Question

A projectile proton with a speed of 500 m/s collides elastically with a target proton initially at rest. The two protons then move along perpendicular paths, with the projectile path at 60° from the original direction. (a) What is the speed of the target proton after the collision? 0 Incorrect: Your answer is incorrect. m/s (b) What is the speed of the projectile proton after the collision?

Answer 1

Answer:

(a) The speed of the target proton after the collision is:$V_{2f} =433(m/s)$, and (b) the speed of the projectile proton after the collision is: $v_{1f}=250(m/s)$.

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:$m_{1} v_{1i} =m_{1} v_{1f}Cos\beta _{1} +m_{2} v_{2f}Cos\beta _{2}$, and y axle:$0=m_{1} v_{1f}Sin\beta _{1}+m_{2} v_{2f}Sin\beta _{2}$. Now replacing the value given as: $v_{1i}=500(m/s)$, $\beta_{1}=+60^{o}$ for the projectile proton and according to the problem $\beta_{1}and\beta_{2}$ are perpendicular so $\beta_{2}=-30^{o}$, and assuming that $m_{1}=m_{2}$, we get for x axle:$500=v_{1f}Cos\beta _{1}+ v_{2f}Cos\beta _{2}$ and y axle: $0=v_{1f}Sin\beta _{1}+v_{2f}Sin\beta _{2}$, then solving for $v_{2f}$, we get:$v_{2f}=-v_{1f}\frac{Sin\beta_{1}}{Sin\beta_{2}}= \sqrt{3}v_{1f}$ and replacing at the first equation we get:$500=\frac{1}{2} v_{1f} +\frac{\sqrt{3}}{2} *\sqrt{3}*v_{1f}$, now solving for $v_{1f}$, we can find the speed of the projectile proton after the collision as:$v_{1f}=250(m/s)$ and $v_{2f}=\sqrt{3}*v_{1f}=433(m/s)$, that is the speed of the target proton after the collision.