# Answer to Question #4193 in Mechanics | Relativity for nantu

Question #4193

A particle makes a head-on-collision with an unknown particle at rest.The proton rebounds back with 4/9 of its initial kinetic energy.Find the ratio of mass of unknown particle to mass of proton assuming the collision to be elastic.

Expert's answer

Let m1 and m2 be the masses of the first (moving) and the second (being in rest) particle,Let also u be the velocity of the first particle, and v1 and v2 be the velocities of particles

after collision.

Since the collision is assumed to be elastic we have the momentum of the system is not changed, so

(A) m

m1 * u

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 * u

v

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

Answer.

Thus the mass of the first particle in 2.2 times greater than the mass of the first particle.

after collision.

Since the collision is assumed to be elastic we have the momentum of the system is not changed, so

(A) m

_{1}*u = m_{1}*v_{1}+ m_{2}*v_{2}Moreover, we also have that the total kinetic energy of both particles is also unchanged, som1 * u

^{2}/ 2 = m1 * v_{1}^{2}/ 2 + m2 * v_{2}^{2}/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 * u

^{2}/ 2 = m1 * v_{1}^{2}/ 2, (5/9) * m1 * u^{2}/ 2 = m2 * v_{2}^{2}/ 2,whencev

_{1}= 2/3 u, v_{2}= 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

Answer.

Thus the mass of the first particle in 2.2 times greater than the mass of the first particle.

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