An alpha-particle of mass m suffers 1-dimensional elastic collision with a nucleus at rest of unknown mass. It is scattered directly backwards losing 64% of its initial kinetic energy. The mass of the nucleus is

Let the initial velocity of alpha-particle be $$v_1$$ and target nucleus mass be $$M$$.

From conservation of linear momentum: $$mv_1 = mv_1' + Mv_2' \quad \text{--- (1)}$$

From coefficient of restitution for elastic collision ($$e=1$$): $$e = \frac{v_2' - v_1'}{v_1} = 1 \implies v_2' = v_1 + v_1' \quad \text{--- (2)}$$

Substituting (2) into (1): $$mv_1 = mv_1' + M(v_1 + v_1') \implies (m - M)v_1 = (m + M)v_1'$$

$$v_1' = \left(\frac{m - M}{m + M}\right)v_1$$

Final kinetic energy of alpha-particle: $$K_f = \frac{1}{2}m(v_1')^2 = \frac{1}{2}m\left(\frac{m - M}{m + M}\right)^2 v_1^2 = \left(\frac{m - M}{m + M}\right)^2 K_i$$

Fractional kinetic energy remaining: $$\frac{K_f}{K_i} = \left(\frac{m - M}{m + M}\right)^2 = 1 - 0.64 = 0.36$$

$$\frac{M - m}{m + M} = \sqrt{0.36} = 0.6 \implies M - m = 0.6m + 0.6M$$

$$0.4M = 1.6m \implies M = 4m$$