BARA AFI 2 Bahan Ajar Siswa SMANBARA Fisika 2
Darsika Aryanta 2017
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BAB I HUKUM NEWTON TENTANG GERAK DAN GRAVITASI
Sekelompok penerjun paying terdiri dari delapan orang sedang melakukan terjun bebas sambil membentuk formasi. Di ketinggian di atas lima ribu kilometer laju jatuh mereka masih lambat karena percepatan gravitasi masih kecil. Sesudah itu semakin lama semagin cepat gerak mereka sehingga susah mempertahankan formasi terjun bebas dan harus segera menarik tali pengikat payung. Mengapa demikian? Menurut Newton percepatan sebanding dengan gaya sedangkan gaya berbanding terbalik dengan kuadrat jarak antara kedua benda. Bagaimana dengan percepatan pada gerak di permukaan bumi? Apakah sebanding dengan gaya? Gaya apakah itu? Pada bab ini akan dikaji hukum-hukum Newton tentang gerak termasuk hukum tentang gravitasi.
1
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BAB 2 HUKUM NEWTON TENTANG GERAK DAN GRAVITASI
KOMPETENSI DASAR 3.8 Menganalisis keteraturan gerak planet dalam tatasurya berdasarkan hukum-hukum Newton 4.8 Menyajikan karya mengenai gerak satelit buatan yang mengorbit bumi, pemanfaatan dan dampak yang ditimbulkannya dari berbagai sumber informasi
Sistem gravitasi pada planet Jupiter, planet terbesar difoto bersama empat satelitnyan antara lain Io (kiri atas), Ganymede satelit terbesarnya (kiri bawah), Europa (tengah), dan Callisto (kanan bawah) menciptakan keteraturan orbit satelit-satelit yang dimiliki Jupiter sebagai planet pemilik satelit terbanyak. Pesawat-pesawat tempur terbang bergerak2di angkasa tidak mengalami gesekan karena tidak bersentuhan dengan benda lain. Gesekan dengan udara relatif kecil dan dapat diabaikan. Lain halnya dengan mobil yang mengalami gaya gesekan antara ban mobil dengan jalan. Demikian pula pesawat tempur yang akan tinggal landas di atas kapal
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A. Hukum Gravitasi Newton Jika malam telah tiba, perhatikanlah bulan di langit! Apakah bulan dalam keadaan diam saja? Apakah bulan jatuh ke bumi? Mengapa? Perhatikan pula situasi sebuah pohon di sekitarmu ? Apakah ada daun pada pohon yang jatuh di bawah pohon ? Mengapa daun yang massanya ringan dapat jatuh ke Bulan dilihat dari Bumi
permukaan bumi, sedang bulan yang massanya jauh lebih besar dibandingkan selembar daun tidak jatuh ke bumi ?
1.
Gaya Gravitasi
Permasalahan di atas telah dikaji oleh Sir Isaac Newton
pada
abad
16
masehi.
