On a smooth horizontal surface there is a wooden

A wooden box with sand of mass M = 50 kg is located on a smooth horizontal surface. A bullet of mass m = 10 g, moving horizontally with a speed V0 = 800 m/s, hits it and gets stuck in it. It is necessary to determine the maximum deformation of the spring holding the box if its stiffness k = 1 kN/m.

To solve the problem we use the law of conservation of momentum. Since the bullet gets stuck in the box, after the collision the system (bullet + box) moves as one. Therefore, we can write:

m * V0 = (M + m) * V

Where V is the speed of the system after the collision.

From this equation we can express the speed V:

V = m * V0 / (M + m)

To determine the maximum deformation of the spring, we use Hooke's law:

F = k * x

Where F is the force acting on the spring, x is its deformation, k is the spring stiffness.

The force acting on the spring is equal to the gravitational force of the box of sand, which is located on the surface of the spring:

F = (M + m) * g

Where g is the acceleration of gravity.

Thus, the maximum deformation of the spring will be equal to:

x = F / k = (M + m) * g / k

Substituting numerical values ​​and solving the equation, we get:

x = ((50 + 0.01) * 9.81) / 1000 = 0.499 mm.

Thus, the maximum deformation of the spring holding the box is 0.499 mm.

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Greetings! As described in the condition of problem 10557, a wooden box with sand of mass M = 50 kg is located on a smooth horizontal surface. A bullet of mass m = 10 g, moving horizontally with a speed V0 = 800 m/s, hits it and gets stuck in it. It is necessary to find the maximum deformation of the spring holding the box if its stiffness k = 1 kN/m.

To solve the problem we can use the law of conservation of momentum. Since the bullet gets stuck in the box, after the collision the system (bullet + box) moves as one. Therefore, we can write:

m * V0 = (M + m) * V

Where V is the speed of the system after the collision.

From this equation we can express the speed V:

V = m * V0 / (M + m)

To determine the maximum deformation of a spring, we can use Hooke's law:

F = k * x

Where F is the force acting on the spring, x is its deformation, k is the spring stiffness.

The force acting on the spring is equal to the gravitational force of the box of sand, which is located on the surface of the spring:

F = (M + m) * g

Where g is the acceleration of gravity.

Thus, the maximum deformation of the spring will be equal to:

x = F / k = (M + m) * g / k

Substituting numerical values ​​and solving the equation, we get:

x = ((50 + 0.01) * 9.81) / 1000 = 0.499 mm.

Thus, the maximum deformation of the spring holding the box is 0.499 mm.

Hope this helps you solve the problem! If you have any additional questions, don't hesitate to reach out.


***


On a smooth horizontal surface there is a wooden box with sand of mass M = 50 kg. A bullet of mass m = 10 g, flying horizontally at a speed V0 = 800 m/s, hits it and gets stuck in it.

To solve the problem, it is necessary to calculate the maximum deformation of the spring holding the box if its stiffness k = 1 kN/m.

The first step is to calculate the momentum of the bullet before it hits the box. Impulse is defined as the product of a body's mass and its speed:

p = m * V0 = 10 g * 800 m/s = 8 N * s

The law of conservation of momentum states that the sum of the momentum of a system of bodies before and after a collision remains unchanged. Since the box with sand is motionless, after the collision the momentum of the system will be equal to the momentum of the bullet:

p' = p = 8 Н * с

To calculate the maximum deformation of a spring, it is necessary to calculate its potential energy, which is equal to the work done by the elastic force when the spring is deformed. The potential energy of the spring is:

Ep = (k * x^2) / 2

where k is the spring stiffness and x is the maximum deformation.

From the law of conservation of energy, we can express the maximum deformation of the spring:

Ep = p'^2 / (2 * M) = (8 N * s)^2 / (2 * 50 kg) = 6,400 J

x = sqrt((2 * Ep) / k) = sqrt((2 * 6,400 J) / 1 kN/m) = 80 mm

Thus, the maximum deformation of the spring holding the box will be 80 mm.


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