Aprojectile proton with a speed of 500 m/s collides elastically with a target proton initially at rest. the two protons then move along perpendicular paths, with the projectile path at 60° from the original direction. (a) what is the speed of the target proton after the collision? 0 incorrect: your answer is incorrect. m/s (b) what is the speed of the projectile proton after the collision?

(a) The speed of the target proton after the collision is:, and (b) the speed of the projectile proton after the collision is: .

Explanation:

We need to apply at the system the conservation of the linear momentum on both directions x and y, and we get for the x axle:, and y axle:. Now replacing the value given as: , for the projectile proton and according to the problem are perpendicular so , and assuming that , we get for x axle: and y axle: , then solving for , we get: and replacing at the first equation we get:, now solving for , we can find the speed of the projectile proton after the collision as: and , that is the speed of the target proton after the collision.

f=ma a/f/m a= 480/60 a = 8m/s/s that is quick wow