Two identical plasticine balls fly in mutually perpendicular courses, the speed of the first is twice the second

Two identical plasticine balls fly in mutually perpendicular courses, the speed of the first is twice the second, and after collision, their speed is 1.5 m / s. Find a lower speed.

Given:

m1 = m2 = m is the mass of plasticine balls;

v = 1.5 m / s – the speed of the balls after collision;

v1 = 2 * v2 – the speed of the first ball before collision is 2 times the speed of the second ball.

It is required to determine v2 (m / s).

According to the law of conservation of momentum (momentum):

m1 * v1 + m2 * v2 = (m1 + m2) * v;

m * v1 + m * v2 = 2 * m * v;

v1 + v2 = 2 * v;

2 * v2 + v2 = 2 * v;

3 * v2 = 2 * v;

v2 = 2 * v / 3 = 2 * 1.5 / 3 = 3/3 = 1 m / s.

Answer: the speed of the second ball was 1 m / s.