a) $\Delta h= 1-\cos50^{o}=0.357\mathrm{m}$

$KE=-mg\Delta h=0.1\times 9.8\times0.357=0.35 \mathrm{J}$

b) All of the kinetic energy is transferred to the second ball so

$KE=0.35\mathrm{J}$

c) KE at top of ramp is $0.35-0.1\times9.8\times0.1=0.252\mathrm{J}$

For a rolling ball $KE=\frac{1}{2}m(1+\frac{2}{5})v^{2}$ so

$v^{2}=\frac{0.252\times 2\times 5}{0.1\times7}=3.6$

$v=1.9\mathrm{m\,s^{-1}}$

d) $L=I\omega$

$I=\frac{2}{5}mr^{2}=\frac{2}{5}\times0.1\times.02^{2}=0.000016\mathrm{kg\,m^{2}}$

$\omega=\frac{v}{r}=\frac{1.9}{0.02}=95\mathrm{s^{-1}}$

$L=0.00152\mathrm{kg\,m^{2}\,s^{-1}}$ in to the page

e) First we need to find the time it takes until the ball hits the ground

$0=0.1+1.9\sin 15^{o}t-\frac{1}{2}gt^{2}$

$4.9t^{2}-0.49t-0.1=0$

$t=0.2 \mathrm{s}$

$x=1.9\times\cos 15^{o}t$

$x=0.37\mathrm{m}$

a) $I=2\times2\times1.2^{2}+2\times3\times(\frac{1}{12}\times1.1^{2}+0.65^{2})+\frac{1}{2}\times50\times0.1^2$

$I=5.76+3.14+0.25=9.15\,\mathrm{kg\,m^{2}}$

b) $\sum \tau =2 \times 100 \times 1.2 =240\,\mathrm{N\,m}$

$\alpha=\frac{\sum \tau}{I}=\frac{240}{9.67}=26.23\,\mathrm{s^{-2}}$

c)$\omega=24.82\times0.5=13.115\mathrm{s^{-1}}=125.24\mathrm{rpm}$

d)$L=I\omega=9.15\times13.115=120\,\mathrm{kg\,m^{2}\,s^{-1}}$

e)$I=2\times5\times0.1^{2}+\frac{1}{2}\times50\times0.1^{2}=0.35\,\mathrm{kg\,m^{2}}$

f) Angular momentum is conserved so

$\omega=\frac{120}{0.35}=342.9\mathrm{s^{-1}}=3274\mathrm{rpm}$

g) $v=\omega r=342.9\times0.1=34.29\,\mathrm{m\,s^{-1}}$

a) Torques around hinge must balance so

$8\times0.25=2m\times 0.5$

$m=\frac{8\times0.25}{2\times0.5}=2\mathrm{kg}$

b) Balance of vertical forces

$8g-4g-F_{Hy}=0$

$F_{Hy}=4g=39.2\mathrm{N}$ up

c) Balance of horizontal forces

$F_{Hx}=0\mathrm{N}$

d) $I=\frac{1}{3}\times8\times0.5^{2}=0.67\,\mathrm{kg\,m^{2}}$

e) For the still attached mass

$2a=T_{2}-2g$

For the pulley

$\frac{1}{2}\times2\times0.15^2\times \frac{a}{0.15}=(T_{1}-T_{2})0.15$

$T_{1}-T_{2}=a$

For the arm

$0.67\times \frac{a}{0.5}=8\times9.8\times0.25-0.5T_{1}$

$2.68a=39.2-T_{1}$

Combining the first two equations

$3a=T_{1}-2g$

And then using it with the third

$2.68a=39.2-3a-19.6$

$5.68a=19.6$

$a=3.45\,\mathrm{ms^{-2}}$ up

e) $T_{1}=3\times3.45+19.6=29.95\,\mathrm{N}$

$T_{2}=T_{1}-a=29.95-3.45=26.5\,\mathrm{N}$