Two identical particles move in mutually perpendicular directions. The modulus of the first particle’s velocity

Two identical particles move in mutually perpendicular directions. The modulus of the first particle’s velocity is v1 = 3.6 m / s. As a result of the collision, the second particle stops, and the first continues to move at a speed, the modulus of which is v ‘= 6.0 m / s. Determine the modulus of velocity of the second particle before the collision.

Since the second particle collides with the first and completely stops, therefore, its kinetic energy is completely transferred to the first particle.

Let’s make an equation. Kinetic energy of the first and second particles before collision:

E1 = m * v1 ^ 2/2;

E2 = m * v2 ^ 2/2.

After collision:

E3 = m * v3 ^ 2/2.

m * v1 ^ 2/2 + m * v2 ^ 2/2 = m * v3 ^ 2/2.

v1 ^ 2 + v2 ^ 2 = v3 ^ 2.

Express v2:

v2 ^ 2 = v3 ^ 2 – v1 ^ 2.

v2 ^ 2 = 6 ^ 2 – 3.6 ^ 2 = (6 – 3.2) * (6 + 3.2) = 2.8 * 9.2 = 25.76.

v2 = 5.1 m / s.

Answer: the speed of the second particle was 5.1 m / s.