Knowns

$\displaystyle m_{A}=950kg$

$\displaystyle m_{B}=2200kg$

$\displaystyle v_{0B}=0m/s$

$\displaystyle d=3.0m$

$\displaystyle \mu =0.8$

Unknowns

$\displaystyle v_{0A}=?$

To solve the problem we must consider two different situations.  The first is when the cars are skidding across the ground.  While they are skidding the force of friction is what is resisting their motion and therefore doing work on the cars to slow them down to a stop.

The second situation is the collision itself when the first car has an initial speed and the second car is stopped.  After the collision, both cars are traveling at the same speed as their bumpers have been locked together.

To begin, we need to find the speed that the cars are skidding across the ground after the bumpers have been interlocked.

The work-kinetic energy theorem states that the work done is equal to the change in the kinetic energy of the object.

$\displaystyle -W=\Delta KE$

Work is directly related to the force times the displacement of the object.

$\displaystyle W=Fd$

In this case, the force that is doing the work is friction.

$\displaystyle F_{f}=\mu F_{N}$

Since the cars are on a level surface the normal force is equal to the force of gravity.

$\displaystyle F_{G=}F_{N}$

The force of gravity is directly related to the mass and the acceleration due to gravity acting on an object.

$\displaystyle F_{G}=mg$

When we put all these equations together we get

$\displaystyle W=\mu mgd$

We also know that kinetic energy is related to the mass and velocity squared.

$\displaystyle KE=\frac{1}{2}mv^{2}$

Therefore our final equation should look like

$\displaystyle -\mu mgd=\frac{1}{2}mv{_{f}}^{2}-\frac{1}{2}mv{_{0}}^{2}$

Notice that the mass falls out of the equation since it is in every term.  Also, note that the final velocity is 0m/s so this will cancel out as well.

$\displaystyle -\mu gd=-\frac{1}{2}v{_{0}}^{2}$

Since both of these terms have a negative we can cancel this out as well.

$\displaystyle \mu gd=\frac{1}{2}v{_{0}}^{2}$

We can now substitute our variables to determine the velocity of both cars just after the crash.

$\displaystyle (0.8)(9.8m/s^{2})(3.0m)=\frac{1}{2}v{_{0}}^{2}$

$\displaystyle 47.04=v{_{0}}^{2}$

$\displaystyle v_{0}=6.9m/s$

This is the velocity of the cars after the collision.  We must now consider our second situation of the collision itself.  During this collision, momentum must be conserved.  The law of conservation of momentum states

$\displaystyle m_{A}v_{0A}+m_{B}v_{0B}=m_{A}v_{fA}+m_{B}v_{fB}$

We can substitute our values for the masses of the cars, the final velocity of the cars after the collision (which we just found), and the initial value of the stopped car.

$\displaystyle (950kg)v_{0A}+(2200kg)(0m/s)=(950kg)(6.9m/s)+(2200kg)(6.9m/s)$

Now we can solve for our missing variable.

$\displaystyle (950kg)v_{0A}=21735kgm/s$

$\displaystyle v_{0A}=22.9m/s$

Knowns

$\displaystyle m_{A}=550kg$

$\displaystyle m_{B}=450kg$

$\displaystyle v_{A0}=3.7m/s$

$\displaystyle v_{B0}=4.5m/s$

Unknowns

$\displaystyle v_{Af}=?$

$\displaystyle v_{Bf}=?$

For elastic collisions, we know that the initial and final velocities are related by the equation

$\displaystyle v_{A0}+v_{Af}=v_{B0}+v_{Bf}$

We also know that the momentum is conserved meaning that

$\displaystyle mv_{A0}+mv_{B0}=mv_{Af}+mv_{Bf}$

Since we have two missing variables and two equations, we can now solve for one of the variables using a system of equations

Let’s get the final velocity of car A by itself from the first equation

$\displaystyle v_{Af}=v_{B0}+v_{Bf}-v_{A0}$

We can now substitute this equation into our momentum equation.

$\displaystyle m_{A}v_{A0}+m_{B}v_{B0}=m_{A}(v_{B0}+v_{Bf}-v_{A0})+m_{B}v_{Bf}$

In our new equation, we only have one missing variable, so let's substitute in values and solve.

