Question

Rutherford fired a beam of alpha particles (helium nuclei) at a thin sheet of gold. An alpha particle was observed to be deflected by 90.0°; its speed was unchanged. The alpha particles used in the experiment had an initial speed of 1.6 ✕ 107 m/s and a mass of 6.7 ✕ 10−27 kg. Assume the alpha particle collided with a gold nucleus that was initially at rest. Find the speed of the nucleus after the collision.

Answer 1

first, we consider the conservation of angular momentum in the x-direction.
m*v _i = M*v_fx

Where m is the mass of the alpha particle and v_i is the initial velocity of the alpha particle. M is the mass of the gold nucleus and v_fx is the velocity of the gold nucleus in the x-direction.
6.7 *10^-27 *1.6*10⁷ = 3.27 *10^-25 * v_fx

v_fx = 3.27*10⁵ m/s

The conservation of angular momentum in the y-direction.
0 = m*v_y + M*v_fy

0 = 6.7*10^-27 *1.6*10⁷ +  3.27 *10^-25 * v_fy

v_fy = -3.27*10⁵  m/s

This is directed along the negative y axis.
The net velocity of the gold nucleus
v = sqrt (v_fx² + v_fy² )
v = sqrt(2) * 3.27*10⁵ m/s

v = 4.62*10⁵ m/s