Answer:
[tex]v_{i}= 19\times 10^5\ m/s[/tex]
Explanation:
given,
scattering angle of alpha particle = 25.0° above its initial direction of motion
oxygen nucleus recoils at = 50.0° below this initial direction.
final speed of the oxygen = 2.08×10⁵ m/s
mass of alpha particle = 4.0 u
mass o oxygen nucleus = 16 u
momentum conservation along x- axis
[tex]m_{a}v_{i} = m_a v_a cos\theta + m_o v_o cos\theta[/tex]
[tex]4v_{i} = 4\times v_a cos25^0 + 16\times 2.08 \times 10^5 cos50^0[/tex]
[tex]v_{i}= \dfrac{3.625\times v_a+ 21.39\times 10^5}{4}[/tex]....(1)
Along y-direction
[tex]0 = m_av_a sin \theta - m_ov_o sin\theta[/tex]
[tex]0 = 4\times v_a sin 25 - 16\times 2.08 \times 10^5 sin50^0[/tex]
[tex]v_a = \dfrac{25.49 \times 10^5}{1.69}[/tex]
[tex]v_a = 15.082\times 10^5\ m/s[/tex]
putting value in equation (1)
[tex]v_{i}= \dfrac{3.625\times 15.082\times 10^5+ 21.39\times 10^5}{4}[/tex]
[tex]v_{i}= 19\times 10^5\ m/s[/tex]
