Question

Деревянный брусок массой 2кг первоначально покоится на абсолютно гладкой горизонтальной поверхности. В брусок попадает и застревает в нем пуля массой 5г. летевшая горизонтально со скоростью 500м.с С какой скоростью станет двигаться брусок с пулей после соударения? ответ дать в еденицах СИ

Answer 1

Закон сохранения импульсов
m2-масса бруска
m1-масса пули
v-скорость и пули и бруска
v1-скорость пули
m1v1=v(m1+m2)
v=m1v1/(m1+m2)
v=0.005*500/2.005
v=1.246 м/с округлить можно до 1.25

Answer 2

Calculation of the Final Velocity of the Block with the Bullet

To calculate the final velocity of the block with the bullet after the collision, we can use the principle of conservation of momentum. According to this principle, the total momentum before the collision is equal to the total momentum after the collision.

The momentum of an object is given by the product of its mass and velocity. Therefore, we can write the equation:

Initial momentum before the collision = Final momentum after the collision

The initial momentum before the collision is the momentum of the block, which is at rest, and the momentum of the bullet, which is moving horizontally with a velocity of 500 m/s.

The final momentum after the collision is the momentum of the block with the bullet.

Let's denote the mass of the block as m1 and the mass of the bullet as m2. The initial velocity of the block is 0 m/s, and the initial velocity of the bullet is 500 m/s.

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

m1 * 0 + m2 * 500 = (m1 + m2) * v

Where v is the final velocity of the block with the bullet.

We are given that the mass of the block is 2 kg and the mass of the bullet is 5 g (0.005 kg).

Substituting the values into the equation, we get:

2 * 0 + 0.005 * 500 = (2 + 0.005) * v

Simplifying the equation, we have:

0.005 * 500 = 2.005 * v

Solving for v, we find:

v = (0.005 * 500) / 2.005

Calculating the value, we get:

v ≈ 0.1247 m/s

Therefore, the block with the bullet will have a final velocity of approximately 0.1247 m/s after the collision.