Question

Вертикально выпущенный снаряд в верхней точке траектории на высоте 2,5 км разорвался на два осколка. Первый осколок массой 1 кг полетел в горизонтальном направлении со скоростью 400м/с. Через какое время t угол между направлениями скоростей осколков будет равен 90 градусов, если масса второго осколка равна 4 кг? Сопротивлением воздуха пренебрегать.

Answer 1

Объяснение:

1кг:400м/с=2,5 ,4:400=0,01, 2,5+0,01=2,51

Answer 2

Problem Analysis

We are given the following information: - A vertically launched projectile reaches its highest point at a height of 2.5 km. - The projectile breaks into two fragments. - The first fragment, with a mass of 1 kg, moves horizontally with a velocity of 400 m/s. - We need to find the time (t) at which the angle between the velocities of the two fragments is 90 degrees, given that the mass of the second fragment is 4 kg.

Solution

To solve this problem, we can use the principles of projectile motion and conservation of momentum.

Let's assume that the vertically launched projectile breaks into two fragments at its highest point. At this point, the vertical component of the velocity of both fragments is zero, and only the horizontal component of the velocity of the first fragment is known (400 m/s).

Since the second fragment has a mass of 4 kg, we need to find its horizontal velocity component. We can use the principle of conservation of momentum to determine this.

According to the principle of conservation of momentum, the total momentum before the explosion is equal to the total momentum after the explosion. Since the vertical component of the velocity is zero for both fragments, we only need to consider the horizontal component.

Let's denote the horizontal velocity component of the second fragment as v2. The horizontal velocity component of the first fragment is given as 400 m/s.

The total momentum before the explosion is the sum of the momentum of the first fragment (m1 * v1) and the momentum of the second fragment (m2 * v2), where m1 is the mass of the first fragment (1 kg) and m2 is the mass of the second fragment (4 kg).

The total momentum after the explosion is the sum of the momentum of the first fragment (m1 * v1) and the momentum of the second fragment (m2 * v2'), where v2' is the horizontal velocity component of the second fragment after the explosion.

Since the total momentum before and after the explosion is the same, we can write the following equation:

m1 * v1 + m2 * v2 = m1 * v1 + m2 * v2'

Simplifying the equation, we get:

m2 * v2 = m2 * v2'

Dividing both sides of the equation by m2, we get:

v2 = v2'

This means that the horizontal velocity component of the second fragment after the explosion is the same as its initial horizontal velocity component.

Now, let's find the time (t) at which the angle between the velocities of the two fragments is 90 degrees. We can use the fact that the angle between two vectors is 90 degrees when their dot product is zero.

The dot product of two vectors is given by the product of their magnitudes and the cosine of the angle between them. In this case, the magnitudes of the velocities of the two fragments are v1 and v2, and the angle between them is 90 degrees.

Therefore, we can write the following equation:

v1 * v2 * cos(90) = 0

Since cos(90) = 0, the equation simplifies to:

v1 * v2 = 0

Substituting the values of v1 (400 m/s) and v2 (v2'), we get:

400 * v2' = 0

Since v2' is the horizontal velocity component of the second fragment after the explosion, we can conclude that it must be zero for the dot product to be zero.

Therefore, the time (t) at which the angle between the velocities of the two fragments is 90 degrees is the time it takes for the second fragment to come to a stop horizontally.

To find this time, we can use the equation of motion for horizontal motion:

s = ut + (1/2)at^2

where s is the displacement, u is the initial velocity, a is the acceleration, and t is the time.

In this case, the displacement (s) is the distance traveled horizontally by the second fragment (which is zero), the initial velocity (u) is the horizontal velocity component of the second fragment (v2'), the acceleration (a) is zero (since we are neglecting air resistance), and we need to find the time (t).

The equation simplifies to:

0 = v2' * t + (1/2) * 0 * t^2

Since the acceleration is zero, the equation becomes:

0 = v2' * t

Since we know that v2' = 0, the equation simplifies to:

0 = 0 * t

This equation is satisfied for any value of t.

Therefore, the time (t) at which the angle between the velocities of the two fragments is 90 degrees is not determined by the given information. It can be any value.

In conclusion, the time (t) at which the angle between the velocities of the two fragments is 90 degrees is not determined by the given information. It can be any value.

Note: The given information does not provide enough information to determine the time (t) at which the angle between the velocities of the two fragments is 90 degrees.