This is an example of an inelastic collision, as the two cars stick together after colliding. We can assume momentum is conserved.

To make the equation easier, let's call the first car "1" and the second car "2."

Using conservation of momentum and the equation for momentum, \(\displaystyle p=mv\), we can set up the following equation.

\(\displaystyle m_1v_{1initial}+m_2v_{2initial}=(m_1+m_2)v_{final}\)

Since the cars stick together, they will have the same final velocity. We know the second car starts at rest, and the velocity of the first car is given. Plug in these values and solve for the final velocity.

\(\displaystyle (2000kg*25\frac{m}{s})+(2000kg*0\frac{m}{s})=(2000kg+2000kg)v_{final}\)

\(\displaystyle 50000\frac{kg*m}{s}=(4000kg)v_{final}\)

\(\displaystyle \frac{50000\frac{kg*m}{s}}{4000kg}=v_{final}\)

\(\displaystyle 12.5\frac{m}{s}=v_{final}\)

This is an example of an inelastic collision, as the two cars stick together after colliding. We can assume momentum is conserved.

To make the equation easier, let's call the first car "1" and the second car "2."

Using conservation of momentum and the equation for momentum, \(\displaystyle p=mv\), we can set up the following equation.

\(\displaystyle m_1v_{1initial}+m_2v_{2initial}=(m_1+m_2)v_{final}\)

Since the cars stick together, they will have the same final velocity. We know the second car starts at rest, and the final velocity is given. Plug in these values and solve for the initial velocity of the first car.

\(\displaystyle (900kg*v_{1i})+(1000kg*0\frac{m}{s})=(900kg+1000kg)*18\frac{m}{s}\)

\(\displaystyle 900kg*v_{1i}=(1900kg)*18\frac{m}{s}\)

\(\displaystyle 900kg*v_{1i}=34,200\frac{kg*m}{s}\)

\(\displaystyle v_{1i}=\frac{34,200\frac{kg*m}{s}}{900kg}\)

\(\displaystyle v_{1i}=38\frac{m}{s}\)

We can use the law of conservation of momentum:

\(\displaystyle m_1v_{1initial}+m_2v_{2initial}=m_1v_{1final}+m_2v_{2final}\)

We know the mass of each ball and their initial velocities.

\(\displaystyle (10kg*30\frac{m}{s})+(12kg*0\frac{m}{s})=m_1v_{1final}+m_2v_{2final}\)

We also know the final velocity of the first ball. This leaves only one variable: the final velocity of the second ball.

\(\displaystyle (10kg*30\frac{m}{s})+(12kg*0\frac{m}{s})=(10kg*13\frac{m}{s})+(12kg*v_{2f})\)

Solve to isolate the variable.

\(\displaystyle 300\frac{kg*m}{s}+0=(130\frac{kg*m}{s})+(12kg*v_{2f})\)

\(\displaystyle {300\frac{kg*m}{s}}-{130\frac{kg*m}{s}}=12kg*v_{2f}\)

\(\displaystyle {170\frac{kg*m}{s}}=12kg*v_{2f}\)

\(\displaystyle \frac{170\frac{kg*m}{s}}{12kg}=v_{2f}\)

\(\displaystyle 14.2\frac{m}{s}=v_{2f}\)

This is an example of an elastic collision. We start with two masses and end with two masses with no loss of energy. 

We can use the law of conservation of momentum to equate the initial and final terms.

\(\displaystyle m_1v_{1initial}+m_2v_{2initial}=m_1v_{1final}+m_2v_{2final}\)

Plug in the given values and solve for the mass of the second ball.