Newton
mengemukakan, bahwa ternyata ada suatu ”gaya pada suatu jarak” yang memungkinkan dua benda atau lebih untuk berinteraksi. Istilah tersebut oleh Michael Faraday, pada abad 18 diubah menjadi istilah ”medan”. Adapun pengertian medan adalah tempat di sekitar
suatu besaran besaran fisis yang masih
dipengaruhi oleh besaran tersebut dalam suatu entitas Sir Isaac Newton (1642 1727)
tertentu. Sebagai contoh, gaya gravitasi akan bekerja pada massa suatu benda yang masih berada dalam
medan gravitasi suatu benda atau planet. Jika medan gravitasi sudah dapat diabaikan, maka sebuah massa yang berada di sekitar besaran benda tersebut tidak dapat dipengaruhi. Dengan demikian, dapatlah kamu pahami, mengapa daun yang massanya lebih kecil dibanding bulan yang massanya jauh lebih besar dapat ditarik bumi. Dalam penelitiannya, Newton menyimpulkan, bahwa gaya gravitasi atau gaya tarik-menarik dapat berlaku secara universal dan sebanding oleh massa masing-
3
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masing benda dan berbanding terbalik dengan kuadrat jarak kedua benda, dan dirumuskan: r
Gambar 1: Diagram gravitasi antara dua buah benda yang terpisah sejauh r
m1.m2
F=G
2
r
F
=
gaya tarik-menarik antara kedua benda (N)
m1
=
massa benda 1 (kg)
m2
=
massa benda 2 (kg)
r
=
jarak antara kedua pusat benda (m)
G
=
tetapan gravitasi universal
Saat itu Newton belum dapat mendefinisikan besar dari G. Nilai G tidak dapat diperoleh dari teori, namun harus melalui eksperimen. Orang yang pertama kali melakukan eksperimen untuk menentukan nilai G adalah Henry Cavendish, dengan menggunakan neraca torsi. Neraca seperti ini kemudian disebut neraca Cavendish. Bola dengan massa yang berbeda, yaitu m dan M yang dapat bergerak bebas pada poros, akan tarik menarik, sehingga akan memuntir serat kuarsa, sehingga cahaya yang 4
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memantul pada cermin pun akan bergeser pada skala. Dengan mengkonversi skala, dan memperhatikan jarak m dan M serta massa m dan M, maka Cavendish menetapkan nilai -11
G sebesar 6,754 x 10
2
2
N.m /kg . Nilai ini kemudian kini dengan perlengkapan yang
lebih canggih disempurnakan, sehingga diperoleh nilai: -11
G = 6,672 x 10
2
2
N.m /kg .
Gaya gravitasi merupakan besaran vektor, sehingga bila suatu benda mengalami gaya tarik gravitasi dari lebih satu benda sumber gravitasi, maka teknik mencari resultannya dipergunakan teknik pencarian resultan vektor. Misalnya dua buah gaya F1 dan F2 yang membentuk sudut
, resultan gayanya dapat ditentukan berdasarkan
persamaan : F F12 F2 2 2F1F2 cos co sα
Gambar :
Contoh : 20
20
1. Jika dua planet masing-masing bermassa 2 x 10 kg dan 4 x 10 kg, mempunyai 5
jarak antara kedua pusat planet sebesar 2 x 10 km. Tentukan besar gaya tarikmenarik antara kedua planet! Penyelesaian : -11
Nilai G jika tidak disebutkan, usahakan untuk dihafalkan sebesar 6,672 x 10 2
2
N.m /kg F=G
m1 .m 2 r
2
-11
F = 6,672 x 10 .
2.10 20 x 4.10 20 (2.10 ( 5
3)
)2
5
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-11
F = 6,672 x 10 .
2.10 20 x 4.10 20 (2.10 8 ) 2
14
F = 1,33 . 10 N
Uji Kompetensi Kerjakan soal-soal berikut dengan benar!
