Insights Blog
-- Browse All Articles --
Physics Articles
Physics Tutorials
Physics Guides
Physics FAQ
Math Articles
Math Tutorials
Math Guides
Math FAQ
Education Articles
Education Guides
Bio/Chem Articles
Technology Guides
Computer Science Tutorials
Forums
Intro Physics Homework Help
Advanced Physics Homework Help
Precalculus Homework Help
Calculus Homework Help
Bio/Chem Homework Help
Engineering Homework Help
Trending
Featured Threads
Log in
Register
What's new
Search
Search
Search titles only
By:
Intro Physics Homework Help
Advanced Physics Homework Help
Precalculus Homework Help
Calculus Homework Help
Bio/Chem Homework Help
Engineering Homework Help
Menu
Log in
Register
Navigation
More options
Contact us
Close Menu
JavaScript is disabled. For a better experience, please enable JavaScript in your browser before proceeding.
You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an
alternative browser
.
Forums
Homework Help
Introductory Physics Homework Help
Distance of Closest Approach
Reply to thread
Message
[QUOTE="Meera.sheeda, post: 4895920, member: 527006"] [h2]Homework Statement [/h2] A proton and an alpha particle (q= +2e, m=4u) are fired directly at one another from far way, each with an initial velocity of 0.01c. What is their distance of closest approach, as measured between their centres? [h2]Homework Equations[/h2] m[SUB]i[/SUB]v[SUB]i[/SUB] = m[SUB]f [/SUB]v[SUB]f[/SUB] and KE[SUB]i [/SUB]+U[SUB]i[/SUB] = KE[SUB]f[/SUB] + U[SUB]f[/SUB] U = (q1 x q2 ) / (4επr) KE= 0.5mv^2 Mass of proton: 1.67E-27 Charge of Proton: 1.692E-19 c = 3E8[h2]The Attempt at a Solution[/h2] I'm sorry this question has been asked before but I didn't understand the explanations in those threads. I tried using the conservation of momentum: (1.67E-27 * 0.01c) - (4 * 1.67E-27 * 0.01c) = -(1.67E-27 * velocity of proton) + (4* 1.67E-27 * velocity of alpha particle) ∴ -9000000= -velocity of proton + 4*velocity of alpha particle but then I don't understand how to find the velocities of the particles individually or how to get this equation into the KE equation. When I tried using the conservation of energy I got: ½*1.67*10^-27*(0.01c)^2 + ½*4*1.67*10^-27*(0.01c)^2 = (3*1.602*10^-19) /4επr ∴ r = 115008.95 which is so far off, as the answer in the back of the book says: 1.93x10^-14 Which equation am I using wrong and how can I fix it? [/QUOTE]
Insert quotes…
Post reply
Forums
Homework Help
Introductory Physics Homework Help
Distance of Closest Approach
Back
Top