Question

Летящий снаряд разорвался на два осколка с одинаковыми массами. модули скоростей осколков v1=300м/с и v2=400м/с, а угол между векторами скоростей осколков равен 90°. найдите модуль скорости v0 снаряда до разрыва

Answer 1

Так как импульсы направлены под углом 90 градусов,то
Р=корень из P1^2+P2^2=2m*V2*m-потому что два осколка равной массы
после сокращения массы получаем
V=(корень из V1^2+ V2^2)/2 =500/2=250 м/с.

Answer 2

Calculation of the Projectile's Velocity Before the Explosion

To find the velocity of the projectile before the explosion, we can use the principle of conservation of momentum. According to this principle, the total momentum before the explosion is equal to the total momentum after the explosion.

Let's denote the mass of each fragment as m1 and m2, and the velocities of the fragments as v1 and v2 respectively. The angle between the velocity vectors of the fragments is given as 90°.

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

m1 * v1 + m2 * v2 = m * v0

Where: - m1 and m2 are the masses of the fragments, - v1 and v2 are the velocities of the fragments, - m is the mass of the projectile before the explosion, and - v0 is the velocity of the projectile before the explosion.

Since the masses of the fragments are equal, we can simplify the equation to:

m * (v1 + v2) = m * v0

Given that the angle between the velocity vectors of the fragments is 90°, we can use the Pythagorean theorem to find the magnitude of the resultant velocity vector (v1 + v2):

(v1 + v2)^2 = v1^2 + v2^2

Substituting this into the equation, we have:

m * (v1^2 + v2^2) = m * v0^2

Simplifying further, we get:

v0^2 = v1^2 + v2^2

Taking the square root of both sides, we find:

v0 = sqrt(v1^2 + v2^2)

Now, let's substitute the given values into the equation to find the magnitude of the velocity of the projectile before the explosion.

Given: - v1 = 300 m/s - v2 = 400 m/s

Substituting these values into the equation, we get:

v0 = sqrt((300 m/s)^2 + (400 m/s)^2)

Calculating this, we find:

v0 ≈ 500 m/s

Therefore, the magnitude of the velocity of the projectile before the explosion is approximately 500 m/s.