Question

З плота масою 200 кг, що пливе озером зі швидкістю 1,44 км/год, у напрямі, протилежному напряму руху плота, зі швидкістю 7,2 км/год стрибає хлопець. Маса хлопця 40 кг. Якої швидкості набирає пліт? Відповідь подати в м/с, округлити до сотих.

Answer 1

Ответ:

0,8м/с

Объяснение:

1,44 км/час = 0,4м/с

7,2км/ч = 2м/с

q¹ - импульс чела

q² - импульс плота

q3 - импульс общий

найти v²(плота)

По закону сохранения импульса q³ = q¹+q²

Также q³ = m(лодки) × v(плота) = 200 × 0,4 = 80 (кг×м/с)

q¹+q² = 80

чел прыгает против направления плота,значит его импульс отрицательный значит q¹ = -(m(чела)×v(чела))

-m(чела)×v(чела) +m(лодки)v²(плота) = 80

-40×2 + 200v²(плота) = 80

-80 + 200v²(плота) = 80

200v²(плота) = 160

v²(плота) = 160/200

v²(плота) = 0,8

Answer 2

To find the speed at which the raft is moving, we can use the principle of conservation of momentum. The total momentum before the boy jumps is equal to the total momentum after the boy jumps.

Let's denote the initial velocity of the raft as Vr and the final velocity of the raft as Vr'. The initial velocity of the boy is 0 since he is on the raft, and his final velocity is Vb.

According to the principle of conservation of momentum:

(mass of raft * initial velocity of raft) + (mass of boy * initial velocity of boy) = (mass of raft * final velocity of raft) + (mass of boy * final velocity of boy)

Substituting the given values:

(200 kg * 1.44 km/h) + (40 kg * 0 km/h) = (200 kg * Vr') + (40 kg * Vb)

Converting the units to m/s:

(200 kg * 1.44 km/h * 1000 m/km * 1 h/3600 s) + (40 kg * 0 km/h * 1000 m/km * 1 h/3600 s) = (200 kg * Vr') + (40 kg * Vb)

Simplifying:

(200 kg * 0.4 m/s) = (200 kg * Vr') + (40 kg * Vb)

80 kg m/s = 200 kg * Vr' + 40 kg * Vb

Now, we need to find the value of Vr' (final velocity of the raft). To do this, we can use the fact that the boy jumps in the opposite direction of the raft with a speed of 7.2 km/h.

Vr' = Vr - Vb

Substituting the given values:

Vr' = 1.44 km/h - 7.2 km/h

Converting the units to m/s:

Vr' = (1.44 km/h - 7.2 km/h) * 1000 m/km * 1 h/3600 s

Simplifying:

Vr' = (-5.76 km/h) * 1000 m/km * 1 h/3600 s

Now, we can substitute the value of Vr' into the equation for momentum conservation:

80 kg m/s = 200 kg * (-5.76 km/h * 1000 m/km * 1 h/3600 s) + 40 kg * Vb

Simplifying:

80 kg m/s = -32 m/s + 40 kg * Vb

Rearranging the equation to solve for Vb:

40 kg * Vb = 80 kg m/s + 32 m/s

Vb = (80 kg m/s + 32 m/s) / 40 kg

Calculating:

Vb = 112 m/s / 40 kg

Vb = 2.8 m/s

Therefore, the boy jumps with a speed of 2.8 m/s.