Question

С лодки массой 200кг,движущейся со скоростью 2м/с,прыгает мальчик массой 50кг в горизонтальном направлении со скоростью 7м/с.Какова скорость лодки после прыжка мальчика , если он прыгает с кормы в сторону противоположную движению лодки. СРОЧНО С ЧЕРТЕЖОМ

Answer 1

M1=200кг, v1=2м/с, m2=50кг, v2= 7м/с. v-?Пусть  лодка с мальчиком движется слева направо. Туда же направим ось ОХ. Тогда в проекции на ось ОХ имеем: (m1+m2)v1= m1v-m2v2. m1v=(m1+m2)v1+m2v2, v=((m1+m2)v1+m2v2)/m1, v=(250*2+50*7)/200 = 850/200 = 4,25 м/с.

Answer 2

Calculation of the Boat's Velocity after the Boy Jumps

To calculate the velocity of the boat after the boy jumps, we can use 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:

Momentum of the boat before the jump = mass of the boat × velocity of the boat

The momentum of the boy before the jump is:

Momentum of the boy before the jump = mass of the boy × velocity of the boy

Since the boy jumps in the opposite direction to the boat's motion, the velocity of the boy is negative. Therefore, the momentum of the boy before the jump is:

Momentum of the boy before the jump = mass of the boy × (−velocity of the boy)

The total momentum before the jump is the sum of the momenta of the boat and the boy:

Total momentum before the jump = Momentum of the boat before the jump + Momentum of the boy before the jump

After the jump, the boat and the boy move together with a common velocity. Let's denote this common velocity as V. The momentum of the boat and the boy after the jump is:

Momentum of the boat and the boy after the jump = (mass of the boat + mass of the boy) × V

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

Total momentum before the jump = Total momentum after the jump

Therefore, we can equate the expressions for the total momentum before and after the jump and solve for V:

Momentum of the boat before the jump + Momentum of the boy before the jump = Momentum of the boat and the boy after the jump

Let's substitute the given values into the equation and solve for V.

Given: - Mass of the boat (m1) = 200 kg - Velocity of the boat (v1) = 2 m/s - Mass of the boy (m2) = 50 kg - Velocity of the boy (v2) = −7 m/s (negative because the boy jumps in the opposite direction)

Substituting the values into the equation:

(m1 × v1) + (m2 × (−v2)) = (m1 + m2) × V

(200 kg × 2 m/s) + (50 kg × (−(−7 m/s))) = (200 kg + 50 kg) × V

400 kg·m/s + 350 kg·m/s = 250 kg × V

750 kg·m/s = 250 kg × V

Now, we can solve for V:

V = (750 kg·m/s) / (250 kg) = 3 m/s

Therefore, the velocity of the boat after the boy jumps is 3 m/s in the direction opposite to its initial motion.

I hope this helps! Let me know if you have any further questions.