1. Dua buah benda masing-masing massanya 10 kg dan 20 kg terpisahkan pada jarak 2 meter satu dengan yang lain. Tentukan gaya gravitasi antara kedua benda itu! -10
2. Gaya tarik gravitasi antara du buah benda bermassa adalah 2,001 x 10 N. Bila massa benda adalah 3 kg dan 9kg. Tentukanlah jarak antara kedua benda itu! 3. Massa sebesar 5 kg terpisah pada jarak 2 meter dari massa yang lain. Gaya gravitasi antara kedua benda adalah sebesar 2,5 x 10-10. Tentukan massa benda yang lain! 4. Tiga buah bola bermassa masing-masing 1kg, 2kg dan 3kg diletakkan pada titik sudut segitiga sama sisi dengan sisi 1 meter. Tentukanlah gaya yang dialami oleh bola bermassa 1 kg dalam susunan ini! 5. Dua buah bola bermassa masing-masing 4 kg terpisah pada jarak 2 3 meter. Tentukanlah gaya tarik gravitasi yang dialami oleh bola bermassa 5 kg yang terletak pada jarak 2 meter dari kedua ked ua massa tersebut! 6. Sebuah bola bermassa 3 kg terletak pada titik pusat sistem sumbu koordinat. Bola lainya yang masing-masing bermassa sebesar 16 kg, 36 kg dan 25 kg terletak pada titik-titik ( 4,0 ), ( 4,5 ) dan ( 0,5 ). Satuan koordinat dalam meter. Tentukanlah gaya yang dialami oleh bola bermassa 3 kg itu! 7. Dua massa masing-masing dari 2kg dan 8 kg terpisah sejauh 1,2 meter. Tentukanlah gaya gravitasi pada massa 1 kg yang terletak pada suatu titik 0,4 meter dari massa 2 kg dan 0,8 meter dari massa 8 kg! 8. Dua buah bermassa 2 kg dan 12,5 kg terpisah pada jarak 7 meter. Tentukanlah letak bola bermassa 6 kg sehingga gaya tarik gravitasi yang dialaminya sama
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9. Dua buah benda bermassa pada saat terpisah sejauh 2 meter saling mengerjakan gaya sebesar 4 g. Bila jarak antaranya di jadikan 4 meter, tentukanlah gaya tarik menarik yang dikerjakan kedua benda itu! 10. Di titik A dan C dari suatu bujur sangkar ABCD ditempatkan massa sebesar 1 kg dan 0,5 kg. Bila gaya tarik menarik antara kedua massa tersebut besarnya 0,5 Gnewton, tentukanlah panjang sisi bujur sangkar tersebut.
2.
Medan Gravitasi
Di samping gaya gravitasi, hukum gravitasi Newton juga menetapkan tentang medan gravitasi disekitar suatu benda atau umumnya sebuah planet. Medan gravitasi ini akan menunjukkan percepatan gravitasi dari suatu benda di sekitar suatu benda atau planet. Adapun besar medan gravitasi atau percepatan gravitasi dirumuskan : g=G g
=
medan
M 2
r
gravitasi
atau
percepatan
gravitasi (m/s2) G
=
tetapan gravitasi universal -11
2
2
= 6,672 x 10 N.m /kg M Gambar 3: Satelit mengorbit bumi berada dalam dalam medan medan gravitasi gravitasi bumi
=
massa dari suatu planet atau benda
=
jarak suatu titik ke pusat planet atau
(kg) r
pusat benda (m) Besar percepatan gravitasi yang dialami semua benda di sebuah permukaan planet adalah sama. Selembar bulu ayam dan segumpal tanah liat dijatuhkan dari ketinggian yang sama dalam tabung hampa akan bersamaan mencapai dasar tabung. Namun bila tabung berisi udara tanah liat akan mencapai dasar tabung lebih dahulu. Hal itu bukan disebabkan karena percepatan gravitasi di tempat tersebut yang berbeda untuk benda
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yang berbeda, namun disebabkan oleh adanya hambatan
Berisi udara
Ham Ha m a uda udara ra
udara di dalam tabung.
Gambar 4: selembar bulu ayam dijatuhkan bersama tanah liat pada dua keadaan berbeda.
Kuat medan gravitasi adalah suatu besaran vektor yang arahnya senantiasa menuju ke pusat benda yang menimbulkannya. Kuat medan gravitasi di suatu titik oleh beberapa benda bermassa diperoleh dengan menjumlahkan vektor-vektor vek tor-vektor medan gravitasi oleh tiap-tiap benda. Kuat medan gravitasi yang disebabkan oleh dua buah benda yang kuat medannya saling membentuk sudut , dapat dinyatakan dengan persamaan : g g12 g 2 2 2g1g 2 cosα
Kerja Kelompok Lakukan percobaan berikut kersama kelompokmu! Tujuan:
Menunjukkan percepatan gravitasi disuatu tempat berlaku sama untuk semua benda. Pelaksanaan:
Gunakan stopwatch untuk mengukur waktu yang diperlukan untuk selembar kertas dan kelereng untuk sampai di permukaan tanah dari ketinggian yang sama. Ulangi percobaan itu berulang-ulang dengan memvariasi bentuk kertas, dengan cara meremas kertas, dari ukuran yang besar, hingga menjadi ukuran yang sangat mampat atau kecil. Perhatikan
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Suatu hal yang perlu diperhatikan dalam membicarakan medan gravitasi atau percepatan gravitasi adalah konsep massa benda dan berat benda tidak sama. Massa benda di mana pun adalah tetap, namun berat benda di berbagai tempat belum b elum tentu sama atau tetap.
