Question

Если мальчик массой 60 кг выпрыгнет со скоростью 3м/c на берег из лодки массой 300кг , какую скорость приобретет лодка?

Answer 1

По закону сохранения импульса: 0=m*v - M*V (знак минус будет потому,ч то направлены скорости в разные стороны)
0=60*3-300*V
V=0,6 м/с 

Answer 2

Calculation of the Boat's Velocity

To calculate the velocity acquired by the boat when the boy jumps out of it, we can use the principle of conservation of momentum. According to this principle, the total momentum before and after the jump should be the same.

The momentum of an object is given by the product of its mass and velocity. Let's denote the mass of the boy as m1 (60 kg) and the mass of the boat as m2 (300 kg). The velocity of the boy before the jump is v1 (3 m/s), and we want to find the velocity of the boat after the jump, denoted as v2.

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

m1 * v1 + m2 * 0 = m1 * 0 + m2 * v2

Since the boy jumps out of the boat, his final velocity is zero. Therefore, the equation simplifies to:

m1 * v1 = m2 * v2

Now we can substitute the given values:

60 kg * 3 m/s = 300 kg * v2

Simplifying the equation, we find:

v2 = (60 kg * 3 m/s) / 300 kg

Calculating this expression, we get:

v2 = 0.6 m/s

Therefore, the boat will acquire a velocity of 0.6 m/s when the boy jumps out of it.