Since the collision is completely elastic, we know that both momentum and kinetic energy are conserved. We can write the following equations (initial momentum and energy of the second ball are neglected since it is not moving:

\(\displaystyle m_1v_i=m_1v_f +m_2v_2\)

\(\displaystyle \frac{1}{2}m_1v_i^2= \frac{1}{2}m_1v_f^2+\frac{1}{2}m_2v_2^2\)

Rearrange the first equation for \(\displaystyle m_2\) and the second for \(\displaystyle v_2\).

\(\displaystyle m_2=\frac{m_1(v_i-v_f)}{v_2}\)

\(\displaystyle v_2^2=\frac{m_1(v_i^2-v_f^2)}{m_2}\)

We can rewrite the second equation as:

\(\displaystyle v_2^2=\frac{m_1(v_i+v_f)(v_i-v_f)}{m_2}\)

Substitute our equation for \(\displaystyle m_2\) into the second equation:

\(\displaystyle v_2^2=\frac{m_1(v_i-v_f)(v_i+v_f)}{\frac{m_1(v_i-v_f)}{v_2}}\)

Rearranging, we get:

\(\displaystyle v_2^2=\frac{m_1(v_i-v_f)(v_i+v_f)}{m_1(v_i-v_f)v_2^{-1}}=\frac{(v_i+v_f)}{v_2^{-1}}\)

\(\displaystyle v_2^2=v_2(v_i+v_f)\)

\(\displaystyle v_2 = (v_i +v_f)\)

Plug in our values for the initial and final velocities:

\(\displaystyle v_2=(2\frac{m}{s}+0.8\frac{m}{s})\)

\(\displaystyle v_2=2.8\frac{m}{s}\)

To solve for mass, we'll use our earlier expression for \(\displaystyle m_2\):

\(\displaystyle m_2=\frac{m_1(v_i-v_f)}{v_2}\)

\(\displaystyle m_2= \frac{(0.1kg)(2\frac{m}{s}-0.8\frac{m}{s})}{2.8\frac{m}{s}}= 0.043kg\)

It does not matter whether the collision is elastic or inelastic (although it would be best to assume that it's inelastic). Momentum is conserved in either type of collision, and is the only value needed for our calculation. Since they come to a standstill, their momentums at the moment of collision are equal and opposite:

\(\displaystyle m_1v_1=m_2v_2\)

Rearrange to solve for \(\displaystyle m_1\):

\(\displaystyle m_1=\frac{m_2v_2}{v_1}\)

Plug in the given values from the question and solve:

\(\displaystyle m_1=\frac{(200kg)(2.3\frac{m}{s})}{2.9\frac{m}{s}}\)

\(\displaystyle m_1= 159 kg\)

Since the collision is completely inelastic, momentum is conserved but energy is not. Furthermore, the two cars stick to each other and travel as one. The equation for conservation of momentum is as follows:

\(\displaystyle p_i = p_f\)

There are two inital masses with different velocities and one final mass with a single velocity. Therefore, we can write:

\(\displaystyle m_1v_1 + m_2v_2 = mv\)

Rearranging for final velocity, we get:

\(\displaystyle v = \frac{m_1v_1 + m_2v_2}{m}\)

At this point, we can denote which direction is positive and which is negative. Since the car traveling west has more momentum, we will consider west to be positive. Substituting our values into the equation, we get:

\(\displaystyle v = \frac{(400)(-20)+ (800)(15)}{400 + 800}\)

\(\displaystyle v = \frac{-800 + 12000}{1200} = \frac{4000}{1200} = 3.33\frac{m}{s}\)

Since this value is positive, the final answer is \(\displaystyle 3.33\frac{m}{s}\) West.