Contoh : 24
1. Sebuah planet bermassa 6 x 10 kg dan berjari-jari 4.000 km. Tentukan percepatan gravitasi di permukaan planet tersebut! Penyelesaian :
g=G
M r 2 -11
g = 6,672 x 10 .
6.10 24 (4.10 6 ) 2
2
g = 25,02 m/s
3. Energi Potensial Gravitasi Benda bermassa m yang terletak diluar bumi, energi potensial grafitasinya pada jarak r dari pusat bumi, dinyatakan dengan persamaan : Ep - G dimana
M.m r
Ep = Energi potensial gravitasi G = Konstanta grafitasi M = massa bumi m = massa benda r = Jarak pusat benda ke pusat bumi.
Tanda negatif (-) berarti jika benda bergerak di bawah pengaruh gaya grafitasi dari jarak
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dipergunakan untuk menambah energi kinetik dengan makin besarnya laju benda waktu bergerak mendekati bumi. Jika mula-mula benda berada di tempat yang jauh tak hingga ( r =
) dengan energi
kinetik sama dengan nol, maka dalam perjalanan mendekati bumi, medan gravitasi merubah energi potensial menjadi energi kinetik. Pada waktu sampai di bumi energi kinetik benda sama dengan energi potensial gravitasi. Jadi : 1 mv 2 2 Dimana
m
G
M.m R
= massa benda.
M
= massa bumi.
R
= jari - jari bumi.
V
= kecepatan benda di permukaan bumi.
4. Potensial Gravitasi Potensial gravitasi didefinisikan didefinisikan sebagai : energi potensial gravitasi per satuan massa. Dapat dinyatakan dengan persamaan :
v Dimana
E p m
v
= potensial gravitasi, satuan : Joule/kg.
E p
= Energi potensial gravitasi, satuan : Joule
m
= massa benda, satuan : kg.
Energi potensial gravitasi gravitasi benda bermassa m’ yang terletak pada jarak r dari pusat massa benda bermassa m dapat kita nyatakan dengan persamaan : Ep - G
m m' r
Bila massa m’ terletak dititik p maka potensial gravitasi gra vitasi di titik p yang dialami oleh
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V
Ep r
G
V G
Dimana
m m' r m'
m r
V
= potensial gravitasi pada jarak r dari massa m
m
= massa benda
r
= jarak tempat yang mengalami potensial gravitasi ke benda.
Potensial gravitasi merupakan besaran skalar, karena itu potensial yang disebabkan oleh berapa benda bermassa merupakan jumlah aljabar dari potensial po tensial gravitasi masing-masing benda bermassa itu, Jadi : Vt = V1 + V2 + V3 + ...... + Vn Beda potensial antara dua titik dalam medan gravitasi didefinisikan sebagai potensial di titik yang satu dikurangi dengan potensial di titik yang lain. Usaha yang dilakukan untuk mengangkut, massa m dari satu titik ke titik lain lewat sembarang lintasan sama dengan massa benda itu kali beda potensial antara kedua titik itu.
WA→B = m (VB - VA)
WA→B = Usaha dari A ke B.
