Answer:
(a) The speed of the target proton after the collision is:[tex]V_{2f} =433(m/s)[/tex], and (b) the speed of the projectile proton after the collision is: [tex]v_{1f}=250(m/s)[/tex].
Explanation:
We need to apply at the system the conservation of the linear momentum on both directions x and y, and we get for the x axle:[tex]m_{1} v_{1i} =m_{1} v_{1f}Cos\beta _{1} +m_{2} v_{2f}Cos\beta _{2}[/tex], and y axle:[tex]0=m_{1} v_{1f}Sin\beta _{1}+m_{2} v_{2f}Sin\beta _{2}[/tex]. Now replacing the value given as: [tex]v_{1i}=500(m/s)[/tex], [tex]\beta_{1}=+60^{o}[/tex] for the projectile proton and according to the problem [tex]\beta_{1}and\beta_{2}[/tex] are perpendicular so [tex]\beta_{2}=-30^{o}[/tex], and assuming that [tex]m_{1}=m_{2}[/tex], we get for x axle:[tex]500=v_{1f}Cos\beta _{1}+ v_{2f}Cos\beta _{2}[/tex] and y axle: [tex]0=v_{1f}Sin\beta _{1}+v_{2f}Sin\beta _{2}[/tex], then solving for [tex]v_{2f}[/tex], we get:[tex]v_{2f}=-v_{1f}\frac{Sin\beta_{1}}{Sin\beta_{2}}= \sqrt{3}v_{1f}[/tex] and replacing at the first equation we get:[tex]500=\frac{1}{2} v_{1f} +\frac{\sqrt{3}}{2} *\sqrt{3}*v_{1f}[/tex], now solving for [tex]v_{1f}[/tex], we can find the speed of the projectile proton after the collision as:[tex]v_{1f}=250(m/s)[/tex] and [tex]v_{2f}=\sqrt{3}*v_{1f}=433(m/s)[/tex], that is the speed of the target proton after the collision.
