Two plasticine balls, moving at speeds of 3 m / s and 5 m / s towards each other, experience an absolutely

| February 5, 2021 | education

Two plasticine balls, moving at speeds of 3 m / s and 5 m / s towards each other, experience an absolutely inelastic impact. What is the velocity of the balls after impact, if the kinetic energy of the first ball before impact was 1.6 times greater than the kinetic energy of the second?

Impulse conservation law:
m1 * V1 + m2 * V2 = m1U1 + m2 * U2, m1, m2 are the masses of the balls; V1, V2 are speeds before impact, U1, U2 are speeds after impact.
According to the beginning. conditions:
U1 = U2, Ek1 / Ek2 = 1.6.
m1 * V1-m2 * V2 = (m1 + m2) U.
Ek1 / Ek2 = 1.6.
Ek1 = (m1 * V1 ^ 2) / 2, V1 = 3 m / s.
Ek2 = (m2 * V2 ^ 2) / 2, V2 = 5 m / s.
((m1 * V1 ^ 2) / 2) / ((m2 * V2 ^ 2) / 2) = 1.6.
(m1 * 9) / (m2 / 25) = 1.6
m1 = 1.6m2 * 25/9 = m2 * 40/9.
Ball speed:
U = (m1 * V1-m2 * V2) / (m1 + m2) = ((m2 * 40/9) * 3-m2 * 5) / (m2 * 40/9 + m2) = ((m2 * 25 / 3) / (m2 * 49/9) = 25 * 9 / (49 * 3) = 1.53 m / s.
Answer: The speed of the balls is 1.53 m / s.



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