Question

Снаряд массой м= 10 кг в верхней точке траектории имеет скорость u= 200 м/с. В этой точке он разорвался на два осколка. Один осколок массой м1= 3кг полетел вперед со скоростью u1=400 м/с пол углом φ1 =60° к горизонту. с какой скоростью u2 и под каким углом к горизонту φ2 полетит второй осколок??

Answer 1

Какой класс?физика..

Answer 2

Problem Analysis

We are given a projectile that weighs 10 kg and has a velocity of 200 m/s at the highest point of its trajectory. The projectile breaks into two fragments, with one fragment weighing 3 kg and flying forward with a velocity of 400 m/s at an angle of 60 degrees to the horizon. We need to find the velocity and angle of the second fragment.

Solution

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

Let's denote the velocity and angle of the second fragment as u2 and φ2, respectively.

Conservation of Momentum

The total momentum before the explosion is equal to the total momentum after the explosion. The momentum of an object is given by the product of its mass and velocity.

Before the explosion: - Mass of the projectile (m) = 10 kg - Velocity of the projectile (u) = 200 m/s

After the explosion: - Mass of the first fragment (m1) = 3 kg - Velocity of the first fragment (u1) = 400 m/s - Mass of the second fragment (m2) = unknown - Velocity of the second fragment (u2) = unknown

Using the conservation of momentum, we can write the equation:

m * u = m1 * u1 + m2 * u2

Substituting the given values:

10 kg * 200 m/s = 3 kg * 400 m/s + m2 * u2

Simplifying the equation:

2000 kg * m/s = 1200 kg * m/s + m2 * u2

m2 * u2 = 2000 kg * m/s - 1200 kg * m/s

m2 * u2 = 800 kg * m/s

Conservation of Kinetic Energy

The total kinetic energy before the explosion is equal to the total kinetic energy after the explosion. The kinetic energy of an object is given by half the product of its mass and the square of its velocity.

Before the explosion: - Mass of the projectile (m) = 10 kg - Velocity of the projectile (u) = 200 m/s

After the explosion: - Mass of the first fragment (m1) = 3 kg - Velocity of the first fragment (u1) = 400 m/s - Mass of the second fragment (m2) = unknown - Velocity of the second fragment (u2) = unknown

Using the conservation of kinetic energy, we can write the equation:

(1/2) * m * u^2 = (1/2) * m1 * u1^2 + (1/2) * m2 * u2^2

Substituting the given values:

(1/2) * 10 kg * (200 m/s)^2 = (1/2) * 3 kg * (400 m/s)^2 + (1/2) * m2 * u2^2

Simplifying the equation:

10000 kg * m^2/s^2 = 1800 kg * m^2/s^2 + (1/2) * m2 * u2^2

(1/2) * m2 * u2^2 = 10000 kg * m^2/s^2 - 1800 kg * m^2/s^2

(1/2) * m2 * u2^2 = 8200 kg * m^2/s^2

m2 * u2^2 = 16400 kg * m^2/s^2

Solving for u2 and φ2

We have two equations:

m2 * u2 = 800 kg * m/s (from conservation of momentum) m2 * u2^2 = 16400 kg * m^2/s^2 (from conservation of kinetic energy)

Dividing the second equation by the first equation:

u2^2 = (16400 kg * m^2/s^2) / (800 kg * m/s)

Simplifying:

u2^2 = 20.5 m^2/s^2

Taking the square root of both sides:

u2 = √(20.5 m^2/s^2)

u2 ≈ 4.53 m/s

Now, we can use the equation for the tangent of an angle to find φ2:

tan(φ2) = (u2 * sin(φ1)) / (u1 + u2 * cos(φ1))

Substituting the given values:

tan(φ2) = (4.53 m/s * sin(60°)) / (400 m/s + 4.53 m/s * cos(60°))

Using a calculator, we can find:

tan(φ2) ≈ 0.087

Taking the inverse tangent of both sides:

φ2 ≈ 4.99°

Answer

The second fragment will fly with a velocity of approximately 4.53 m/s and at an angle of approximately 4.99 degrees to the horizon.

Note: The calculations provided above are based on the given information and assumptions.