Two billiard balls with masses of m1 = 300g and m2 = 400g are rolling towards each other at speeds

| September 6, 2021 | education

Two billiard balls with masses of m1 = 300g and m2 = 400g are rolling towards each other at speeds v1 = v2 = 13 meters per second. find the velocities of the balls after an absolutely elastic impact.

Data: mbsh1 – mass of the first billiard ball (mbsh1 = 300 g = 0.3 kg); mbsh2 is the mass of the second billiard ball (mbsh2 = 400 g = 0.4 kg); oncoming traffic; Vbsh1 = Vbsh2 = V – speed of billiard balls before impact (V = 13 m / s); absolutely elastic impact.

1) The law of conservation of momentum: mbsh1 * V – mbsh2 * V = -mbsh1 * U1 + mbsh2 * U2.

0.3 * 13 – 0.4 * 13 = -0.3 * U1 + 0.4 * U2.

3.9 – 5.2 = -0.3 * U1 + 0.4 * U2.

-1.3 = -0.3 * U1 + 0.4 * U2.

U1 = (0.4U2 + 1.3) / 0.3.

2) The law of conservation of energy: mbsh1 * V2 / 2 + mbsh2 * V2 / 2 = mbsh1 * U1 ^ 2/2 + mbsh2 * U2 ^ 2/2.

0.3 * 13 ^ 2 + 0.4 * 132 = 0.3 * ((0.4U2 + 1.3) / 0.3) 2 + 0.4 * U2 ^ 2.

118.3 = ((0.4U2 + 1.3) ^ 2 / 0.3) + 0.4 * U2 ^ 2 | * 0.3.

35.49 = 0.16U2 ^ 2 + 1.04U2 + 1.69 + 0.12 * U2 ^ 2.

0.28U ^ 22 + 1.04U2 – 33.8 = 0 | : 0.28.

D = 38.9376 and U21 ≈ 9.29 m / s (right, the speed of the second billiard ball); U2 ^ 2 = -13 m / s (wrong).

3) The speed of the first billiard ball: U1 = (0.4U2 + 1.3) / 0.3 = (0.4 * 9.29 + 1.3) / 0.3 = 16.72 m / s.

Answer: The speed of the first billiard ball is 16.72 m / s; the second – 9.29 m / s.



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