Bola A dan B yang bermassa 100 gram dan 200 gram berada pada posisi seperti gambar berikut.    Apabila kedua bola menggelinding pada bidang licin dan saling bertumbukan lenting sempurna, kecepatan masing-masing bola dan arahnya setelah tumbukan adalah...(g=9,8 m/s2)

Pertanyaan

Bola A dan B yang bermassa 100 gram dan 200 gram berada pada posisi seperti gambar berikut.

  

Apabila kedua bola menggelinding pada bidang licin dan saling bertumbukan lenting sempurna, kecepatan masing-masing bola dan arahnya setelah tumbukan adalah...begin mathsize 14px style open parentheses g equals 9 comma 8 space straight m divided by straight s squared close parentheses end style 

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S. Syifa

Master Teacher

Jawaban terverifikasi

Pembahasan

Diketahui:

begin mathsize 12px style m subscript A equals 0 comma 1 space k g end style

size 12px m subscript size 12px B size 12px equals size 12px 0 size 12px comma size 12px 2 size 12px space size 12px k size 12px g

size 12px g size 12px equals size 12px 9 size 12px comma size 12px 8 size 12px space size 12px m size 12px divided by size 12px s to the power of size 12px 2

size 12px h subscript size 12px A size 12px equals size 12px h subscript size 12px B size 12px equals size 12px 3 size 12px comma size 12px 6 size 12px space size 12px m

Pembahasan:

Kecepatan A sebelum tumbukan

begin mathsize 12px style m subscript A. g. h subscript A equals 1 half. m subscript A. v subscript A squared end style

begin mathsize 12px style v subscript A equals square root of 2. g. h subscript A end root end style

begin mathsize 12px style v subscript A equals square root of 2. left parenthesis 9 comma 8 space m divided by s squared right parenthesis. left parenthesis 3 comma 6 space m right parenthesis end root end style

begin mathsize 12px style v subscript A equals 8 comma 4 space m divided by s end style (berlawanan begin mathsize 12px style v subscript A end style)

Kecepatan B sebelum tumbukan

size 12px m subscript B size 12px. size 12px g size 12px. size 12px h subscript B size 12px equals size 12px 1 over size 12px 2 size 12px. size 12px m subscript B size 12px. size 12px v subscript B to the power of size 12px 2

size 12px v subscript B size 12px equals square root of size 12px 2 size 12px. size 12px g size 12px. size 12px h subscript B end root

size 12px v subscript B size 12px equals square root of size 12px 2 size 12px. size 12px left parenthesis size 12px 9 size 12px comma size 12px 8 size 12px space size 12px m size 12px divided by size 12px s to the power of size 12px 2 size 12px right parenthesis size 12px. size 12px left parenthesis size 12px 3 size 12px comma size 12px 6 size 12px space size 12px m size 12px right parenthesis end root

size 12px v subscript B size 12px equals size 12px minus size 12px 8 size 12px comma size 12px 4 size 12px space size 12px m size 12px divided by size 12px s 

Tumbukan lenting sempurna

begin mathsize 12px style e equals 1 end style

begin mathsize 12px style negative fraction numerator v subscript B apostrophe minus v subscript A apostrophe over denominator v subscript B minus v subscript A end fraction equals 1 end style

size 12px minus fraction numerator size 12px v subscript size 12px B size 12px apostrophe size 12px minus size 12px v subscript size 12px A size 12px apostrophe over denominator size 12px minus size 12px 8 size 12px comma size 12px 4 size 12px minus size 12px 8 size 12px comma size 12px 4 end fraction size 12px equals size 12px 1

begin mathsize 12px style v subscript B apostrophe minus v subscript A apostrophe equals 16 comma 8 end style

begin mathsize 12px style v subscript B apostrophe equals 16 comma 8 plus v subscript A apostrophe end style ... (1)

Hukum kekekalan momentum

begin mathsize 12px style p equals p apostrophe end style

begin mathsize 12px style m subscript A. v subscript A plus m subscript B. v subscript B equals m subscript A. v subscript A apostrophe plus m subscript B. v subscript B apostrophe end style

begin mathsize 12px style 0 comma 1.8 comma 4 plus 0 comma 2. left parenthesis negative 8 comma 4 right parenthesis equals 0 comma 1. v subscript A apostrophe plus 0 comma 2. v subscript B apostrophe end style ... (2)

Substitusikan pers. 1 ke pers. 2

begin mathsize 12px style 0 comma 84 minus 1 comma 68 equals 0 comma 1. v subscript A apostrophe plus 0 comma 2. left parenthesis 16 comma 8 plus v subscript A apostrophe right parenthesis end style

begin mathsize 12px style negative 0 comma 84 equals 0 comma 1. v subscript A apostrophe plus 3 comma 36 plus 0 comma 2. v subscript A apostrophe end style

begin mathsize 12px style negative 0 comma 3. v subscript A apostrophe equals 3 comma 36 plus 0 comma 84 end style

begin mathsize 12px style v subscript A apostrophe equals fraction numerator 4 comma 2 over denominator negative 0 comma 3 end fraction end style

begin mathsize 12px style v subscript A equals negative 14 space m divided by s end style (begin mathsize 12px style v subscript A apostrophe end style sebesar begin mathsize 12px style 14 space m divided by s end style ke kiri)

Substitusikan nilai begin mathsize 12px style v subscript A apostrophe end style ke pers. 1

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begin mathsize 12px style v subscript B apostrophe equals 16 comma 8 plus left parenthesis negative 14 right parenthesis end style

begin mathsize 12px style v subscript B apostrophe equals 2 comma 8 space m divided by s end style (begin mathsize 12px style v subscript B apostrophe end style sebesar begin mathsize 12px style 2 comma 8 space m divided by s end style ke kanan)

Jadi, jawaban yang tepat adalah begin mathsize 12px style v subscript A apostrophe end style sebesar begin mathsize 12px style 14 space m divided by s end style ke kiri dan begin mathsize 12px style v subscript B apostrophe end style sebesar begin mathsize 12px style 2 comma 8 space m divided by s end style ke kanan (E)

 

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