Answer to Question #4193 in Mechanics | Relativity for nantu
Since the collision is assumed to be elastic we have the momentum of the system is not changed, so
(A) m1*u = m1*v1 + m2*v2Moreover, we also have that the total kinetic energy of both particles is also unchanged, so
m1 * u2 / 2 = m1 * v12 / 2 + m2 * v22 /2By assumption the kinetic energy of the first particle after collision is equal to 4/9 of its initial kinetic energy,
whence the kinetic energy of the second particle after collision is equal to 1-4/9=5/9 of the initial kinetic energy
of the first particle, so (4/9) * m1 * u2 / 2 = m1 * v12 / 2, (5/9) * m1 * u2 / 2 = m2 * v22 / 2,whence
v1 = 2/3 u, v2 = sqrt(5)/3 u,Substituing these values to the formula (A) we get:
m1*u = m1*u*2/3 + m2*u*sqrt(5)/3
Contracting by u we get
m1 = m1*2/3 + m2*sqrt(5)/3
m1/3 = m2*sqrt(5)/3
m1 / m2 = sqrt(5) = 2.236
Thus the mass of the first particle in 2.2 times greater than the mass of the first particle.
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