Halliday, Fundamentals of Physics, 10e US l System Announcements unread) PRINTER

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Halliday, Fundamentals of Physics, 10e US l System Announcements unread) PRINTER VERSnow .BACx NECT Chapter 09, Problem o75 A projectile proton with a speed of 1 m/s collides elastically with a target proton initially at rest. The two protons then move along perpendicular paths, with the projectile path at 36° from the original direction. A what are the speeds of Ca) the target proton and Ch) the projectie Proton? (a) Number (b) Number Units Click if you would like to show work for this question: open show work SHOW HINT Go TUTORIAL LINK TO TEXT By accessing this Question Assistance, you will learn while you earn points on the Point Potential policy set by yourinstructor. based Question A o of 3 used SAVE FOR LATER ANSWR Earn Maximum Points available only if you answer this question correctly in two attempts less. OLPalio G2000-20izobn wiley Asions unc, Al Rights Reserved. A Division of lohn WileLA sense unc. Version 4.22.1.2

Explanation / Answer

Using conservation in momentum about the original direction:

m(1400 m/s) = m(v1)cos(36) + m(v2)cos(54)

(v1)cos(36) + (v2)cos(54) = 1400 m/s

Using conservation in momentum perpendicular to the original direction:

m(0 m/s) = m(v1)sin(36) - m(v2)sin(54)

(v1)sin(36) - (v2)sin(54) = 0 m/s

Thus,

v2 = v1*sin(36)/sin(54)

v2 = v1*0.72654

Plugging this into the first equation:

(v1)(0.8090) + (v1*0.72654)*0.58778 = 1400 m/s

(0.8090)(v1) + (0.427)(v1) = 1400 m/s

1.236(v1) = 1400 m/s

v1 = 1132.68 m/s

Thus, v2 = v1*0.72654

v2 = 822.91 m/s

the target proton's speed is 822.91 m/s

And, the projectile proton's speed is 1132.68 m/s

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