трение не учитывается.

Answer 1

 m1Vо1+mгVо2=0 (так как до броска скорость первого человека и груза равны 0).Следовательно после броска согласно закону сохранения импульса 
m1V1+mгVг=0  откуда V1=-mгVг/m1=-10*5/70=-5/7[м/c] Знак "-" показывает, что скорость первого человека противоположна скорости груза.
mг*Vг=(mг+m2)V  откуда V=mг*Vг/(mг+m2)=10*5/(10+80)=50/90=5/9[м/с]

Answer 2

Problem Analysis

We have two people on roller skates facing each other. The mass of the first person is m1 = 70 kg, and the mass of the second person is m2 = 80 kg. The first person throws a weight with a mass of m = 10 kg at the second person with a horizontal velocity of V = 5 m/s relative to the ground. We need to determine the velocities of both people after the throw, assuming no friction is present.

Solution

To solve this problem, we can apply the principle of conservation of linear momentum. According to this principle, the total momentum before the throw is equal to the total momentum after the throw.

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

Before the throw, the first person has a momentum of m1 * v1, where v1 is their initial velocity. The second person has a momentum of m2 * v2, where v2 is their initial velocity.

After the throw, the first person's momentum is (m1 * v1) - (m * V), where V is the velocity of the thrown weight relative to the ground. The second person's momentum is (m2 * v2) + (m * V), as they catch the weight.

Since the total momentum before and after the throw must be equal, we can set up the following equation:

(m1 * v1) + (m2 * v2) = (m1 * v1') + (m2 * v2') + (m * V),

where v1' and v2' are the velocities of the first and second person after the throw, respectively.

Now we can solve this equation to find the velocities of both people after the throw.

Calculation

Substituting the given values into the equation, we have:

(70 kg * v1) + (80 kg * v2) = (70 kg * v1') + (80 kg * v2') + (10 kg * 5 m/s).

Simplifying the equation, we get:

70v1 + 80v2 = 70v1' + 80v2' + 50.

Since the people are initially at rest (v1 = v2 = 0), the equation becomes:

0 + 0 = 70v1' + 80v2' + 50.

Simplifying further, we have:

70v1' + 80v2' = -50.

Now we have a system of equations with two unknowns. To solve it, we need another equation. Let's consider the conservation of mass:

m1 + m2 = m1' + m2' + m,

where m1' and m2' are the masses of the first and second person after the throw, respectively.

Since the masses of the people do not change, the equation becomes:

70 kg + 80 kg = m1' + m2' + 10 kg.

Simplifying, we have:

150 kg = m1' + m2' + 10 kg.

Now we have two equations:

70v1' + 80v2' = -50, m1' + m2' = 140 kg.

We can solve this system of equations to find the velocities of both people after the throw and the sum of their masses.

Solution

To solve the system of equations, we can use the method of substitution. Let's solve the second equation for m1':

m1' = 140 kg - m2'.

Substituting this expression into the first equation, we have:

70v1' + 80v2' = -50.

Substituting the value of m1' from the second equation, we get:

70(140 kg - m2') + 80v2' = -50.

Simplifying, we have:

9800 kg - 70m2' + 80v2' = -50.

Rearranging the equation, we get:

70m2' - 80v2' = 9850 kg.

Now we have a system of equations with one unknown:

70m2' - 80v2' = 9850 kg, m1' + m2' = 140 kg.

We can solve this system of equations to find the velocities of both people after the throw and the mass of the second person.

Solution

To solve the system of equations, we can use the method of substitution. Let's solve the second equation for m1':

m1' = 140 kg - m2'.

Substituting this expression into the first equation, we have:

70m2' - 80v2' = 9850 kg.

Now we have a single equation with one unknown. We can solve it to find the velocity of the second person after catching the weight.

Calculation

Solving the equation, we have:

70m2' - 80v2' = 9850 kg.

Let's solve for v2':

80v2' = 70m2' - 9850 kg.

Dividing both sides by 80, we get:

v2' = (70m2' - 9850 kg) / 80.

Now we have the velocity of the second person after catching the weight in terms of the mass of the second person.

Calculation

To find the velocity of the first person after the throw, we can substitute the value of v2' into the second equation:

m1' + m2' = 140 kg.

Substituting the expression for m1' from the second equation, we have:

(140 kg - m2') + m2' = 140 kg.

Simplifying, we get:

140 kg - m2' + m2' = 140 kg.

The mass of the first person cancels out, and we are left with:

140 kg = 140 kg.

This equation is true for any value of m2'. Therefore, the mass of the second person does not affect the velocity of the first person after the throw.

Answer

The velocity of the first person after the throw is 5 m/s (the same as the velocity of the thrown weight), and the velocity of the second person after catching the weight is given by the equation:

v2' = (70m2' - 9850 kg) / 80.

The mass of the second person does not affect the velocity of the first person after the throw.