Question

Пуля массой 5г летит горизонтально со скоростью 300 м/с и пробивает насквозь деревянный брусок массой 200 г ,лежащий на столе .После этого брусок приходит в движение ,а пуля вылетает из него со скоростью 100м/с .Определите кинетическую энергию бруска сразу после вылета пули .

Answer 1

1. закон сохранения импульса m1v1 + m2v2 = m1v3 + m2v4 / 0.005 * 300 + 0.2 * 0 = V * 0.2 + 0.0005 * 100 / V = 5 м/с
2. Ек = mv^2/2 = 0.2 *25/2 = 2.5

Answer 2

Initial Parameters

A bullet with a mass of 5g is traveling horizontally at a speed of 300 m/s and penetrates a wooden block with a mass of 200g lying on a table. After this, the block starts moving, and the bullet exits it with a speed of 100 m/s.

Solution

To determine the kinetic energy of the block immediately after the bullet exits, we can use the principle of conservation of momentum and energy.

First, let's calculate the initial kinetic energy of the bullet and the block using the given data.

The initial kinetic energy of the bullet can be calculated using the formula: \[ KE_{bullet} = \frac{1}{2} m_{bullet} v_{bullet}^2 \]

Where: - \( m_{bullet} = 5g = 0.005kg \) (mass of the bullet) - \( v_{bullet} = 300 m/s \) (speed of the bullet)

The initial kinetic energy of the block can be calculated using the formula: \[ KE_{block} = \frac{1}{2} m_{block} v_{block}^2 \]

Where: - \( m_{block} = 200g = 0.2kg \) (mass of the block) - \( v_{block} = 0 \) m/s (initial speed of the block)

Let's calculate the initial kinetic energy of the bullet and the block using the given data.

Calculations

The initial kinetic energy of the bullet: \[ KE_{bullet} = \frac{1}{2} \times 0.005 \times (300)^2 \]

The initial kinetic energy of the block: \[ KE_{block} = \frac{1}{2} \times 0.2 \times (0)^2 \]

Results

The initial kinetic energy of the bullet is calculated to be \( KE_{bullet} = 225J \). The initial kinetic energy of the block is calculated to be \( KE_{block} = 0J \), as the initial speed of the block is zero.

Therefore, immediately after the bullet exits the block, the kinetic energy of the block is \( 225J \).