Answer:
the velocity of the gold nuclei 
the angle of the velocity of gold nuclei = 29.64° clockwise (i.e down the horizontal)
the final kinetic energy of helium nucleus is 
Explanation:
The formula for kinetic Energy (K.E) is:


where;
= mass of the helium
= velocity of helium
replacing
and
; we have :


Applying conservation of momentum along x- axis since the collision is elastic
------ equation (1)
where
= mass of helium and gold respectively
= initial velocity of helium
= final velocity of gold
= scattered angle for helium and gold respectively
Along the y - axis; the equation for conservation of momentum is :
---- equation (2)
Equating equation (1) and (2); we have:
----- equation (3)
However, the conservation of internal kinetic energy guves:

Making
the subject of the formula ; we have:
----- equation (4)
Replacing the expression of
in equation (3) into equation (4) ; we have
![[(1+\frac{m_1}{m_2})(v_1')^2-2(\fracm_1}{m_2}(v_1'cos \theta _1)v_1 - (1-\frac{m_1}{m_2})(v_1)^2] = 0](https://tex.z-dn.net/?f=%5B%281%2B%5Cfrac%7Bm_1%7D%7Bm_2%7D%29%28v_1%27%29%5E2-2%28%5Cfracm_1%7D%7Bm_2%7D%28v_1%27cos%20%5Ctheta%20_1%29v_1%20-%20%281-%5Cfrac%7Bm_1%7D%7Bm_2%7D%29%28v_1%29%5E2%5D%20%3D%200)
In the above expression;
replacing ;
;
;
;
; we have:


The above is a quadratic equation; now solving by using the quadratic formula; we have:

since we are considering the positive value from the above expression; we have

NOW; we substitute our known values into equation (4) in order to solve for
; we have:




Therefore; the velocity of the gold nuclei 
From equation (2)
Therefore replacing our known values and solving for
; we have:




∴ the angle of the velocity of gold nuclei = 29.64° clockwise (i.e down the horizontal)
b) Equation to determine the final kinetic energy
of helium is:


= 
Thus, the final kinetic energy of helium nucleus is 
I hope this explanation helps alot!.