The core, flying horizontally at a speed of 20 m / s, exploded into two fragments weighing 5 kg and 10 kg.

The core, flying horizontally at a speed of 20 m / s, exploded into two fragments weighing 5 kg and 10 kg. The speed of the smaller fragment is 90 m / s and is directed in the same way as the speed of the nucleus prior to rupture. Find the speed and direction of movement of the large shard.

m1 = 5 kg.

m2 = 10 kg.

Vsn = 20 m / s.

V1 “= 90 m / s.

V2 “=?

Let us write the law of conservation of momentum for a closed-loop system: (m1 + m2) * Vсн = m1 * V1 “+ m2 * V2”, where m1, m2 is the mass of the corresponding fragments, Vсн is the speed of the nucleus before rupture, V1 “, V2” is the speed of the fragments after the nucleus burst.

m2 * V2 “= (m1 + m2) * Vсн – m1 * V1”.

The speed of the second fragment after rupture will be expressed by the formula: V2n “= ((m1 + m2) * Vcn – m1 * V1”) / m2.

V2 “= ((5 kg + 10 kg) * 20 m / s – 5 kg * 90 m / s) / 10 kg = – 15 m / s.

The sign “-” means that the velocity of the second fragment is directed in the opposite direction of the movement of the nucleus.

Answer: the velocity of the second fragment after the rupture will be directed in the opposite direction of the movement of the nucleus and will be V2 “= 15 m / s.