Physics momentum

Question

A neutron of mass u traveling with a speed 5.00x10⁵ m/s collides with a stationary alpha particle of mass 4u.  The neutron scatters off the alpha particle at an angle 65˚ with speed 4.33x10⁵ m/s

 

Determine

a) The components of the neutron’s velocity after the collision

 

b)  The direction and speed of the alpha particle after the collision

 

I have a few problems. How is it possible to know the components of the neutron after the collision without actually calculating part b) I am assuming since the masses are different they ARE NOT perpendicular after the collision.

 

or is part a) simply  V_x = 4.33x10⁵ m/s x cos(65)   and V_{y =}  4.33x10⁵ m/s x sin (65)

 

My attempt at part b) I got 1.26 x 10⁵ m/s at a direction of 51 degrees. 

 

 

Answer 1

(a)

 

You are correct.

 

v_x = (4.33 x 10⁵ m/s) cos(65°) ≅ 1.83 x 10⁵ m/s 

v_y = (4.33 x 10⁵ m/s) sin(65°) ≅ 3.92 x 10⁵ m/s

 

(b)

 

Note the speeds are low enough to ignore relativistic effects.  Apply conservation of linear momentum.

 

v₁ = initial neutron speed (assumed to be along x)

v_x = final neutron speed along x

v_y = final neutron speed along y

w_x = final alpha particle speed along x

w_y = final alpha particle speed along y

 

Conservation along x-direction:

 

uv₁ = uv_x + (4u)w_x ⇒ w_x = (v₁ - v_x)/4 = (5 - 1.83)/4 x 10⁵ m/s ≅ 0.793 x 10⁵ m/s

 

Conservation along y-direction:

0 = uv_y + (4u)w_y ⇒ w_y = -v_y/4 = -(3.92 x 10⁵ m/s) / 4 ≅ -9.80 x 10⁴ m/s

 

w = √(w_x² + w_y²) =√[(0.793 x 10⁵)² + (-9.80 x 10⁴)²] = 1.26 x 10⁵ m/s

 

θ = tan^-1(w_y/w_x) = tan^-1(-9.8 x 10⁴ / 0.793 x 10⁵) = (360 - 51.0)° = 309°

 

I got the same speed as you for the alpha particle, but a different direction.  Note the alpha particle moves in the +x and -y directions, which puts its velocity vector direction between 270° and 360°, so I think you put the angle in the wrong quadrant.  Check my math and let me know what you think.