Question

A projectile 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?

Answer 1

(a) The speed of the target proton after the collision is:
`V_(2f) =433(m/s)`, and (b) the speed of the projectile proton after the collision is:
`v_(1f)=250(m/s)`.

Step-by-step 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:
`m_(1) v_(1i) =m_(1) v_(1f)Cos\beta _(1) +m_(2) v_(2f)Cos\beta _(2)`, and y axle:
`0=m_(1) v_(1f)Sin\beta _(1)+m_(2) v_(2f)Sin\beta _(2)`. Now replacing the value given as:
`v_(1i)=500(m/s)`,
`\beta_(1)=+60^(o)` for the projectile proton and according to the problem
`\beta_(1)and\beta_(2)` are perpendicular so
`\beta_(2)=-30^(o)`, and assuming that
`m_(1)=m_(2)`, we get for x axle:
`500=v_(1f)Cos\beta _(1)+ v_(2f)Cos\beta _(2)` and y axle:
`0=v_(1f)Sin\beta _(1)+v_(2f)Sin\beta _(2)`, then solving for
`v_(2f)`, we get:
`v_(2f)=-v_(1f)(Sin\beta_(1))/(Sin\beta_(2))= √(3)v_(1f)` and replacing at the first equation we get:
`500=(1)/(2) v_(1f) +(√(3))/(2) *√(3)*v_(1f)`, now solving for
`v_(1f)`, we can find the speed of the projectile proton after the collision as:
`v_(1f)=250(m/s)` and
`v_(2f)=√(3)*v_(1f)=433(m/s)`, that is the speed of the target proton after the collision.