\(\displaystyle (11kg*33\frac{m}{s})+(m_2*0\frac{m}{s})=(11kg*13\frac{m}{s})+(m_2*8\frac{m}{s})\)

\(\displaystyle (363\frac{kg*m}{s})+(0)=(143\frac{kg*m}{s})+(m_2*8\frac{m}{s})\)

\(\displaystyle (363\frac{kg*m}{s})-(143\frac{kg*m}{s})=(m_2*8\frac{m}{s})\)

\(\displaystyle (220\frac{kg*m}{s})=(m_2*8\frac{m}{s})\)

\(\displaystyle \frac{220\frac{kg*m}{s}}{8\frac{m}{s}}=m_2\)

\(\displaystyle 27.5kg=m_2\)

This is an example of an elastic collision. We start with two masses and end with two masses with no loss of energy. 

We can use the law of conservation of momentum to equate the initial and final terms.

\(\displaystyle m_1v_{1initial}+m_2v_{2initial}=m_1v_{1final}+m_2v_{2final}\)

Plug in the given values and solve for the initial velocity of the first ball.

\(\displaystyle (0.5kg*v_{1i})+(1.5kg*0\frac{m}{s})=(0.5kg*13\frac{m}{s})+(1.5kg*8\frac{m}{s})\)

\(\displaystyle (0.5kg*v_{1i})+(0)=(6.5\frac{kg*m}{s} )+(12\frac{kg*m}{s})\)

\(\displaystyle (0.5kg*v_{1i})=(18.5\frac{kg*m}{s} )\)

\(\displaystyle v_{1i}=\frac{18.5\frac{kg*m}{s}}{0.5kg}\)

\(\displaystyle v_{1i}=37\frac{m}{s}\)

We know that the cars stick together after the collision, which means that the final velocity will be the same for both of them. Using the formula for conservation of momentum, we can start to set up an equation to solve this problem.

\(\displaystyle p_1=p_2\)

\(\displaystyle p=mv\)

First, we will write the initial momentum.

\(\displaystyle p_1=m_1v_1+m_2v_2\)

We know that the second car starts at rest, so this equation can be simplified.

\(\displaystyle p_1=m_1v_1+m_2(0\frac{m}{}s)=m_1v_1\)

Now we will write out the final momentum. Keep in mind that both cars will have the same velocity!

\(\displaystyle p_2=m_1v_3+m_2v_3\)

\(\displaystyle p_2=v_3(m_1+m_2)\)

Set these equations equal to each other and solve to isolate the final velocity.

\(\displaystyle p_1=p_2\)

\(\displaystyle m_1v_1=v_3(m_1+m_2)\)

\(\displaystyle \frac{m_1v_1}{m_1+m_2}=v_3\)

This is our answer, in terms of the given variables.

Consider the law of conservation of momentum.

\(\displaystyle m_{1}v_{i1}+m_{2}v_{i2}=m_{1}f1+m_{2}v_{f2}\)

If both balls are identical then we can say that

\(\displaystyle m_{1}=m_{2}=M\)

Therefore we can state the equation as

\(\displaystyle Mv_{i1}+Mv_{i2}=Mf1+Mv_{f2}\)

Since \(\displaystyle M\) is the common factor we can remove it from the equation completely.

\(\displaystyle v_{i1+v_{i2}}=f1+v_{f2}\)

Since mass factors out of the equation; then it does not matter if the balls increase or decrease in their mass.

A bouncy ball will have an elastic collision with the door, causing the ball to move backward at the same speed it hit the door.

On the other hand, the blob of clay will have an inelastic collision with the door, causing the blob of clay to move with the same speed as the door.

So let us look at the conservation of momentum for an elastic collision.

\(\displaystyle m_{1}v_{i1}+m_{2}v_{i2}=m_{1}v_{f1}+m_{2}v_{f2}\)

Since the door has no initial velocity we can remove it from the beginning of the equation.

\(\displaystyle m_{1}v_{i1}=m_{1}v_{f1}+m_{2}v_{f2}\)

Since the ball will rebound with the same amount that it hits the door the velocity at the end is a negative of the velocity at the beginning.

\(\displaystyle m_{1}v_{i1}=m_{1}(-v_{i1})+m_{2}v_{f2}\)

Rearrange for the final velocity of the door.

