xample 9.9 Proton-Proton Collision A proton collides elastically with another pr

Question

xample 9.9 Proton-Proton Collision A proton collides elastically with another proton that is initially at rest. The incoming proton has an initial speed of 3.30 x 105 m/s and makes a glancing collision with the second proton as in the figure At close separations, the protons exert a repulsive electrostatic force on each other.) After the collision, one proton moves off at an angle of 30.8 to the original direction of motion and the second deflects at an angle of p to the same axis. Find the final speeds of the two protons and the angle p.

Explanation / Answer

v1i = (4.4 x 106) i + 0 j

v2i = 0 i + 0 j

v1f = v1 Cos17 i + v1 sin17 j

v2f = ?

using conservation of momentum

m1 v1i + m2 v2i = m1 v1f + m2 v2f

since m1 = m2

v1i + v2i = v1f + v2f

(4.4 x 106) i + 0 j + 0 i + 0 j = v1 Cos17 i + v1 sin17 j + v2f

(4.4 x 106 - v1 Cos17) i - v1 sin17 j = v2f   

|v2f | = sqrt((4.4 x 106 - v1 Cos17)2 + (- v1 sin17)2)                         eq-1

using conservation of kinetic energy

m1 v21i + m2 v22i = m1 v21f + m2 v22f

(4.4 x 106)2 + 02 = ((v1 Cos17)2 + (v1 Sin17)2) + v22f

using eq-1

(4.4 x 106)2 = ((v1 Cos17)2 + (v1 Sin17)2) + (4.4 x 106 - v1 Cos17)2 + (- v1 sin17)2

v1 = 4.21 x 106 m/s

using eq-1

|v2f | = sqrt((4.4 x 106 - v1 Cos17)2 + (- v1 sin17)2) = sqrt((4.4 x 106 - (4.21 x 106)Cos17)2 + ((4.21 x 106) sin17)2)

|v2f | = 1.28 x 106 m/s

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