Question

Шар, массой 200г, движущийся со скоростью 8м/с , сталкивается с неподвижным шаром массой 400г, и отлетает назад со скоростью 4 м/с. Найдите скорость второго шара после упругого удара

Answer 1

 По закону сохранения импульса: m1*v1+m2*v2=m1*v3+m2*v4, (m - масса; v1,2-скорость до взаимодействия; v3,4-скорость после взаимодействия). Получается 200*8+400*0=200*4+400*v;
1600=800+400*v
800=400*v;
v=2м/с.

Answer 2

Calculation of the Second Ball's Velocity After the Elastic Collision

To find the velocity of the second ball after the elastic collision, we can use the principle of conservation of momentum. According to this principle, the total momentum before the collision is equal to the total momentum after the collision.

Let's denote the mass of the first ball as m1 (200g) and its initial velocity as V1 (8m/s). The mass of the second ball is m2 (400g), and we need to find its final velocity, denoted as V2.

Using the conservation of momentum equation, we have:

m1 * V1 = (m1 + m2) * V2 Substituting the given values:

200g * 8m/s = (200g + 400g) * V2

Simplifying the equation:

1600g*m/s = 600g * V2

Dividing both sides of the equation by 600g:

V2 = (1600g*m/s) / 600g

Simplifying further:

V2 = 2.67 m/s

Therefore, the velocity of the second ball after the elastic collision is 2.67 m/s.

Please note that the calculations assume an elastic collision, where kinetic energy is conserved.