textbf{Concept:}

The magnitude of each canoe from the example is $55,kg.m/s$ that is used with the definition of momentum to find the velocity of each canoe and then it is checked if the total velocity after and before push remains the same.

textbf{Solution:}
First the velocity of the canoe 1 is to be calculated

$$vec{p}=mvec{v}Rightarrowvec{v_1}=frac{vec{p_1}}{m_1}=frac{-55,kg.m/s}{130,kg}=boxed{-0.42,m/s}$$

Secondly the velocity of canoe2 is calculated

$$vec{p}=mvec{v}Rightarrowvec{v_2}=frac{vec{p_2}}{m_2}=frac{55,kg.m/s}{250,kg}=boxed{0.22,m/s}$$

Verify if the velocity is conserved

$$vec{V}_{total,,before}=vec{V}_{total,,after}$$

$$0=vec{V}_{1}+vec{V}_{2}$$
$$0=-0.42+0.22,m/s$$
$$boxed{bf color{blue}0ne-0.20,m/s}mbox{hence it is not conserved}$$

$$0ne-0.20,m/smbox{ hence not conserved}$$

textbf{Concept:}
Each canoe was initially at rest hence started with zero momentum. we need to find the change in moment with momentum-impulse theorem, $vec{F}Delta t=vec{p}_f-vec{v}_i$\

textbf{Solution:}
Solving for $vec{F}Delta t=vec{p}_f-vec{v}_i$ for $vec{p}_f$, substituting the numerical values for canoe 1 with $vec{p}_{1,i}=0$ and $vec{F}_1=-56N$ gives
$$vec{p}_{1,f}=vec{F}Delta t+vec{p}_{1,i}$$
$$=(-56,N)(1.3,s)+0$$
$$boxed{bf color{blue}vec{p}_{1,f}=-73,kg.m/s}mbox{ (-x direction)}$$

Now solve $vec{F}Delta t=vec{p}_f-vec{p_i} for vec{p}_f.$ Substituting values fo $vec{p}_{2,i}=0$ and $vec{F}_{2}=56,N,{+x, direction}$ gives

$$vec{p}_{2,f}=vec{F}Delta t+vec{p}_{2,i}$$
$$=(56,N)(1.3,s)+0$$
$$boxed{bf color{blue}vec{p}_{2,f}=73,kg.m/s}mbox{..(+x, direction)}$$

$$vec{p}_{1,f}=-73,kg.m/smbox{..(-x direction)}$$
$$vec{p}_{2,f}=73,kg.m/smbox{..(+x direction)}$$

$$vec{p}_{1,f}=-73,kg.m/smbox{..(-x direction)}$$
$$vec{p}_{2,f}=73,kg.m/smbox{..(+x direction)}$$

$textbf{Given: }$

The mass of the first skater is $m_{1} = 45 mathrm{~kg}$. The initial speed of the two skaters is equal to zero. The final speed of the first skater is $v_{f,1} = 0.62 mathrm{~m/s}$. The final speed of the second skater is $v_{f,2} = 0.89 mathrm{~m/s}$.

$textbf{Required: }$

Finding the mass of the second skater.

$textbf{Calculation: }$

According to the Momentum-Impulse Theorem:

$$
begin{align*}
vec{I} &= vec{F} ~ Delta t \
&= Delta vec{p} \
&= left( vec{p_{f}} – vec{p_{i}} right) \
&= left( m ~ vec{v_{f}} – m ~ vec{v_{i}} right) \
&= m ~ left( vec{v_{f}} – vec{v_{i}} right) \
end{align*}
$$

According to conservation law of momentum, the impulse is equal to zero.

$$
begin{align*}
Delta p_{t} &= Delta p_{1} + Delta p_{2} \
&= 0 \
&= m_{1} ~ left( vec{v_{f,1}} – vec{v_{i,1}} right) + m_{2} ~ left( vec{v_{f,2}} – vec{v_{i,2}} right) \
&= m_{1} ~ left( vec{v_{f,1}} – 0right) + m_{2} ~ left( vec{v_{f,2}} – 0 right) \
end{align*}
$$

Rearrange and solve for the mass of the second skater:

$$
begin{align*}
m_{2} &= – dfrac{ m_{1} ~ v_{f,1} }{ v_{f,2} } \
&= – dfrac{ 45 mathrm{~kg} times 0.62 mathrm{~m/s} }{ – 0.89 mathrm{~m/s} } \
&= 31.348 mathrm{~kg}
end{align*}
$$

So, the mass of the second skater is $31.348 mathrm{~kg}$.

The mass of the second skater is $31.348 mathrm{~kg}$.

$$
textbf{Concept:}
$$

The total momentum of the astronaut and the satellite should remain zero because there is no friction just like it was zero in the start. The speed of the astronaut will be determined using the law of conservation of momentum and will be multiplied by time to find the distance.

$$
textbf{Solution:}
$$

Assuming that motion of satellite is in negative x-direction

$$
p_a+p_s=0=m_a v_a+m_s v_s
$$

$$
v_a=frac{-m_sv_s}{m_a}t=-frac{(1200,kg)(-0.14,m/s)}{(92,kg)}
$$

$$
boxed{bf color{#4257b2}v_a=-1.826,m/s}
$$

Now find the distance to the shuttle

$$
d=v_a t = -frac{-m_sv_s}{m_a}t=-(1.826,m/s)(7.5,s)
$$

$$
d=boxed{bf color{#4257b2}14,m}
$$

$$
d=14,m
$$