1. Jawaban: 24 m

1. Perpindahan = luas di bawah kurva selama $v>0$ (sampai $v=0$ di $t=6$ s).
2. Segitiga $[0,2]=\frac12(2)(6)=6$; persegi $[2,4]=6(2)=12$; segitiga $[4,6]=\frac12(2)(6)=6$.
3. Setelah $t=6$ s kecepatan negatif → perpindahan berkurang. Maksimum $=6+12+6=24$ m.

2. Jawaban: 36,87°

1. $H/R=\tan\theta/4 \Rightarrow \tan\theta=4H/R=4(15)/80=0{,}75$.
2. $\theta=\arctan 0{,}75 = 36{,}87^\circ$.

3. Jawaban: 13,17 m

1. Kecepatan: benda 1 $(v_1,-gt)$, benda 2 $(-v_2,-gt)$. Tegak lurus $\Rightarrow -v_1v_2+g^2t^2=0 \Rightarrow t=\sqrt{v_1v_2}/g=\sqrt{60}/10=0{,}775$ s.
2. Komponen vertikal sama → jarak $=(v_1+v_2)t=17(0{,}775)=13{,}17$ m.

4. Jawaban: 2

1. $a/b = 1/\sin(\pi/n)$ untuk $n$ partikel pada poligon beraturan.
2. Untuk $n=6$: $1/\sin 30^\circ = 2$.

5. Jawaban: 0,75

1. Di kerangka kereta ada gaya semu $ma$ horizontal. Kesetimbangan sepanjang bidang licin:
2. $ma\cos\alpha=mg\sin\alpha \Rightarrow a/g=\tan\alpha=(3/5)/(4/5)=0{,}75$.

6. Jawaban: 58,3°

1. $a=g(\sin\alpha-\mu\cos\alpha)$, jarak $s=\ell/\cos\alpha$; $t=\sqrt{2s/a}$ minimum $\Leftrightarrow \cos\alpha(\sin\alpha-\mu\cos\alpha)$ maksimum.
2. $\Rightarrow \tan 2\alpha=-1/\mu \Rightarrow \alpha=45^\circ+\tfrac12\arctan\mu = 45^\circ+\tfrac12\arctan 0{,}5 = 58{,}3^\circ$.

7. Jawaban: 23,1 m/s

1. $v_{maks}=\sqrt{gr(\tan\theta+\mu)/(1-\mu\tan\theta)}$.
2. $=\sqrt{(10)(60)(0{,}466+0{,}3)/(1-0{,}3\cdot0{,}466)}=\sqrt{534{,}6}=23{,}1$ m/s.

8. Jawaban: 0,2

1. Dengan $T_1=2T_2$, $a_2+a_3=-2a_1$: $m_1g-T_1=m_1a_1$; $m_2g-T_2=m_2a_2$; $m_3g-T_2=m_3a_3$.
2. Substitusi $m_1=4, m_2=1, m_3=2$ → $a_1/g = 1/5 = 0{,}2$ ($m_1$ turun).

9. Jawaban: 0,75 m/s²

1. $a=[F-mg\sin\alpha-\mu_k mg\cos\alpha]/m$.
2. $=[70-8(10)(0{,}6)-0{,}25(8)(10)(0{,}8)]/8=(70-48-16)/8=0{,}75$ m/s$^2$.

10. Jawaban: 25 J

1. $\vec{\Delta r}=(4\hat{i}+3\hat{j})$ m; $W=\vec{F}\cdot\vec{\Delta r}=(4)(4)+(3)(3)=25$ J.

11. Jawaban: 20 cm

1. Kubus bawah lepas saat pegas teregang $x_0=mg/k$.
2. Kekekalan energi: $\tfrac12 k\Delta\ell^2=mg(\Delta\ell+x_0)+\tfrac12 kx_0^2 \Rightarrow \Delta\ell=3mg/k=3(0{,}4)(10)/60=0{,}20$ m $=20$ cm.

12. Jawaban: 557,9 km

1. $\eta(-dM)=\mu Mg\,dx \Rightarrow x=\dfrac{\eta}{\mu g}\ln\dfrac{m+m_b}{m}$.
2. Dengan $m=4m_b$ (rasio $5/4$): $x=\dfrac{2{,}5\times10^6}{(0{,}1)(10)}\ln 1{,}25 = 557{,}9$ km.

13. Jawaban: 90 Ns

1. Aksi–reaksi sama besar → impuls tiap perahu sama: $p=F\Delta t=(60)(1{,}5)=90$ Ns.

14. Jawaban: 4 kg

1. Transfer energi maksimum lewat dua tumbukan elastik berurutan saat $m_2=\sqrt{m_1m_3}=\sqrt{8\cdot2}=4$ kg.

15. Jawaban: 9798 Ns

1. Laju di dasar $v=\sqrt{2g\sin\alpha\,\ell}=\sqrt{96}=9{,}80$ m/s.
2. Komponen momentum peluru sepanjang bidang $p\cos\alpha = Mv \Rightarrow p=Mv/\cos\alpha=800(9{,}80)/0{,}8=9798$ Ns.

16. Jawaban: 9,11 N·m

1. $\vec{\tau}=\vec{r}\times\vec{F}=(-7,\,3,\,5)$ N·m.
2. $|\vec{\tau}|=\sqrt{49+9+25}=\sqrt{83}=9{,}11$ N·m.

17. Jawaban: 60 N

1. Torsi terhadap tumpuan kiri: $F_{kanan}\cdot L = W\cdot 2L/3 \Rightarrow F_{kanan}=2W/3=60$ N (kiri menahan 30 N).

18. Jawaban: 1 rad/s

1. Presesi giroskop: $\Omega=\tau/(I\omega)$; $\tau=m_1g\ell+m_2g\ell/2=10+10=20$ N·m; $I\omega=4\cdot5=20$.
2. $\Omega=20/20=1$ rad/s.

19. Jawaban: 1,54 s

1. Silinder pejal dalam permukaan cekung: $\omega^2=\dfrac{2g}{3(R-r)}$.
2. $T=2\pi\sqrt{3(R-r)/(2g)}=2\pi\sqrt{3(0{,}4)/20}=2\pi\sqrt{0{,}06}=1{,}54$ s.

20. Jawaban: 43,87 N/m

1. Setimbang awal: pegas menahan $3mg$, $L_0=3mg/k$. Setelah putus, $m$ ber-SHM sekitar $L_{eq}=mg/k$, amplitudo $A=2mg/k$; capai langit-langit saat $\cos\omega t=-\tfrac12 \Rightarrow t_m=\dfrac{2\pi}{3}\sqrt{m/k}$.
2. $M$ jatuh bebas $t_M=\sqrt{2h/g}$. Samakan: $k=\dfrac{2\pi^2mg}{9h}=\dfrac{40\pi^2}{9}=43{,}87$ N/m.