Impulse can be written as either of two popular expressions:

\(\displaystyle I = F\Delta t = m\Delta v\)

From the problem statement, we can determine the velocity of the marble as it hits the floor, allowing us to use the latter expression. To determining the velocity of the marble, we can use the equation for conservation of energy:

\(\displaystyle E=U_i+K_i = U_f + K_f\)

Assuming the final height is zero, we can eliminate initial kinetic energy and final potential energy. Therefore, we can write:

\(\displaystyle mgh_i = \frac{1}{2}mv_f^2\)

Canceling out mass and rearranging for final velocity, we get:

\(\displaystyle v_f = \sqrt{2gh_i}\)

We know these variables, allowing us to solve for the velocity:

\(\displaystyle v_f = \sqrt{2(10\frac{m}{s^2})(1.2m)} = 4.9\frac{m}{s}\)

Plugging this value into the expression for impulse, we get:

\(\displaystyle I = (0.1kg)(4.9\frac{m}{s})= 0.49 N\cdot s\)

To calculate the momentum of the block, we first need to know the velocity of the block. This can be found using the equation for the conservation of momentum:

\(\displaystyle E = U_i+K_i = U_f +K_f\)

If we assume that the final height is zero, we can eliminate initial kinetic energy and final potential energy, getting:

\(\displaystyle U_i = K_f\)

Substituting expressions for each term, we get:

\(\displaystyle mgh_i = \frac{1}{2}mv_f^2\)

Cancel out mass and rearrange to solve for velocity:

\(\displaystyle v_f = \sqrt{2gh_i} = \sqrt{2(10\frac{m}{s^2})(h_i)}\)

We can use the horizontal distance traveled and the angle of the slope to determine the initial height:

\(\displaystyle tan(A) = \frac{h_i}{horizontal}\)

\(\displaystyle h_i = tan(A)\cdot horizontal\)

\(\displaystyle h_i = tan(50^{\circ})*10m = 11.9m\)

Now that we have the initial height, we can solve for final velocity:

\(\displaystyle v_f = \sqrt{2(10\frac{m}{s^2})(11.9m)} = 15.4\frac{m}{s}\)

Finally, we can now use the equation for momentum to solve the problem:

\(\displaystyle p = m\Delta v\)

\(\displaystyle p = (20kg)(15.4\frac{m}{s})\)

\(\displaystyle p = 309 \frac{kg\cdot m}{s}\)

Momentum is always conserved. Equation for conservation of momentum: 

\(\displaystyle m_{1}v_{1}+m_{2}v_{2}=(m_{1}+m_{2})v_{total}\)

There is only one velocity on the right since the two astronauts grab onto each other, thus they move together at the same velocity. Solve.

\(\displaystyle (70kg\cdot15\frac{m}{s})+(60kg(-5\frac{m}{s}))=(70kg+60kg)v_{total}\)

\(\displaystyle v_{total}=5.77\frac{m}{s}\)

The equation for momentum is:

\(\displaystyle p=mv\)

To maintain conservation of momentum, a new state must have the same momentum as a previous state:

\(\displaystyle m_{1i}v_{1i}+m_{2i}v_{2i}=m_{1f}v_{1f}+m_{2f}v_{2f}\)

Since the rock and the bean bag move together after the collision, \(\displaystyle v_{1f}=v_{2f}\). And since the bean bag is initially stationary, \(\displaystyle v_{2i}=0\)

Plug in known values and solve.

\(\displaystyle 5kg(12\frac{m}{s})+20kg(0)=(5kg+20kg)(v_f)\)

\(\displaystyle 60\frac{kgm}{s}=25kg(v_f)\)

\(\displaystyle v_f=2.4\frac{m}{s}\)

To solve this problem, we need to consider the change in the ball's momentum. To do so, we'll use the following equation.

\(\displaystyle J=m\Delta v=F_{avg}t\)

Rearrange the above equation to solve for the average force.

\(\displaystyle F_{avg}=\frac{m\Delta v}{t}=\frac{(0.15 kg)(40 \frac{m}{s})}{0.7 s}\)

\(\displaystyle F_{avg}=8.57N\)

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.