Question

Из лодки,движущейся к берегу со скоростью 0,5м/с,на берег прыгнул человек со скоростью 2м/с относительно берега.С какой скоростью будет двигаться лодка после прыжка человека,если масса человека 80кг,а масса лодки 120кг?

Answer 1

по закону сохранения импульса:

0.5*(80+120)=120x+80*2

x=2.16 м/с

Answer 2

Calculation of the Final Velocity of the Boat

To determine the final velocity of the boat after the person jumps off, we can apply the principle of conservation of momentum. According to this principle, the total momentum before the jump is equal to the total momentum after the jump.

The momentum of an object is given by the product of its mass and velocity. Therefore, the momentum of the boat before the jump is equal to the momentum of the boat and the person after the jump.

Let's denote the initial velocity of the boat as v_boat_initial, the final velocity of the boat as v_boat_final, the mass of the boat as m_boat, the mass of the person as m_person, and the velocity of the person relative to the shore as v_person_relative.

The momentum before the jump is given by:

momentum_before_jump = (m_boat * v_boat_initial) + (m_person * v_person_relative)

The momentum after the jump is given by:

momentum_after_jump = (m_boat * v_boat_final)

According to the principle of conservation of momentum, these two quantities are equal:

(m_boat * v_boat_initial) + (m_person * v_person_relative) = (m_boat * v_boat_final)

We can rearrange this equation to solve for v_boat_final:

v_boat_final = (m_boat * v_boat_initial + m_person * v_person_relative) / m_boat

Now, let's substitute the given values into the equation and calculate the final velocity of the boat.

Given: - Initial velocity of the boat (v_boat_initial) = 0.5 m/s - Velocity of the person relative to the shore (v_person_relative) = 2 m/s - Mass of the boat (m_boat) = 120 kg - Mass of the person (m_person) = 80 kg

Substituting these values into the equation, we get:

v_boat_final = (120 kg * 0.5 m/s + 80 kg * 2 m/s) / 120 kg

Calculating this expression, we find:

v_boat_final = (60 kg*m/s + 160 kg*m/s) / 120 kg

v_boat_final = 220 kg*m/s / 120 kg

v_boat_final ≈ 1.83 m/s

Therefore, the boat will move with a final velocity of approximately 1.83 m/s after the person jumps off.