An explosion breaks a rock into three parts in a horizontal plane. Two of them go off at right angles to each other. The first part of mass 1 kg moves with a speed of 12 m/s and the second part of mass 2 kg moves with 8 m/s speed. If the third part flies off with 4 m/s speed, then its mass is?

Let the three parts have masses m1, m2, and m3, and velocities v1, v2, and v3 respectively.
According to the law of conservation of linear momentum, the total momentum before the explosion is equal to the total momentum after the explosion. Since the rock is initially at rest, the total initial momentum is zero.
Therefore, m1v1 + m2v2 + m3v3 = 0
Given:
m1 = 1 kg
v1 = 12 m/s
m2 = 2 kg
v2 = 8 m/s
v3 = 4 m/s
Let's assume that v1 and v2 are perpendicular to each other. We can represent the momentum vectors in the x and y directions.
In the x-direction:
m1v1x + m2v2x + m3v3x = 0
1(12) + 2(0) + m3(v3x) = 0
12 + m3v3x = 0
v3x = -12/m3
In the y-direction:
m1v1y + m2v2y + m3v3y = 0
1(0) + 2(8) + m3(v3y) = 0
16 + m3v3y = 0
v3y = -16/m3
The magnitude of v3 is given as 4 m/s.
Therefore, √(v3x² + v3y²) = 4
√((-12/m3)² + (-16/m3)²) = 4
√(144/m3² + 256/m3²) = 4
√(400/m3²) = 4
20/m3 = 4
m3 = 20/4 = 5 kg
Therefore, the mass of the third part is 5 kg.