Question

A proton collides elastically with another proton that is initially at rest. The incoming proton

has an initial speed of 3.5 x 105m/s and makes a glancing collision with the second. (At

close separations, the protons exert a repulsive electrostatic force on each other.) After the

collisions, one proton moves off at an angle of 37o to the original direction of motion and

the second deflects at an angle of φ to the same axis. Find the final speeds of the two

protons and the angle φ. [12]

Answer 1

$v_{1i}=v_{1f}cos \theta + v_{2f}cos \phi \\ 0 = v_{1f} sin \theta -v_{2f} sin \phi \\ v_{1i}^2 = v_{1f}^2 +v_{2f}^2 \\ v_{2f}cos \phi= v_{1i} -v_{1f} cos \theta \\ v_{2f}sin \phi = v_{1f}sin \theta \\ v_{2f}^2cos^2 \phi + v_{2f}^2sin^2 \phi = v_{1i}^2 -2v_{1i}v_{1f} cos \theta + v_{1f}^2cos^2 \theta + v_{1f}^2sin^2 \theta \\ v_{2f}^2 = v_{1i}^2 -2v_{1i}v_{1f}cos \theta + v_{1f}^2 \\ v_{1f}^2 + (v_{1i}^2 -2v_{1i}v_{1f}cos \theta + v_{1f}^2)=v_{1i}^2 \\ v_{1f}^2 -v_{1i}v_{1f}cos \theta =0 \\ v_{2f}=v_{1i} cos \theta = (3.50 \times 10^5 \;m/s)cos 37.0º = 2.80 \times 10^5 \;m/s \\ v_{2f} = \sqrt{v_{1i}^2 -v_{1f}^2} = \sqrt{ (3.50 \times 10^5 \;m/s)^2 – (2.80 \times 10^5 \;m/s)^2 } \\ = 2.11 \times 10^5 \; m/s \\ \phi = sin^{-1} \frac{v_{1f}sin \theta}{v_{2f}} \\ = sin^{-1} \frac{(2.80 \times 10^5 \; m/s)sin 37.0º}{(2.11 \times 10^5 \;m/s)} \\ = 53.0º$