Answer:
v2'=14435.02 m/s
Explanation:
Linear momentum and energy must conserve before and after the collision:
[tex]p_b=p_a\\E_b=E_a\\[/tex]
by taking into account both x and y component of the linear momentum we have:
[tex]m_pv_1=m_pv'_1cos(30\°)+m_pv'_2cos(\alpha)\\\\0=m_pv'_1sin(30\°)+m_pv'_2sin(\alpha)[/tex]
Where v1, v2, v1' and v2' are velocities of the first and second proton before and after the collision respectively. mp is the mass of the proton.
For the kinetic energy we have:
[tex]\frac{1}{2}m_pv_1^2=\frac{1}{2}m_pv'_1^2+\frac{1}{2}m_pv'_2^2[/tex]
by replacing in the first and second equation equation we obtain:
[tex]v'_2cos(\alpha)=v_1-v_1'cos(30\°)=(50000\frac{m}{s})-(43300\frac{m}{s})cos(30\°)=12501.1\frac{m}{s}\\\\v_2'sin(\alpha)=v_1'sin(30\°)=(43300\frac{m}{s})sin(30\°)=21650\frac{m}{s}[/tex]
by dividing these equations we get:
[tex]\frac{v_2'sin(\alpha)}{v_2'cos(\alpha)}=\frac{21650}{12501.1}\\\\tan(\alpha)=1.731\\\\\alpha=tan^{-1}(1.731)59.69\°\approx60\°[/tex]
Finally, by replacing in some of the equations for v2':
[tex]v_2'cos(30\°)=12501.1\\\\v_2'=\frac{12501.1}{cos(30\°)}=14435.02\frac{m}{s}[/tex]
hope this helps!!
