Question

Тележка, масса которой 120 кг, движется по рельсам без трения со скоростью 6 м/с. С тележки соскакивает человек массой 80 кг под углом 30 градусов к направлению её движения. Скорость тележки уменьшается при этом до 5 м/c . Какой была скорость человека во время прыжка относительно земли?

Answer 1

Из закона сохранения импульса
(M + m)V = MU + muCosα
можно выразить искомую скорость u
u = ((M + m)V - MU)/(mCosα) = ((120 + 80)·6 - 120·5)/(80·0.866) = 600/69.28 = 8.66 м/с

Answer 2

Calculation of the Person's Velocity Relative to the Ground

To calculate the velocity of the person relative to the ground during the jump, we need to consider the conservation of momentum. The initial momentum of the system (the person and the cart) is equal to the final momentum after the person jumps off.

The initial momentum of the system is given by the product of the mass of the cart and its velocity:

Initial momentum of the system = mass of the cart × velocity of the cart

The final momentum of the system is the sum of the momentum of the cart and the momentum of the person after the jump:

Final momentum of the system = (mass of the cart × final velocity of the cart) + (mass of the person × final velocity of the person)

Since there is no external force acting on the system, the total momentum is conserved. Therefore, the initial momentum of the system is equal to the final momentum of the system:

Initial momentum of the system = Final momentum of the system

Let's calculate the velocities using the given information:

- Mass of the cart (m1) = 120 kg - Velocity of the cart before the person jumps (v1) = 6 m/s - Mass of the person (m2) = 80 kg - Angle of the jump (θ) = 30 degrees - Final velocity of the cart after the person jumps (v2) = 5 m/s

Using the conservation of momentum equation, we can solve for the final velocity of the person (v2_person):

m1 × v1 = (m1 × v2) + (m2 × v2_person)

Substituting the given values:

120 kg × 6 m/s = (120 kg × 5 m/s) + (80 kg × v2_person)

Simplifying the equation:

720 kg·m/s = 600 kg·m/s + 80 kg × v2_person

120 kg·m/s = 80 kg × v2_person

v2_person = 120 kg·m/s / 80 kg

v2_person = 1.5 m/s

Therefore, the velocity of the person relative to the ground during the jump is 1.5 m/s.