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3. Suatu benda yang massanya 10 kg berada pada suatu tempat yang memiliki energi potensial gravitasi yang besarnya sama dengan 5 x 108 joule. Tentukanlah potensial gravitasi yang dialami oleh benda itu. 4. Tentukanlah potensial gravitasi pada suatu titik yang terletak 2 meter dari suatu benda bermassa 25 kg. 5. Pada gambar di bawah ini, massa m1 = 0,3 kg dan massa m2 = 0,1 kg.
a. Tentukanlah potensial gravitasi yang disebabkan oleh massa m1 dan m2 dititik O dan dititik A. b. Berapakah usaha yang dilakukan untuk mengangkut massa m = 0,01 0, 01 kg dari titik A ke titik O -5 G J/kg. 6. Dua massa masing-masing 0,2 kg dan 0,8 kg terpisah sejauh 0,12 meter. a. Tentukan potensial gravitasi pada titik 0,04 meter dari massa 0,2 kg dan 0,08 meter dari massa 0,8 kg. b. Berapa usaha yang diperlukan untuk memindahkan massa sebesar 1 kg dari titik jauh tak hingga kesuatu titik yang terletak 0,08 meter dari massa 0,8 kg.
3.
Hukum-Hukum Keppler
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Semua planet bergerak pada lintasan elips mengitari matahari dengan matahari berada di salah satu fokus elips. Hukum I ini dapat menjelaskan akan lintasan planet yang berbentuk elips, namun belum dapat menjelaskan kedudukan planet terhadap matahari, maka muncullah hukum II Kepler.
b.
Hukum II Kepler
Suatu
garis
menghubungkan
khayal matahari
yang dengan
planet, menyapu luas juring yang sama dalam selang waktu yang sama.
c.
Hukum III Kepler
Perbandingan kuadrat periode terhadap pangkat tiga dari setengah sumbu panjang elips adalah sama untuk semua planet. Hukum III Kepler dapat dirumuskan :
T 2 R
T
=
3
=k
atau
T 1
2
R1
3
=
kala revolusi suatu plenet (s atau tahun)
T 2
2
R 2 3
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1. Sebuah planet mempunyai kala revolusi terhadap Matahari sebesar 4 tahun. Tentukan jarak planet tersebut terhadap Matahari! Penyelesaian :
Jika nilai pembanding dari planet lain tidak diketahui, gunakan nilai yang dimiliki bumi. T 1
2
R1
3
12 13
=
=
R 2 =
T 2
2
R 2
3
42 R 2 3 3
sehingga
16
R 2 = 2,5 sa (satuan astronomis = 150 juta km)
2. Jika dua benda mengalami gaya tarik gravitasi 400 N, maka tentukan gaya gravitasinya kini, jika jarak kedua benda dijadikan ½ kali semula! Penyelesaian :
F 2 F 1
F 2 F 1 F
G
= G
=
m1 .m 2 r 22 m1 .m 2 2
r 1
r 12 r 22 r 2
semua nilai tetap, kecuali besaran jarak yang berubah
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Penyelesaian : w2 w1 w2 w1
w2 w1
=
=
m. g 2
w2
m. g 1
w1
=
g 2
w2
g 1
2500
G
= G
r 12 2
r 2
2
r 1
(3r 1 ) 2
M 2
r 2
M
w2 =
2500 9
= 277,78 N
2
r 1
4. Dua benda masing-masing bermassa 2500 kg dan 900 kg pada jarak 10 m. Tentukan letak benda ketiga di antara benda pertama dan kedua, jika benda ketiga yang bermassa 4500 kg mengalami gaya gravitasi nol! Penyelesaian: Langkah 1: Gambarkan posisi atau uraian gayanya:
F13
=
gaya tarik menarik antara benda 1 dan 3
F23
=
gaya tarik menarik antara benda 2 dan 3
Jika gaya yang dialami benda ketiga nol, maka besar F13 = F23 Langkah 2:
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2500 (10 x ) 2
=
50 (10 x)
=
900 x 2
30 x
50 x = 300 – 300 – 30 30 x 80 x = 300 x = 3,75 m Uji Kompetensi Kerjakan soal-soal berikut dengan benar!