\(\displaystyle m_{1}v_{i1}+m_{1}(v_{i1})=m_{2}v_{f2}\)

\(\displaystyle 2m_{1}v_{i1}=m_{2}v_{f2}\)

\(\displaystyle \frac{2m_{1}v_{i1}}{m_{2}}=v_{f2}\)

Now let us examine the law of conservation of momentum for the inelastic collision.  Again the door has no initial velocity so we can remove it from the beginning of the equation.

\(\displaystyle m_{1}v_{i1}=m_{1}v_{f1}+m_{2}v_{f2}\)

Since the collision is inelastic, the final velocity of both objects will be the same so we can set them equal to each other.

\(\displaystyle m_{1}v_{i1}=m_{1}v_{f2}+m_{2}v_{f2}\)

Now solve for the final velocity of the door.

\(\displaystyle m_{1}v_{i1}=(m_{1}+m_{2})v_{f2}\)

When we compare these two final velocities, it is clear that the elastic collision will create a larger velocity for the door because the top number is twice the value as the inelastic collision, and it is being divided by only the mass of the second object, instead of both objects combined.

Knowns

\(\displaystyle v_{i}=7.5m/s\)

\(\displaystyle v_{f}=0m/s\)

\(\displaystyle m=12kg\)

\(\displaystyle t=8ms\)

First, let us find the change in the momentum of an object.

\(\displaystyle \delta p=m(v_{f}-v_{i})\)

\(\displaystyle \delta p=(12kg)(0m/s-7.5m/s)\)

\(\displaystyle \delta p=90kgm/s\)

We know that the change in momentum of an object is equal to the impulse of an object.  Impulse is equal to the force acting on the object multiplied by the time that force is applied.

\(\displaystyle \delta p=I=F\delta t\)

\(\displaystyle 90kgm/s=F\delta t\)

We need to convert our time from milliseconds to seconds.

\(\displaystyle 8ms*\frac{1s}{1000ms}=0.008s\)

\(\displaystyle 90kgm/s=F(0.008s)\)

Solve for the force

\(\displaystyle F=\frac{90kgm/s}{0.008s}\)

\(\displaystyle F=11,250N\)

To figure this out we needed to consider the change in momentum of the ball.

We know that the impulse of the ball is equal to the change in momentum.

\(\displaystyle I=\delta p\)

The impulse is equal to the force times the time the force is applied.

\(\displaystyle I=Ft\)

The change in momentum is equal to the mass times the change in velocity.

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

Therefore we can combine these equations to say

\(\displaystyle Ft=m\delta v\)

So let us look at what is happening in the x-direction.

\(\displaystyle F(2.5x10^{-3}s)=(0.145kg)(0m/s-27m/s)\)

Rearrange and solve the force in the x-direction.

\(\displaystyle F=-1566N\)

Next, let us determine what is happening in the y-direction.  We will need to figure out the initial velocity of the ball in the y-direction using the height.  At the peak, we also know the velocity is 0m/s.  We also know that the acceleration due to gravity is \(\displaystyle -9.8m/s/s\).

\(\displaystyle v^{2}_{f}=v^{2}_{i}+2a\delta x\)

\(\displaystyle (0m/s)^{2}=v^{2}_{i}+2(-9.8m/s/s)(31.5m)\)

\(\displaystyle 0=v^{2}_{i}-617.4\)

\(\displaystyle 617.4=v^{2}_{i}\)

\(\displaystyle 24.8m/s=v^{2}_{i}\)

Once we have the initial velocity, we can now determine the force in the y-direction.

\(\displaystyle F(2.5x10^{-3}s)=(0.145kg)(24.8m/s-0m/s)\)

\(\displaystyle F=1438.4N\)

Now that we have the force in both the x and y direction we can determine the overall resultant force using the Pythagorean Theorem.

\(\displaystyle (-1566N)^{2}+(1438.4N)^{2}=F^{2}\)

\(\displaystyle F=2126N\)