Two balls of the same size, weighing 0.4 and 1 kg., Fly horizontally towards each other at speeds of 5 and 6 m / s, respectively. After a frontal impact, they fly in opposite directions, and the second ball has a speed of 0.2 m / s. What is the speed of the first ball after hitting 2. A body weighing 50 kg falls from a height of 15 m. Determine the value of potential and kinetic energy at a distance of 6 m from the earth’s surface.
m1 = 0.4 kg.
m2 = 1 kg.
V1 = 5 m / s.
V2 = 6 m / s.
V2 “= 0.2 m / s.
m1 * V1 + m2 * V2 = m1 * V1 “+ m2 * V2” – the law of conservation of momentum in vector form.
m1 * V1 – m2 * V2 = – m1 * V1 “+ m2 * V2”.
V1 “= (- m1 * V1 + m2 * V2 + m2 * V2”) / m1.
V1 “= (- 0.4 kg * 5 m / s + 1 kg * 6 m / s + 1 kg * 0.2 m / s) / 0.4 kg = 10.5 m / s.
Answer: the first ball will move at a speed V1 “= 10.5 m / s.
m = 50 kg.
g = 10 m / s2.
h = 15 m.
h1 = 6 m.
En1 = m * g * h1.
Ep1 = 50 kg * 10 m / s2 * 6 m = 3000 J.
Ek1 = En – En1 = m * g * h – m * g * h1 = m * g * (h – h1).
Ek1 = 50 kg * 10 m / s2 * (15 m – 6 m) = 4500 J.
Answer: En1 = 3000 J, Ek1 = 4500 J.
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