Question

A proton, moving with a velocity of v_iî, collides elastically with another proton that is initially at rest. Assuming that after the collision the speed of the initially moving proton is 1.40 times the speed of the proton initially at rest, find the following.

(a) the speed of each proton after the collision in terms of v_i

initially moving proton? xv_i

initially at rest proton?

(b) the direction of the velocity vectors after the collision (assume that the initially moving proton scatters toward the positive y direction

Initially moving proton in degrees relative to +x direction?

Initially at rest proton in degrees relative to +x direction?

Answer 1

(a) If the proton collides eleastically, then kinetic energy is conerved. If the mass of the protons be m and the speed of the target proton after the collision be v. Then A proton, moving with a velocity of v;(i vector), collides elastically with another proton that is initially at rest v = 0 mv = 5 m (1.4v)? + mv v} = (1.96+1)v2 v2 = 2.96v2 v; = 1.72v Therefore the speed of the rest proton is, v=1.4v, = 1.4(1.73v) = 2.41v (b) The impulses between the bodies have a sent the bodies off at angles and to x-axis, from the conservation of momentum along the x-axis is, mv; = n Viscos(x)+m, V2, cos(x)---(1) Mass cancel because the masses are equal V; = 1, cos()+ V2cos(@) 1.72 cos 2.41 =1.1075 Q = cos(0.713) = 44.46° with +x axis The initially proton scatters toward the positive y direction is, 0 = 90-44.46° = 45.5° with +x asis