# The projectile, flying at a speed of 500 m / s, exploded into two fragments

**The projectile, flying at a speed of 500 m / s, exploded into two fragments. The speed of the first fragment weighing 5 kg increased by 200 m / s in the direction of movement of the projectile. Determine the speed of the second shard if its mass is 4 kg.**

m1 = 5 kg.

m2 = 4 kg.

v = 500 m / s.

Δv1 = 200 m / s.

v2 -?

To solve this problem, we will use the momentum conservation law.

psn = p1 + p2, where psn is the impulse of the projectile before rupture, p1 is the impulse of the first fragment, p2 is the impulse of the second fragment.

The momentum of a body p is a vector physical quantity equal to the product of the body’s velocity V by its mass m: p = m * V.

The impulse of the projectile will be determined by the formula: pcn = (m1 + m2) * v.

The impulses of the fragments will be determined by the formulas: p1 = m1 * v1, p2 = m2 * v2.

Let’s write the law of conservation of momentum: (m1 + m2) * v = m1 * v1 + m2 * v2.

m2 * v2 = (m1 + m2) * v – m1 * v1;

v2 = ((m1 + m2) * v – m1 * v1) / m2.

Since the velocity of the first fragment has increased by Δv1 = 200 m / s, then its velocity after rupture will be v1 = v + Δv1.

v1 = 500 m / s + 200 m / s = 700 m / s.

v2 = ((5 kg + 4 kg) * 500 m / s – 5 kg * 700 m / s) / 4 kg = 250 m / s.

Answer: the speed of the second fragment will be v2 = 250 m / s.