Question

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?

Answer 1

(a) 445.87 m/s

(b) 420.14 m/s

Step-by-step 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