Question

In the figure below, projectile particle 1 is an alpha particle and target particle 2 is an oxygen nucleus. The alpha particle is scattered at angle θ1 = 62.0° and the oxygen nucleus recoils with speed 1.30 105 m/s at angle θ2 = 55.0°. In atomic mass units, the mass of an alpha particle is 4.00 u and the mass of an oxygen nucleus is 16.0 u.

(a) Find the final speed of the alpha particle.

(b) Find the initial speed of the alpha particle.

I am looking for numerical answers as well as explanation.

Answer 1

angle θ₁ =62⁰
final velocity of the particle 2 , v_2f = 1.30 10⁵ m/s,
angle θ₂ = 55°
mass of the particle 1 , m₁ = 4.00 u,
mass of the particle 2 , m₂ = 16.0 u.
from law of conservation of momentum ,
for y direction :
                 0 = -m₁v_1f sinθ₁ +m₂v_2fsinθ₂
              v_1f = (m₂/m₁)v_2f(sinθ₂/sinθ₁)

                     = (16.0/4.00)*(1.30 10⁵ m/s)(sin55⁰/sin62⁰)
                     = 4.82 10⁵m/s,
for x direction :
         m₁v_1i = m₁v_1fcosθ₁ + m₂v_2fcosθ₂
          v_1i = v_1f cosθ₁ +(m₂/m₁)v_2f cosθ₂

                = (4.8210⁵)(cos62⁰) + (16.0/4.00)(1.30 10⁵)(cos55⁰)
                = 5.24 10⁵ m/s,