Question

В проплывающую под мостом лодку массой m1=150 кг вертикально опускают груз массой m2=50 кг. Модуль начальной скорости движения лодки v1=4 м/с. Найдите модуль конечной скорости движения лодки.

Answer 1

150*4=600
600/200=3-ответ

Answer 2

Problem Analysis

We are given a boat with a mass of m1 = 150 kg and a vertically lowered cargo with a mass of m2 = 50 kg. The initial velocity of the boat is v1 = 4 m/s. We need to find the final velocity of the boat.

Solution

To solve this problem, we can apply the principle of conservation of momentum. According to this principle, the total momentum before an event is equal to the total momentum after the event, assuming no external forces are acting.

The momentum of an object is given by the product of its mass and velocity: momentum = mass * velocity.

Before the cargo is dropped, the boat and the cargo are moving together with an initial velocity of v1. After the cargo is dropped, the boat will experience a change in velocity due to the change in momentum caused by the dropped cargo.

Let's assume the final velocity of the boat is v2. The momentum of the boat before the cargo is dropped is given by m1 * v1, and the momentum of the cargo is given by m2 * 0 (since it is at rest before being dropped).

According to the principle of conservation of momentum, the total momentum before the cargo is dropped is equal to the total momentum after the cargo is dropped:

m1 * v1 = (m1 + m2) * v2

Now we can solve for v2:

v2 = (m1 * v1) / (m1 + m2)

Substituting the given values:

v2 = (150 kg * 4 m/s) / (150 kg + 50 kg)

Calculating the result:

v2 = (600 kg*m/s) / 200 kg

v2 = 3 m/s

Answer

The module of the final velocity of the boat is 3 m/s.