A proton of mass mp collides with a heavy particle at rest
Ramji80906461:
Hi
Answers
Answered by
2
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.
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) 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
Answer.
Thus the mass of the first particle in 2.2 times greater than the mass of the first particle
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) 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
Answer.
Thus the mass of the first particle in 2.2 times greater than the mass of the first particle
Answered by
79
Answer:
Explanation:
Please look into the picture for the answer.
Hope it helps.
Attachments:
Similar questions
Sociology,
6 months ago
Political Science,
6 months ago
Math,
6 months ago
Economy,
1 year ago
Math,
1 year ago