$\displaystyle P_{total}=P_1+P_2.......$

The train and car are our only two objects in the system.

The initial momentum of the car is zero.

So the only momentum that will contribute is that of the train.

$\displaystyle p=mv$

$\displaystyle p=1.8\times10^7\cdot15$

Plugging in our values, we get $\displaystyle 2.7\times10^8 kg\frac{m}{s}$

We will need to use conservation of momentum to solve this problem.

$\displaystyle M_1V_1+M_2V_2=M_{final}V_{final}$

Where $\displaystyle M_1$ and $\displaystyle V_1$ refer to the train, and $\displaystyle M_2$ and $\displaystyle V_2$ refer to the car. $\displaystyle M_{final}$and $\displaystyle V_{final}$ refer to the final state of both the train and the car.

Rearranging using algebra......

$\displaystyle \frac{M_1V_1+M_2V_2}{M_{final}}=V_{final}$

$\displaystyle \frac{1.8\times10^7\cdot15+1300\cdot0}{1.8\times10^7}=V_{final}$

Plugging in our values, we get $\displaystyle 15 \frac{m}{s}$.

We will need to use conservation of momentum to solve this problem.

$\displaystyle M_1V_1+M_2V_2=M_{final}V_{final}$

Where $\displaystyle M_1$ and $\displaystyle V_1$ refer to the train, and $\displaystyle M_2$ and $\displaystyle V_2$ refer to the car. $\displaystyle M_{final}$and $\displaystyle V_{final}$ refer to the final state of both the train and the car.

Rearranging using algebra......

$\displaystyle \frac{M_1V_1+M_2V_2}{M_{final}}=V_{final}$

$\displaystyle \frac{1.8\times10^7\cdot15+1300\cdot0}{1.8\times10^7}=V_{final}$

Plugging in our values, we get $\displaystyle 15 \frac{m}{s}$.

We will need to use conservation of momentum to solve this problem.

$\displaystyle M_1V_1+M_2V_2=M_{final}V_{final}$

Where $\displaystyle M_1$ and $\displaystyle V_1$ refer to the train, and $\displaystyle M_2$ and $\displaystyle V_2$ refer to the car. $\displaystyle M_{final}$and $\displaystyle V_{final}$ refer to the final state of both the train and the car.

Rearranging using algebra......

$\displaystyle \frac{M_1V_1+M_2V_2}{M_{final}}=V_{final}$

$\displaystyle \frac{1.8\times10^7\cdot15+1300\cdot0}{1.8\times10^7}=V_{final}$

Plugging in our values, we get $\displaystyle 15 \frac{m}{s}$.

Then, we will need to find out final momentum of the car.

$\displaystyle M_{car}V_{final}=P_{final}$

$\displaystyle 1300kg\cdot15\frac{m}{s}=P_{final}$

$\displaystyle M_{car}V_{final}=P_{final}$

 $\displaystyle \Delta P=P_{final}-P_{initial}$

Since our intial momentum of the car was $\displaystyle 0$, our change in momentum will be equal to the $\displaystyle P_{final}$.

 $\displaystyle \Delta P=P_{final}$

We will use the definition of impulse, which is the change in momentum:

$\displaystyle F\cdot \Delta t= \Delta P$

We will use substitution:

$\displaystyle F\cdot \Delta t= P_{final}$

Plugging in our values, we get $\displaystyle 1.95\times10^6 N$

Determine the change in the velocity of the car:

$\displaystyle 35\frac{m}{s}-10\frac{m}{s}=25\frac{m}{s}$

Calculate the average acceleration:

$\displaystyle \frac{25\frac{m}{s}}{2 s}=\frac{\Delta v}{\Delta s}=a$

$\displaystyle 12.5\frac{m}{s^2}=a$

Use the definition of force:

$\displaystyle F=ma$

$\displaystyle F=1110kg*12.5\frac{m}{s^2}$

$\displaystyle F=1.4*10^4N$

First, break the airborne time into two pieces, the ascent and descent.

