Question

An alpha particle collides with an oxygen nucleus, initially at rest. The alpha particle is scattered at an angle of 66.0° above its initial direction of motion, and the oxygen nucleus recoils at an angle of 42.0° below this initial direction. The final speed of the oxygen nucleus is 1.84×10⁵m/s. (The mass of an alpha particle is 4.0 u, and the mass of an oxygen nucleus is 16 u.)

1. What is the final speed of the alpha particle?

2. What was the initial speed of the alpha particle?

Answer 1

We apply the conservation of linear momentum to the x and y axes respectively.

We are givenv,,-1.84x10° m/s. a 64.0° and θ 51.0°. Thus, we are left with two unknowns and two equations, which can be readily solved (a) We solve for the final alpha particle speed using the y-momentum equation: ,v, sine, (16.0) ( 1.84 ×10°)sin(42.0。) m-m sina, (4,00) sin( 66,0。) 二1540x105 m/s if (4.00)sin (66.0 (b) Plugging our result from part (a) into the x-momentum equation produces the initial alpha particle speed li 11m li (4.00) ( 5.40x10) cos 66.0)(16.0) 1.84x10) cos (42.0) 4.00 17.66×105 m/s

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