Jawaban:

Dua buah benda memiliki energi kinetik yang sama. Jika massa benda pertama dua kali massa benda kedua, maka perbandingan antara momentum benda pertama dan kedua adalah...undefined

begin mathsize 14px style 1 space colon space 2 end style

begin mathsize 14px style 2 space colon space 1 end style

begin mathsize 14px style square root of 2 space colon space 2 end style

begin mathsize 14px style 2 space colon space square root of 2 end style

begin mathsize 14px style 3 space colon space 2 end style

Jawaban:

Diketahui :

begin mathsize 14px style E K subscript italic 1 space equals italic space E K subscript italic 2 m subscript italic 1 space equals space 2 space. space m subscript italic 2 end style

Untuk menentukan perbandingan momentum benda pertama dan benda kedua dapat digunakan persamaan sebagai berikut.

Menentukan perbandingan kecepatan benda pertama dan benda kedua

begin mathsize 14px style E K subscript italic 1 space equals space E K subscript italic 2 1 half space. space m subscript italic 1 space. space v subscript italic 1 to the power of italic 2 space equals space 1 half space. space m subscript italic 2 space. space v subscript italic 2 to the power of italic 2 space m subscript italic 1 space. space v subscript italic 1 to the power of italic 2 space equals space m subscript italic 2 space. space v subscript italic 2 to the power of italic 2 2 space m subscript italic 2 space. space v subscript italic 1 to the power of italic 2 space equals space m subscript italic 2 space. space v subscript italic 2 to the power of italic 2 2 space. space v subscript italic 1 to the power of italic 2 space equals space v subscript italic 2 to the power of italic 2 v subscript italic 1 to the power of italic 2 space equals space 1 half space. space v subscript italic 2 to the power of italic 2 v subscript italic 1 space equals space square root of bevelled 1 half end root space v subscript italic 2 end style

Menentukan perbandingan momentum benda pertama dan benda kedua

begin mathsize 14px style P subscript italic 1 over P subscript italic 2 space equals space fraction numerator m subscript italic 1 space. space v subscript italic 1 over denominator m subscript italic 2 space. space v subscript italic 2 end fraction space equals space fraction numerator 2 space. space m subscript italic 2 space. space square root of begin display style bevelled 1 half end style end root space v subscript italic 2 over denominator m subscript italic 2 space. space v subscript italic 2 end fraction space equals space fraction numerator 2 over denominator square root of 2 end fraction end style

Jadi, jawaban yang benar adalah D.undefined