Question

Пуля, летящая горизонтально, попадает в шар, подвешенный на легком жестком стержне, и застревает в нем. Масса пули 5г, масса шара 0,5кг. Скорость пули 500 м/с. При какой предельной длине стержня (расстояние точки подвеса до центра шара) шар от удара пули поднимается до верхней точки окружности

Answer 1

Пусть L - длина стержня, M=0,5 кг - масса шара и m=0,005 кг- масса пули, v =500 м/с - скорость движения пули. Шар с пулей поднимутся до верхней точки окружности, то есть на высоту h=2*L, при условии, что потенциальная энергия шара с пулей в верхней точке E1=(M+m)*g*2*L будет равна их кинетической энергии E2=(M+m)*v1²/2 в нижней точке. Скорость v1 определим из закона сохранения импульса: m*v=(M+m)*v1, откуда v1=m*v/(M+m)=0,005*500/(0,505)≈4,95 м/с. Из равенства 2*(M+m)*g*L=(M+m)*v1²/2, или 2*g*L=v1²/2 находим L=v1²/(4*g). Полагая g≈10 м/с², окончательно получаем L≈(4,95)²/40≈0,61 м. Ответ: L≈0,61 м. 

Answer 2

Calculation of the Maximum Length of the Rod

To determine the maximum length of the rod (the distance from the suspension point to the center of the sphere) at which the sphere will rise to the highest point of the circle after being hit by the bullet, we can use the principle of conservation of momentum.

The momentum of the bullet before the collision is equal to the momentum of the bullet and the sphere after the collision. Since the bullet gets embedded in the sphere, the combined mass of the bullet and the sphere after the collision is equal to the mass of the sphere alone.

Let's denote the maximum length of the rod as L.

The momentum of the bullet before the collision is given by:

momentum_before = mass_bullet * velocity_bullet The momentum of the bullet and the sphere after the collision is given by:

momentum_after = mass_sphere * velocity_after_collision Since the bullet gets embedded in the sphere, the velocity of the bullet after the collision is zero. Therefore, the momentum after the collision is equal to the momentum of the sphere alone:

momentum_after = mass_sphere * velocity_sphere Equating the two momenta, we have:

mass_bullet * velocity_bullet = mass_sphere * velocity_sphere

Substituting the given values:

5g * 500 m/s = 0.5kg * velocity_sphere

Simplifying the equation:

velocity_sphere = (5g * 500 m/s) / 0.5kg

Now, we can calculate the velocity of the sphere after the collision.

Next, we can use the conservation of energy to determine the maximum length of the rod. The initial kinetic energy of the bullet is equal to the final potential energy of the sphere at the highest point of the circle.

The initial kinetic energy of the bullet is given by:

kinetic_energy_bullet = (1/2) * mass_bullet * velocity_bullet^2 The final potential energy of the sphere at the highest point of the circle is given by:

potential_energy_sphere = mass_sphere * g * height Since the sphere rises to the highest point of the circle, the height is equal to the radius of the circle.

Equating the initial kinetic energy and the final potential energy, we have:

(1/2) * mass_bullet * velocity_bullet^2 = mass_sphere * g * height

Substituting the given values:

(1/2) * 5g * (500 m/s)^2 = 0.5kg * 9.8 m/s^2 * height

Simplifying the equation:

height = [(1/2) * 5g * (500 m/s)^2] / (0.5kg * 9.8 m/s^2)

Now, we can calculate the height of the sphere at the highest point of the circle.

Finally, we can use the height and the radius of the circle to determine the maximum length of the rod:

L = height + radius

Substituting the calculated values, we can find the maximum length of the rod.

Please note that the calculations provided above are based on the given information and assumptions. If there are any additional factors or considerations, they should be taken into account for a more accurate analysis.

Let me know if you need further assistance!