Question

A projectile proton with a speed of 1100 m/s collides elastically with a target proton initially at rest. The two protons then move along perpendicular paths, with the projectile path at 45° from the original direction. After the collision, what are the speeds of (a) the target proton and (b) the projectile proton?

Answer 1

Answer:

So, the target proton's speed is  777.82 m/s

And, the projectile proton's speed is  777.82 m/s

Explanation:

as per the system, it conserves the linear momentum,

so along x axis :

Mp V1 (i) = Mp V1 (f) cos θ1 + Mp V2 (f) cos θ2

along y axis :

0 = -Mp V1 (f) sin θ1 + Mp V2 (f) sin θ2

let us assume before collision it was moving on positive x axis, hence target angle will be θ2 = 45° from x axis

V2(f) = V1 (i) sin θ1 / ( cosθ2 sin θ1 + cos θ1 sin θ2)

      =  1100 * sin 45 / ( cos 45 sin 45 + cos 45 sin 45 )

     =  1100 * 0.7071 /( 0.7071 * 0.7071 +0 .7071 * 0.7071 )

     = 777.82 /( 0.5 + 0.5)

    = 777.82 m/s

(b) 

 the speed of projectile , V1 (f) = sinθ2 * V2(f) / sinθ1

                                                              = sin 45 * 777.82 / sin 45

                                                              = 777.82 m/s

So, the target proton's speed is  777.82 m/s

And, the projectile proton's speed is  777.82 m/s