contestada

A projectile proton with a speed of 610 m/s collides elastically with a target proton initially at rest. The two protons then move along perpendicular paths, with the projectile path at 47° from the original direction. After the collision, what are the speeds of (a) the target proton and (b) the projectile proton?

Respuesta :

Answer:

(a)   445.87 m/s

(b)   420.14 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 = 43° from x axis

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

       =  610 * sin 47 / ( cos 43 sin 47 + cos 47 sin 43 )

      =  610 * .7313 /( .7313 * .7313 + .68199 * .68199 )

      = 446.093 /( .53538 + .46511)

     = 445.87 m/s

(b)

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

                                                               = sin 43 * 445.87 / sin 47

                                                               = 420.14 m/s

Otras preguntas