A small ball weighing M = 52g hangs on a weightless inextensible thread l = 72 cm long.

A small ball weighing M = 52g hangs on a weightless inextensible thread l = 72 cm long. A bullet with mass m = 8.0 g, flying at a speed v, hits the ball and gets stuck in it. If the bullet velocity was directed along the diameter of the ball, then the ball will make a complete revolution in a circle in the vertical plane with the minimum value of the bullet velocity modulus v equal to … m / s

l = 72 cm = 0.72 m.

M = 52 g = 0.052 kg.

m = 8 g = 0.008 kg.

g = 9.8 m / s2.

V -?

Let us express the speed of the ball V1 immediately after the bullet hits it according to the law of conservation of momentum: m * V = (M + m) * V1.

V1 = m * V / (M + m).

According to the law of conservation of total mechanical energy: (M + m) * V1 ^ 2/2 = (M + m) * V2 ^ 2/2 + (M + m) * g * 2 * l, where V2 is the speed of the ball with the bullet at the top of the trajectory.

V1 ^ 2/2 = V2 ^ 2/2 + g * 2 * l.

V1 ^ 2 = V2 ^ 2 + g * 4 * l.

At the upper point of the trajectory, the centripetal acceleration a must be equal to the acceleration of gravity g: a = g.

g = V2 ^ 2 / l.

V ^ 22 = g * l.

V1 ^ 2 = g * l + g * 4 * l = 5 * g * l.

m ^ 2 * V ^ 2 / (M + m) ^ 2 = 5 * g * l.

V ^ 2 = 5 * g * l * (M + m) ^ 2 / m ^ 2.

V = (M + m) * √ (5 * g * l) / m.

V = (0.052 kg + 0.008 kg) * √ (5 * 9.8 m / s2 * 0.72 m) / 0.008 kg = 44.5 m / s.

Answer: before hitting the ball, the bullet velocity was V = 44.5 m / s.