Sebuah bola 1 bermassa 2 kg bergerak dengan kelaju...

Sebuah bola 1 bermassa 2 kg bergerak dengan kelajuan 8 m/s menumbuk lenting sempurna bola 2 bermassa 3 kg yang sedang bergerak dengan kelajuan 5 m/s. Jika setelah tumbukan kedua bola bergerak searah, maka besar kecepatan bola 2 setelah tumbukan adalah...

Jawaban:

Pada kedua bola terjadi Hukum Kekekalan Momentum, dan didapatkan perhitungan :

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell bold italic m subscript bold 1 bold italic v subscript bold 1 bold space bold plus bold space bold italic m subscript bold 2 bold italic v subscript bold 2 bold space end cell bold equals cell bold space bold italic m subscript bold 1 bold italic v subscript bold 1 bold apostrophe bold space bold plus bold space bold italic m subscript bold 2 bold italic v subscript bold 2 bold apostrophe end cell row cell 2 space kg open parentheses 8 space straight m divided by straight s close parentheses plus 3 space kg open parentheses negative 5 space straight m divided by straight s close parentheses space end cell equals cell space 2 space kg space v subscript 1 apostrophe plus 3 space kg space v subscript 2 apostrophe end cell row cell 16 space minus space 15 space end cell equals cell space 2 v subscript 1 apostrophe space plus space 3 v subscript 2 apostrophe end cell row cell bold 1 bold space end cell bold equals cell bold space bold 2 bold italic v subscript bold 1 bold apostrophe bold space bold plus bold space bold 3 bold italic v subscript bold 2 bold apostrophe bold space space space space bold space bold. bold. bold. bold. bold. bold left parenthesis bold 1 bold right parenthesis end cell end table end style$

*tanda negatif (-) v₂ menunjukkan arah yang berlawanan

Pada bola terjadi tumbukan lenting sempurna dengan nilai koefisien restitusi adalah 1. Maka didapatkan :

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell e space end cell equals cell space minus space open parentheses fraction numerator v subscript 2 apostrophe space minus space v subscript 1 apostrophe over denominator v subscript 2 space minus space v subscript 1 end fraction close parentheses end cell row cell 1 space end cell equals cell space minus space open parentheses fraction numerator v subscript 2 apostrophe space minus space v subscript 1 apostrophe over denominator 8 space straight m divided by straight s space minus space 5 space straight m divided by straight s end fraction close parentheses end cell row cell 1 space end cell equals cell space minus space open parentheses fraction numerator v subscript 2 apostrophe space minus space v subscript 1 apostrophe over denominator 3 end fraction close parentheses end cell row cell 3 space end cell equals cell space minus space v subscript 2 apostrophe space plus space v subscript 1 apostrophe end cell row cell bold 3 bold space end cell bold equals cell bold space bold italic v subscript bold 1 bold apostrophe bold space bold minus bold space bold italic v subscript bold 2 bold apostrophe bold space bold space bold space bold space bold space bold. bold. bold. bold. bold. bold left parenthesis bold 2 bold right parenthesis end cell end table end style$

Eliminasi persamaan 1 dan persamaan 2

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell 2 v subscript 1 apostrophe space plus space 3 v subscript 2 apostrophe space end cell equals cell space 1 space open vertical bar cross times space 1 close vertical bar end cell row cell v subscript 1 apostrophe space minus space v subscript 2 apostrophe space end cell equals cell space 3 space open vertical bar cross times space 2 close vertical bar end cell row blank blank blank row cell 2 v subscript 1 apostrophe space plus space 3 v subscript 2 apostrophe space end cell equals cell space 1 end cell row cell 2 v subscript 1 apostrophe space minus space 2 v subscript 2 apostrophe space end cell equals cell space 6 space space space space bold asterisk times bold dikurangkan end cell row cell 5 v subscript 2 apostrophe space end cell equals cell space minus 5 end cell row cell v subscript 2 apostrophe space end cell equals cell space fraction numerator negative 5 over denominator 5 end fraction end cell row cell v subscript 2 apostrophe space end cell equals cell space minus 1 space straight m divided by straight s end cell row cell bold italic v subscript bold 2 bold apostrophe bold space end cell bold equals cell bold space bold 1 bold space bold m bold divided by bold s end cell end table end style$

Jadi, besar kecepatan bola 2 setelah tumbukan adalah 1 m/s.

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