A proton, moving with a velocity of v_i i cap, collides elastically with another

Question

A proton, moving with a velocity of v_i i cap, collides elastically with another proton that is initially at rest. Assuming that after the collision the speed of the initially moving proton is 0.80 times the speed of the proton initially at rest, find the following. the speed of each proton after the collision in terms of v-i initially moving proton Your response differs from the correct answer by more than 10%. Double check your calculations. x v_i initially at rest proton Your response differs from the correct answer by more than 10%. Double check your calculations. x v_i the direction of the velocity vectors after the collision (assume that the initially moving proton scatters toward the positive y direction) initially moving proton degree relative to the +x direction initially at rest proton degree relative to the +x direction

Explanation / Answer

a)

If the protons collide elastically, then kinetic energy is conserved. Let the mass of the protons be m and the speed of the target proton after the collision be v. In that case,

KEi = KEf

(1/2)m (vi)^2 = (1/2)m(0.80vf)^2 + (1/2)m vf^2

The halves cancel, as do the masses, so

vi^2 = 0.64vf^2 + vf^2

vi^2 = 1.64vf^2

vf^2 = vi^2 / 1.64

vf = sqrt(vi^2 / 1.64)

vf = vi / (sqrt(1.64))

b)

  Let vf 1 be the final velocity for the incident proton, and vf2 be the final velocity for the proton initially at rest. Conserving momentum in the j direction

Piy = 0 = Pf y = mpvf1y +mpvf2y

vf 1y = vf 2y

So the protons have equal magnitude speeds in the j direction. Because the speed of the particles are equal, the magnitude of their speeds in the i direction should also be equal |v f 1x| = |vf 2x|. Conserving momentum in the i direction.

Pix = mpvi = Pfx = mpvf 1x +mpvf 2x = 2mpvfx

vfx = vi / 2

Using the Pythagorean theorem to solve for the magnitude of vf y

vf^2 = vi^2/ 2 =( vfx)^2 +(vfy)^2 = vi^2/4 + (vfy)^2

vfy = visqrt(1/2 - 1/4) = vf/2 = vfx

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