When exploded, the stone breaks into three pieces. The first fragment weighing 1 kg flies horizontally at a speed of 12 m / s

When exploded, the stone breaks into three pieces. The first fragment weighing 1 kg flies horizontally at a speed of 12 m / s, and the second fragment weighing 2 kg at a speed of 8 m / s perpendicular to the direction of movement of the first piece. The third fragment flies off at a speed of 40 m / s. What is the mass of the third fragment and in what direction is it flying with respect to the horizon?

m1 = 1 kg.
V1 = 12 m / s.
m2 = 2 kg.
V2 = 8 m / s.
V3 = 40 m / s.
m3 -?
∠α -?
Before the rupture, the total impulse of the stone was equal to 0. According to the law of conservation of momentum, and after the explosion, the total impulse of the parts of the stone should be equal to 0.
m1 * V1 + m2 * V2 + m3 * V3 = 0 – vector.
Since the impulses of the first m1 * V1 and the second body m2 * V2 are perpendicular, the impulse of the third body will be the hypotenuse of a right-angled triangle with legs m1 * V1 and m2 * V2.
By the Pythagorean theorem: m3 * V3 = √ ((m1 * V1) 2 + (m2 * V2) 2).
m3 = √ ((m1 * V1) 2 + (m2 * V2) 2) / V3.
m3 = √ ((1 kg * 12 m / s) 2 + (2 kg * 8 m / s) 2) / 40 m / s = 0.5 kg.
ОХ: m1 * V1 – m3 * V3 * cosα = 0.
OU: m2 * V2 – m3 * V3 * sinα = 0.
cosα = m1 * V1 / m3 * V3.
∠α = arccos (m1 * V1 / m3 * V3).
∠α = arccos (1 kg * 12 m / s / 0.5 kg * 40 m / s) = arccos0.6 = 53 °.
Answer: m3 = 0.5 kg, ∠α = 53 °.