$$
I=sum_i^nm_ir_i^2
$$

Angular momentum has to be conserved if there is no external torque.

And to find out if angular velocity changes we can use equation for conservation of angular momentum:

$$
L=Iomega
$$

In this case there is no external forces, that means that angular momentum is conserved and we can conclude that angular momentum $textbf{stays constant}$.

We know that angular momentum is constant and that moment of inertia decrease, hence angular velocity $textbf{increase}$.

Answer is $textbf{B}$.

$$
begin{align*}
m_1v_1+m_2v_2=m_1v_1’+m_2v_2’\
end{align*}
$$

$$
begin{align*}
40cdot2+10cdot0&=40v’+10v’\
v’&=frac{40cdot2}{40+10}
end{align*}
$$

$$
boxed{v’=1.6,,rm m/s}
$$

The speed of the ice-skater and the sled after they collide is 1.6 m/s and the answer is $textbf{C}$.

$$
Delta L=L_2-L_1
$$

$$
Delta L=3.5-7
$$

$$
boxed{Delta L=-3.5,,rm kgm^2/s}
$$

The angular impulse on each wheel is $-3.5,,rm kgm^2/s$, the answer is $textbf{D}$.

$$
begin{align*}
m_1v_1+m_2v_2=m_1v_1’+m_2v_2’\
end{align*}
$$

$$
begin{align*}
45cdot0+5v_2&=45cdot0.5+5cdot0.5\
v_2&=frac{45cdot0.5+5cdot0.5}{5}
end{align*}
$$

$$
boxed{v_2=5,,rm m/s}
$$

The speed of the ball at the moment just before the skater caught it is 5 m/s. The answer is $textbf{D}$.

$$
p_{t}-p_{r}=m_tv_t-m_rv_r
$$

$$
p_t-p_r=3cdot10^{3}cdot1-50cdot3
$$

$$
boxed{p_t-p_r=2850,,rm kgm/s}
$$

The difference in momentum is $2850,,rm kgm/s$, the answer is $textbf{C}$.

$$
L=Iomega
$$

we know moment of inertia, so we just put that in, but in this case we have to put angular velocity in terms of linear speed because linear velocity is the same for both gears but in opposite direction.

$$
L=frac{1}{2}mr^2cdotfrac{v}{r}
$$

We can calculate ration with equaiton:

$$
begin{align*}
frac{L_l}{L_s}&=frac{frac{1}{2}m_lr_lv_l}{frac{1}{2}m_sr_sv_s}\
frac{L_l}{L_s}&=frac{m_lr_lv_l}{m_sr_sv_s}\
end{align*}
$$

$$
frac{L_l}{L_s}=frac{4m_scdot2r_scdot1v_s}{m_sr_sv_s}
$$

$$
L_l=-8L_s
$$

The answer is $textbf{C}$.

$$
Ft=mvrightarrow m=frac{Ft}{v}
$$

$$
m=frac{0.8}{4}
$$

$$
boxed{m=0.2,,rm kg}
$$

The answer is $textbf{A}$.

$$
Ft=p_2-p_1=-mv
$$

$$
Ft=-12cdot20
$$

$$
boxed{Ft=-240,,rm kgm/s}
$$