1. Dua buah benda masing-masing massanya 10 kg dan 20 kg terpisahkan pada jarak 2 meter satu dengan yang lain. Tentukan gaya gravitasi antara kedua benda itu! -10
2. Gaya tarik gravitasi antara du buah benda bermassa adalah 2,001 x 10 N. Bila massa benda adalah 3 kg dan 9kg. Tentukanlah jarak antara kedua benda itu! 3. Massa sebesar 5 kg terpisah pada jarak 2 meter dari massa yang lain. Gaya gravitasi antara kedua benda adalah sebesar 2,5 x 10-10. Tentukan massa benda yang lain!
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8. Dua buah bermassa 2 kg dan 12,5 kg terpisah pada jarak 7 meter. Tentukanlah letak bola bermassa 6 kg sehingga gaya tarik gravitasi yang dialaminya sama dengan nol! 9. Dua buah benda bermassa pada saat terpisah sejauh 2 meter saling mengerjakan gaya sebesar 4 g. Bila jarak antaranya di jadikan 4 meter, tentukanlah gaya tarik menarik yang dikerjakan kedua benda itu! 10. Di titik A dan C dari suatu bujur sangkar ABCD ditempatkan massa sebesar 1 kg dan 0,5 kg. Bila gaya tarik menarik antara kedua massa tersebut besarnya 0,5 G newton, tentukanlah panjang sisi bujur sangkar tersebut.
Ulangan BAB 1 Soal-soal Pilihan Ganda
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2
a. 2,34 G m/s 2
2
d. 25 G m/s
2
b. 4 G m/s
e. 34,2 G m/s 2
c. 16 G m/s 4.
2
Jika percepatan gravitasi di permukaan planet 5 m/s , maka percepatan gravitasi pada ketinggian satu kali jari-jari planet dari permukaan planet adalah ... . 2
d. 5 m/s
2
e. 6,25 m/s
a. 1,25 m/s b. 2,25 m/s
2 2
2
c. 4,25 m/s 5.
Dua benda berada pada garis lurus dan berjarak 0,4 m. Jika massa 1 = 10 kg, kg, dan massa 2 = 40 kg, maka letak benda ketiga dengan massa 3 = 1 kg terhadap massa 1 adalah ... . a. 0,33 m
d. 0,133 m
b. 0,29 m
e. 0,003 m
c. 0,27 m 6.
Dua planet masing-masing bermassa m1 = 25 juta ton dan m2 = 400 juta ton, pada jarak 2 juta kilo meter. Letak suatu titik yang memiliki medan gravitasi nol adalah
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9.
Faktor-faktor yang mempengaruhi gaya tarik-menarik dua benda di angkasa adalah ... . (1) massa masing-masing benda (2) jenis dari masing-masing benda (3) jarak kedua benda (4) intensitas dari masing-masing benda Dari pernyataan di atas yang benar adalah ... . a. (1), (2), dan (3)
d. (4) saja
b. (1) dan (3)
e. semua benar
c. (2) dan (4) 10.
Jika suatu planet memiliki memiliki kala revolusi 8 tahun, maka jarak planet ke matahari adalah ... . a. ¼ sa
d. 4 sa
b. ½ sa
e. 8 sa
c. 2 sa 11.
Menurut hukum kedua Kepler, maka dalam waktu yang sama, garis garis khayal yang menghubungkan suatu planet dengan matahari dalam orbit planet, akan
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Soal-soal Uraian Jawablah dengan singkat dan jelas !
1.
Jika jarak dua planet dijadikan 2 kali semula, maka tentukan gaya gravitasi antara keduanya kini, relatif terhadap gaya gravitasinya semula!
2.
8
8
Dua planet masing-masing bermassa 4 . 10 kg dan 6 . 10 kg pada jarak 2.000 km. -11
2
2
Jika G = 6,672 . 10 N.m /kg , maka tentukan gaya gravitasi antara kedua planet! 3.
