Projectile particle 1 is an alpha particle and target particle 2 is an oxygen nu

Question

Projectile particle 1 is an alpha particle and target particle 2 is an oxygen nucleus. The alpha particle is scattered at angle

1 = 68.0° and the oxygen nucleus recoils with speed 1.10 × 105 m/s and at angle 2 = 45.0°. In atomic mass units, the mass of the alpha particle is 4.00 u and the mass of the oxygen nucleus is 16.0 u. What are the (a) final and (b) initial speeds of the alpha particle?

There is a figure with a particle 1 running along the x-axis. It collides with particle 2 at the (x,y) intercept of (0,0). Particle 1 goes in a negative y direction and positve x direction at a 68 degree angle from the x-axis, and particle 2 goes in a positve y direction and a positive x direction at a 45 degree angle from the x-axis.

Explanation / Answer

You can solve for v1 and u by projecting the vector equation onto two axes to get 2 scalar equations.
Let the x-axis be in the same direction as initial velocity u, y axis be 90° to x axis. The projections of vector m1.u will be (m1.u, 0). Keep in mind that the plane xy is oriented and that 1 and 2 must be counted algebraically; if 1 > 0 then 2 < 0 and vice versa.

Conservation of momentum:
vector (m1.u) = vector (m1.v1) + vector (m2.v2)

Projection on y-axis:
0 = m1.v1.sin1 + m2.v2.sin2
v1 = -v2(m2/m1)(sin2/sin1)
v1 = -1.10×10^5 x (16/4)×[sin45°/sin(-68°)]
v1 = 3.36 ×10^5 m/s

Projection on x-axis:
m1.u = m1.v1.cos1 + m2.v2.cos2
u = v1.cos1 + v2.(m2/m1).cos2
u = 3.36×10^5 cos(-68°) + 1.1×10^5 ×(16/4)×cos45°
u = 4.37 ×10^5 m/s

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