To solve this problem we need to use the relationship between force and impulse, which is given by the following equation:

$\displaystyle F=\frac{\Delta P}{t}$

This equation represents that the rate of change of momentum with respect to time is equal to the net force that causes said change in momentum. Thus:

$\displaystyle F=\frac{P_f-P_i}{t}=\frac{mV_f-mV_i}{t}$

Note that Joe must have an initial velocity of $\displaystyle 0\frac{m}{s}$ before he begins to apply the upwards Force that accelerates him upwards, therefore our equation simplifies to:

$\displaystyle F=\frac{mV_f}{t}$

Solve for $\displaystyle V_f$:

$\displaystyle V_f=\frac{Ft}{m}=\frac{(100N)(0.9s)}{90kg}=\frac{90kg\cdot \frac{m}{s}}{90kg}=1\frac{m}{s}$

Say that, when we hit the ground, we have a velocity $\displaystyle v$, which is predetermined by whatever happens before the impact. When we hit the ground you will experience a force for some time. This force will cause the acceleration that reduces our velocity to zero and gets us to stop. Note that, regardless of how much time it takes us to stop, the change in momentum (impulse) is fixed, since it directly depends on how much our velocity changes:

$\displaystyle P_i=mv$

$\displaystyle P_f=\frac{0kg\cdot m}{s}$ (since we come to a stop)

Note that the initial momentum does not depend on the impact force nor on how much time it takes to stop. The initial momentum depends on the velocity we have when we first hit the ground. This velocity is given by whatever happened before we hit the ground, which no longer concerns us since we only care about what happens from the moment we first hit the ground till the moment we stop. Yes, the time that passes for you to stop is very small, but it is impossible for it to be zero. So we have that the change in momentum (impulse) is a constant:

$\displaystyle \Delta P=-mv= constant$, since $\displaystyle v$ is predetermined. 

Remember that any change in momentum for a given mass occurs because its velocity changes. The velocity of the mass changes due to an acceleration and an acceleration is caused by a force. This gives us a relationship between force and impulse:

$\displaystyle F=\Delt\frac{\Delta P}{t}$

In our scenario, $\displaystyle F$ would be the impact force that stops us and $\displaystyle t$ the time it takes us to stop. From the equation above, it is easy to see that, since $\displaystyle \Delta P$ is fixed, when $\displaystyle t$ gets larger $\displaystyle F$ gets smaller, and the other way around. Therefore, we bend our knees to effectively increase the time it takes us to stop. Thus, diminishing the impact force as to avoid hurting ourselves.

In order to solve this problem we need to use the relationship between force and impulse:

$\displaystyle F=\frac{\Delta P}{t}$ 

Since the ball is moving with a Velocity of $\displaystyle -50\frac{m}{s}$, we have that

$\displaystyle \Delta P=P_f-P_i=mV_f-mV_i$

$\displaystyle \Delta P=(0.5kg)(0 \frac{m}{s})-(0.5kg)(-50\frac{m}{s})=\frac{25kg\cdot m}{s}$

Note that the final velocity of the ball is zero since it comes to a stop. We want the force, $\displaystyle F$, experienced by the person's hand to be less than or equal to the maximum impact force

 $\displaystyle F_i_m_p=20,000N$

Mathematically:

$\displaystyle F\leq F_i_m_p$

$\displaystyle \frac{1}{F}\geq \frac{1}{F_i_m_p}$

Use the impulse equation and solve for time:

$\displaystyle t=\frac{\Delta P}{F}\geq \frac{\Delta P}{F_i_m_p}$

$\displaystyle t\geq \frac{\frac{25kg\cdot m}{s}}{20000N}=1.25*10^{-3}s$

Impulse is defined as the change in momentum. 

$\displaystyle J=\mid p_{final}-p_{initial}\mid$

Where $\displaystyle J$ is the impulse, $\displaystyle p_{final}$ is final momentum and $\displaystyle p_{initial}$ is the initial momentum. Recall:

$\displaystyle p=mv$

Where $\displaystyle m$ is mass and $\displaystyle v$ is velocity. 

Our bowling ball has a mass of 10kg, with initial velocity of $\displaystyle 2 \frac{m}{s}$ and final velocity of $\displaystyle 1 \frac{m}{s}$. 

