Question

A proton, moving with a velocity of vii, collides elastically with another proton that is initially at rest. Assuming that after the collision the speed of the initially moving proton is 1.60 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 _______________ x v_i

initially at rest proton  _______________ x v_i

(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 _______________ degrees relative to +x-axis

initially at rest proton  _______________ degrees relative to +x-axis

Answer 1

For the elastic collision, we can write:

1/2m_pv_i²=1/2m_pv₂² + 1/2m_pvf²  and we know that v_f = 1.60v₂

1/2m_pv_i²=1/2m_pv₂² + 1/2m_p(1.60v₂)²  

v_i² = 3.56 v₂²

v₂= v_i / (3.56)^1/2

v₂ = 0.529 v_i

v_f= 0.846 vi

And the speeds are:

initially moving proton 0.846 x v_i

initially at rest proton 0.529 x v_i

b) Also, we know that the momentum is conserved too. Since momentum is a vector, it must be conserved in all directions.

Consider, then, the direction perpendicular to the initial direction of travel of the projectile proton. That must be 0. So whatever momentum the projectile proton has after the collision perpendicular to its initial direction of travel must be canceled by the momentum of the target proton perpendicular to the initial direction of travel after the collision.

If the initially moving proton scatters towards the +y direction, its velocity is 90 degrees relative to the +x axis, and the velocity of the initially at rest proton is 270 degress (-90 deg) relative to the + x axis.

initially moving proton 90 degrees relative to +x-axis

initially at rest proton  270 degrees relative to +x-axis