A proton, moving with a velocity of v_1I^, collides elastically with another pro

Question

A proton, moving with a velocity of v_1I^, collides elastically with another proton that is initially at rest. Assuming that after the collision the speed of the initially moving proton is 2.30 times the speed of the proton initially at rest, find the following. (a) the speed of each proton after the collision in terms of v_1 (b) the direction of the velocity vectors after the collision (assume that the initially moving proton scatters toward the positive y direction) so you want the initially moving proton and the initially at rest proton relative to the positive z-axis.

Explanation / Answer

v1i = initial velocity of first proton before collision = Vi

v2i = initial velocity of second proton before collision = 0

v1f = final velocity of first proton after collision

v2f = final velocity of first proton after collision

given that : v1f = 2.3 v2f    eq-1

m = mass of each proton

using conservation of kinetic energy

m v21i + m v22i = m v21f + m v22f

using eq-1

V2i + 0 = 5.29 v22f + v22f

v2f = 0.398 Vi

using eq-1

v1f = 2.3 v2f

v1f = 2.3 (0.398 Vi)

v1f = 0.915 Vi

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.