$\displaystyle p_{final}=m*v_{final}=10kg*1\frac{m}{s}=10 N\cdot s$

$\displaystyle p_{initial}=m*v_{initial}=10kg*2\frac{m}{s}= 20 N\cdot s$

Plug in these two values into the equation for impulse and solve.

$\displaystyle J=\mid 20-10\mid N\cdot s= 10N\cdot s$

Impulse is the change in momentum. So all that is needed for this problem is to solve for the change in momentum. 

$\displaystyle \Delta P = (mv)_{final}-(mv)_{initial}$

Note!!!! momentum is a vector quantity. This means that we must accound for the change in the ball's direction. This can be done by defining one of the directions negative.

For convience, I'll define the initial velocity as negative. Plugging in numbers we get

$\displaystyle 100 \: N\cdot s$.

We never accounted for the time of the collision. We would have needed to do this if the problem asked for the force the bat applied. 

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

To solve this problem, we will use conservation of momentum.  The initial momentum of the system must be equal to the final momentum of the system if no external forces act on it.  It is important to note the directions and signs  of the velocities.  From this information, we may write:

$\displaystyle P_i = P_f\Rightarrow \frac{m_{bullet}\left( v_{bullet,i} -v_{bullet,f}\right )}{m_{block}}=v_{block,f}$

$\displaystyle v_{block,f}=\frac{.07kg}{7kg} \left( 400 \frac{m}{s}- (-300 \frac{m}{s})\right) = 7 \frac{m}{s}$

To solve this problem, we must first note that this is a collision problem.  Secondly, we must be careful with the signs of the velocities associated with each vehicle.  

$\displaystyle P_i=P_f\Rightarrow m_1v_1+m_2v_2=\left ( m_1+m_2 \right ) v_f$

Because of the parameters of the problem, it is easy to see that the total momentum of the system initially is equal to zero.  Therefore, the final momentum of the system must also be zero.

The key is that both carts will receive the same impulse ($\displaystyle F\,\Delta t$) from the spring, but in opposite directions. Since impulse is change in momentum, find the change in light cart's momentum:

$\displaystyle \Delta p=p_{final}-p_{initial}=mv_f-mv_i =1kg\times 5\frac{m}{s}-1kg\times2\frac{m}{s}=3\frac{kg\cdot m}{s}$

This is directed to the right. So the change in the momentum of the heavier cart is:

$\displaystyle \Delta p = -3\frac{kg\cdot m}{s}$, to the left. Solve for its final momentum:

$\displaystyle p_f=p_i+\Delta p=3kg\times2\frac{m}{s}-3\frac{kg \cdot m}{s}=3\frac{kg\cdot m}{s}$

Use this to find its velocity:

$\displaystyle p=mv\rightarrow 3\frac{kg\cdot m}{s}=3kg\times v$

$\displaystyle v=1\frac{m}{s}$ 

The positive sign means that it is moving to the right.

We could also use momentum conservation. Find total momentum before:

$\displaystyle p=mv=4kg\times 2\frac{m}{s}=8\frac{kg\cdot m}{s}$ 

The less massive cart has:

$\displaystyle p=mv=1kg\times5\frac{m}{s}=5\frac{kg\cdot m}{s}$ after the explosion, so the more massive cart has the other $\displaystyle 3\frac{kg\cdot m}{s}$

Due to conservation of momentum, the initial momentum must equal the final momentum of the system. Both billiard balls are of equal mass, and since the collision is elastic than the billiard balls will simply exchange momentum. This is a problem that is best to think about before starting to solve any equations because sometimes the correct answer is one you can deduce without any calculations. Therefore:

$\displaystyle v_{1f}=-2v$

$\displaystyle v_{2f}=v$

We begin by writing down the definition of an object's linear momentum $\displaystyle \vec{p}$

$\displaystyle \vec{p}=m\vec{v}$

$\displaystyle \vec p=30kg\left(5\hat{i}-2\hat{j} \right )\frac{m}{s}$

$\displaystyle \vec p=\left( 150\hat{i}-60\hat{j}\right )kg\frac{m}{s}$

We then find the magnitude of the momentum by taking the square root of the sum of squares of its components.

$\displaystyle \left| \vec{p}\right| = \sqrt{p_x^2+p_y^2}$

$\displaystyle \left|\vec p\right|=\sqrt{150^2+60^2}kg\frac{m}{s}=162kg\frac{m}{s}$