Two non-elastic balls weighing 6 kg and 4 kg are moving at speeds of 8 m / s and 3 m / s, respectively

| September 2, 2021 | education

Two non-elastic balls weighing 6 kg and 4 kg are moving at speeds of 8 m / s and 3 m / s, respectively, directed along one straight line. With what speed they will move after a non-elastic collision, if a) the first catches up with the second b) if they move towards each other.

V1 = 8 m / s.

V2 = 3 m / s.

m1 = 6 kg.

m2 = 4 kg.

V “-?

Let us use the law of conservation of momentum in vector form in the collision of balls: p1 + p2 = p1 “+ p2”, where p1, p2 are the impulses before the impact, p1 “, p2” are the impulses after the impact.

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

The law of conservation of momentum for an inelastic impact will have the form: m1 * V1 + m2 * V2 = m1 * V “+ m2 * V”.

m1 * V1 + m2 * V2 = (m1 * + m2) * V “.

a) The balls are moving towards each other.

m1 * V1 – m2 * V2 = (m1 + m2) * V “.

V “= (m1 * V1 – m2 * V2) / (m1 + m2).

V “= (6 kg * 8 m / s – 4 kg * 3 m / s) / (6 kg + 4 kg) = 3.6 m / s.

The velocity after impact will be directed towards the ball of mass m1.

b) The balls move in one direction.

m1 * V1 + m2 * V2 = (m1 + m2) * V “.

V “= (m1 * V1 + m2 * V2) / (m1 + m2).

V “= (6 kg * 8 m / s + 4 kg * 3 m / s) / (6 kg + 4 kg) = 6 m / s.

Answer: V “= 3.6 m / s, V” = 6 m / s.



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