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-iii! 22% (a Fri 11:11 AM Q E Firefox File Edit View History Bookmarks Tools Windo Help e o e MasteringPhysics: Bonus C Physics question I Chegg X 0 a 0 https Session masteringphysics.com /myct em view?assignmentProble c Ca Search Most Visited Physics 2750 Spring 2016 (University Physics Signed in as Juwan Mahaney Help close Bookmarks Help Sign Out Bonus Homework Short Answer Question 8.1 previous l 10 of 28 l next Bookmarks Bookmarks rted Be Short Answer Question 8.1 myZou: P Sig Bb Login Part A N Netflix A particle of mass 5.34 x 10 27 kg, moving at 6.2 x 105 m/s, strikes an identical particle which is initially at rest. After the interaction, the particles (which can't be All-N distinguished) are observed moving at angles 57.7 and 32.3°, both angles being measured with respect to the original direction of motion. What are the final speeds of the particles? files posted 3 Print Sm Login Pocket m/s at 32.3 Subm My Answers Give U Part B Grade m/s at 57.7 Subm My Answers Give U Continue Provide Feedback 29

Explanation / Answer

we need to use conservation of momentum in the horizontal direction and vertical direction

now let the initial velocity be u and masses be m1 and m2

let the final velocities be v1 and v2

now,

in the vertical direction we will have,

initial momentum= final momentum

0 = m1*v1*sin(57.7) + m2*v2*sin(32.3)

=>v1= -v2* 0.632 ....................(1)

now in the horizontal direction using conservation of mometum we have,

m1*u= m1*v1*cos(57.7) + m2*v2* cos(32.3)

=>u=v1*cos(57.7) + v2* cos(32.3)

now putting the values of u and v1

we get,

=>6.2x105 = (-v2* 0.632)* 0.534 + v2*

=>6.2x105 = -0.3377* v2 + 0.845*v2

=>v2= 1.22 x106 m/s   ( at 32.3)

now v1 = -v2* 0.632 = 7.724 x105 m/s (at 57 degrees)

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