Question

Ernest Rutherford (the first New Zealander to be awarded the Nobel Prize in chemistry) demonstrated that nuclei were very small and dense by scattering helium-4 nuclei ("He) from gold-197 (9Au). See the fiqure below. The energy of the incoming helium nucleus was 8.20 x 10, and the masses of the helium and gold nuclei were 6.68 x 10 and 3.29 x 10 197). If a helium nucleus scatters to an angle of 120° during an elastic collision with a gold nucleus, calculate the helium nucleus's final speed kg, respectively (note that their mass ratio is 4 to m/s Calculate the final velocity (magnitude and direction) of the gold nucleus. (Assume the positive x direction is the direction in which the helium nucleus is initially traveling, and that it scatters 120° clockwise from the +x-axis.) magnitude m/s °(counterciockwise from the +x-axis) direction Au 120 Hea Before collision After collision

Answer 1

@ formula for kinetic energy, KE= mv2 V = a KE 268-2x10-13) 668x10-at = 1,566X18 mis Apply conservation of momentum, Miriami Vicosost Mara' coses O=miri smo itmanal singa mini-miricoser mara' cosea -mirisino, ma la Singa (mini-mistoso,)²+(-mi risine,) _m² (val) cos²02 the cra!) Sinda Cada una (CM) - aviv, Cosoit(v.))) mive = 1 m (vi) ² + 1 ma (va) 2 (172-671)2) = ma (x)?av, vicose, +(2002) (item cm)2 -21 ma) Crikoso), -01- )(6)=0 (H S 40 )(0)2(6 al Serio (1-560 x 10°Os 120°) -use x 10 ) = (102) ² - 3. 129x105 vi- 2.4x1014 = = [1.549x107 ms 320x10-25 ((1.566X 107) (1549%207)2) = va? v2 = (3.27 x 105m/s.

Sinea = -668x10-*(1,549X107) sim 1200 (329x1095)(3:27X105) 02 –56.40 02= 360° - 564 = (303.60 ccw from +x-axis