In a bowl, suspended on the thread, the mass of which is 5kg, falls the bullet weighing 2g

In a bowl, suspended on the thread, the mass of which is 5kg, falls the bullet weighing 2g, flying from the horizontal speed. 1000m / s.proyya through the ball, it continues to move in the same direction at a speed of 500m / c. What maximum height will raise the ball?

Denote by v1 and v2 – the velocity of the bullet before entering the ball and after that, through M, M bullet mass and ball, respectively. According to the law of energy conservation, equality is true:

E1 + W1 = E2 + W2, where E1 and E2 are the kinetic energies of the bullet before after the interaction, W1 and W2 are the potential energies of the ball. We accept W1 for 0. Then we obtain the equation:

M * (v1) ^ 2/2 = M * (V2) ^ 2/2 + M * G * H, H is height.

h = m / 2 * (V1 ^ 2 – V2 ^ 2) / M * g.

H = 0.002 / 2 * (1000 ^ 2 – 500 ^ 2) / 5 * 10 = 7.5 m.

Answer: The desired height is 7.5 m.