Ascent:

The ball will need to decelerate from $\displaystyle 15\frac{m}{s}$ to $\displaystyle 0 \frac{m}{s}$.

Use acceleration due to gravity.

$\displaystyle 15\frac{m}{s}-9.8\frac{m}{s^2}*\Delta t=0\frac{m}{s}$

Solve for $\displaystyle \Delta t$

$\displaystyle \Delta t=1.5s$

Descent:

Due to parabolic motion, the ball will have the same magnitude of velocity when it returns to it's height as when it launched, albeit in the opposite direction.

$\displaystyle 0\frac{m}{s}-9.8\frac{m}{s^2}*\Delta t=-15\frac{m}{s}$

Solving for $\displaystyle \Delta t$

$\displaystyle \Delta t=1.5s$

Add the times together to get the total airtime.

$\displaystyle \Delta t_{airtime}=3.0s$

First, break the airborne time into two pieces, the ascent and descent.

Ascent:

The ball will need to decelerate from $\displaystyle 15\frac{m}{s}$ to $\displaystyle 0 \frac{m}{s}$.

Use acceleration due to gravity.

$\displaystyle 15\frac{m}{s}-9.8\frac{m}{s^2}*\Delta t=0\frac{m}{s}$

Solve for $\displaystyle \Delta t$

$\displaystyle \Delta t=1.5s$

Descent:

Due to parabolic motion, the ball will have the same magnitude of velocity when it returns to it's height as when it launched, albeit in the opposite direction.

$\displaystyle 0\frac{m}{s}-9.8\frac{m}{s^2}*\Delta t=-15\frac{m}{s}$

Solve for $\displaystyle \Delta t$

$\displaystyle \Delta t=1.5s$

Add the times together.

$\displaystyle \Delta t_{airtime}=3.0s$

Multiply the airtime by the velocity of the cart.

$\displaystyle 3.0s*\frac{3m}{s}=d$

$\displaystyle 9m=d$

Using conservation of momentum:

$\displaystyle P_i=P_f$

The initial momentum can be taken as zero from our reference frame.

$\displaystyle 0\frac{kg*m}{s}=P_{astro}+P_{shotput}$

Plug in values:

$\displaystyle 0\frac{kg*m}{s}=100kg*v_a+8kg*-4\frac{m}{s}$

Note: The sign of the velocity for the shot put is negative because it will be in the opposite direction as the astronaut. The question is asking for magnitude though, so we know it will be positive regardless of the direction since magnitude does not take into account the direction of motion.

$\displaystyle v_a=.32\frac{m}{s}$

We use the equation: 

$\displaystyle \Delta p = F \Delta t$ 

Where the change in momentum is $\displaystyle \Delta p$.

The force, $\displaystyle F$, is $\displaystyle 100N$ and t is the time during which the force is exerted ($\displaystyle \Delta t=0.1s$) as given by the problem statement, we plug in to the formula:

$\displaystyle \Delta p=(100N)*(0.1s)=10Ns$

Use conservation of momentum

$\displaystyle P_i=P_f$

$\displaystyle m_{1i}v_{1i}+m_{2i}v_{2i}=m_{1,2f}v_{1,2f}$

Plug in values:

$\displaystyle (167*10^3)*4+(67*10^3)*0=(234*10^3)v_{1,2f}$

Solve for $\displaystyle v_{1,2f}$:

$\displaystyle v_{1,2f}=\frac{167*10^3*4}{234*10^3}$

$\displaystyle v_{1,2f}=2.85\frac{km}{hr}$

Definition of impulse:

$\displaystyle J=\Delta P=m(v_f-v_i)$

Convert $\displaystyle \frac{km}{hr}$ to $\displaystyle \frac{m}{s}$

$\displaystyle 2.85\frac{km}{hr}*\frac{1hr}{3600s}*\frac{1000m}{1km}=.79\frac{m}{s}$

Plug in values:

$\displaystyle J=\Delta P=67*10^3(.79-0)$

$\displaystyle J=52.3*10^3\frac{kg*m}{s}$