Jika terdapat sebuah meteor pada jarak 100.000 km dari pusat sebuah planet yang 10
-11
bermassa 5 . 10 kg dan G = 6,672 . 10
2
2
N.m /kg , maka tentukan medan
gravitasi yang dialami meteor tersebut! 4.
2
Jika medan gravitasi di permukaan bumi 10 m/s , maka tentukan medan gravitasi pada ketinggian 4 kali jari-jari bumi dari pusat bumi!
5.
10
8
6
Dua planet masing-masing bermassa 1.10 kg dan 4.10 kg pada jarak 5.10 km. 6
Tentukan letak sebuah planet bermassa 2.10 kg yang mengalami gaya gravitasi 8
nol di antara kedua planet, planet , relatif terhadap planet bermassa 4.10 kg! 6.
2
Jika medan gravitasi di bulan 1 m/s , maka tentukan medan gravitasi pada jarak satu kali jari-jari bulan dari permukaan bulan!
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BAB 2 IMPULS DAN MOMENTUM
KOMPETENSI DASAR
3.10 Menerapkan konsep momentum dan impuls, serta hukum kekekalan momentum dalam kehidupan sehari-hari 4.10 Menyajikan hasil pengujian penerapan hukum kekekalan momentum, misalnya bola jatuh bebas ke lantai dan roket sederhana
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Gerbang Tahukah Kamu bagaimana roket bisa meluncur meninggalkan bumi menuju ruang angkasa? Konsep fisika fisika apa saja yang berperan pada proses peluncuran roket? Untuk
menjawab
pertanyaan
diatas,
kalian
harus
memahami konsep Impuls dan momentum. Momentum adalah suatu kuantitas yang tersimpan. Bila dua objek bertumbukan, mementum tiap benda mungkin berubah, tetapi momentum sistem seluruhnya tetap konstan.
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Kerja Berpasangan. Kerjakan soal-soal berikut bersama teman terdekatmu!
1. Sebuah bola 800 gr ditendang dengan gaya 200 N. Sesaat setelah ditendang ditendang bola mempunyai kelajuan 144 km/jam.
Tentukan momentum bola sesaat sesaat setelah setelah
ditendang! 2. Sebuah benda bergerak dipercepat dari keadaan diam. Hubungan momentum dengan kelajuan benda dapat dibuat grafik di bawah ini : Ρ (kg m/s) x
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Contoh :
1. Sebuah bola massa 800 gram ditendang dengan gaya 400 N. Jika kaki dan bolah bersentuhan selama 0,5 sekon, tentukan Impuls pada peristiwa tersebut.
Diketahui: m = 0,8 kg F = 400 N ∆t = 0,5 S Ditanya :
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b) gaya penghambatan, c) lama waktunya (untuk perlambatan), d) impuls tumbukannya! Penyelesaian :
400 m/det V=0 0,1 m
VP
= 400 m/det
x
= 0,1 m
a) a b) F
= ....... ? = ....... ?
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a)
I
=
F dt
=
(103 t + 10) dt
=
10 . ½ t + 10 t
=
10 . ½ (0,1) + 10 . 0,1
3 3
2
2
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b. Gaya yang diberikan pemukul pada bola, jika bola bersentuhan dengan pemukul selama 20 ms.
Kerja Mandiri
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Contoh
peristiwa yang menunjukkan hukum kekekalan momentum adalah adanya
momentum suatu granat sebelum meledak sama dengan jumlah momentum seluruh pecahan granat setelah meledak.
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1. Seorang anak naik skate board yang massanya massanya 5 kg dengan kelajuan 5 m/s. Jika massa anak 25 kg, tentukan kecepatan skate board pada saat : a. orang melompat ke depan dengan kelajuan 2 m/s b. orang melompat ke belakang dengan kelajuan 2 m/s
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mobil-mobil sebelum bertabrakan, dengan menganggap tidak ada gesekan dengan tanah. Kecepatan benda-benda sesudah tumbukan adalah sama. Pada tumbukan tidak elastis berlaku momentum kekal dan energi kinetik tidak kekal.
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