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?

Question

Elise Tran

Physics

10 April, 08:53

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?

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Answers (1)

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Coco Chanel · Answer 1

(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