$\displaystyle (550kg)(3.7m/s)+(450kg)(4.5m/s)=(550kg)(4.5m/s+v_{Bf}-3.7m/s)+(450kg)(v_{Bf})$

$\displaystyle 4060=550(0.8+v_{Bf})+450(v_{Bf})$

$\displaystyle 4060=440+550v_{Bf}+450v_{Bf}$

$\displaystyle 3620=1000v_{Bf}$

$\displaystyle v_{Bf}=3.6m/s$

We can now plug this value back into our equation for $\displaystyle v_{Af}$

$\displaystyle v_{Af}=v_{B0}+v_{Bf}-v_{A0}$

$\displaystyle v_{Af}=4.5m/s+3.6m/s-3.7m/s$

$\displaystyle v_{Af}=4.4m/s$

The fastest way to solve a problem like this is with momentum.

Remember that momentum is equal to mass times velocity: $\displaystyle p=mv$. We can rewrite this equation in terms of force.

$\displaystyle mv=(kg)(\frac{m}{s})=(kg)(\frac{m}{s})*(\frac{s}{s})$

$\displaystyle \frac{kg*m}{s^2}*s=N*s=Ft$

Using this transformation, we can see that momentum is also equal to force times time.

$\displaystyle \Delta p=F\Delta t$ can also be thought of as $\displaystyle (p_{final}-p_{initial})=F\Delta t$.

Expand this equation to include our given values.

$\displaystyle (mv_{final}-mv_{initial})=F\Delta t$

Since the object is not moving at the end, its final velocity is zero. Plug in the given values and solve for the force.

$\displaystyle (50kg* 0\frac{m}{s})-(50kg* 12\frac{m}{s})=F(10s)$

$\displaystyle (-50kg* 12\frac{m}{s})=F(10s)$

$\displaystyle (-600\frac{kg\cdot m}{s})=F(10s)$

$\displaystyle \frac{(-600\frac{kg\cdot m}{s})}{10s}}=F$

$\displaystyle -60N=F$

We would expect the answer to be negative because the force of friction acts in the direction opposite to the initial velocity.

The fastest way to solve a problem like this is with momentum.

Remember that momentum is equal to mass times velocity: $\displaystyle p=mv$. We can rewrite this equation in terms of force.

$\displaystyle mv=(kg)(\frac{m}{s})=(kg)(\frac{m}{s})*(\frac{s}{s})$

$\displaystyle \frac{kg*m}{s^2}*s=N*s=Ft$

Using this transformation, we can see that momentum is also equal to force times time.

$\displaystyle \Delta p=F\Delta t$ can also be thought of as $\displaystyle (p_{final}-p_{initial})=F\Delta t$.

Expand this equation to include our given values.

$\displaystyle (mv_{final}-mv_{initial})=F\Delta t$

Since the hammer is not moving at the end, its final velocity is zero. Plug in the given values and solve for the force.

$\displaystyle (2.5kg* 0\frac{m}{s})-(2.5kg*20\frac{m}{s})=(F)(0.2s)$

$\displaystyle (-2.5kg* 20\frac{m}{s})=(F)(0.2s)$

$\displaystyle (-50\frac{kg \cdot m}{s})=(F)(0.2s)$

$\displaystyle \frac{-50\frac{kg \cdot m}{s}}{0.2s}=F$

$\displaystyle -250N=F$

This equation solves for the force of the nail on the hammer, as we were looking purely at the momentum of the hammer.

According to Newton's third law, $\displaystyle F_{hammer}=-F_{nail}$. This means that if the nail exerts $\displaystyle -250N$ of force on the hammer, then the hammer must exert $\displaystyle 250N$ of force on the nail.

Our answer will be $\displaystyle 250N$.

The fastest way to solve a problem like this is with momentum.

Remember that momentum is equal to mass times velocity: $\displaystyle p=mv$. We can rewrite this equation in terms of force.

$\displaystyle mv=(kg)(\frac{m}{s})=(kg)(\frac{m}{s})*(\frac{s}{s})$

$\displaystyle \frac{kg*m}{s^2}*s=N*s=Ft$

Using this transformation, we can see that momentum is also equal to force times time.

$\displaystyle \Delta p=F\Delta t$ can also be thought of as $\displaystyle (p_{final}-p_{initial})=F\Delta t$.

Expand this equation to include our given values.

$\displaystyle (mv_{final}-mv_{initial})=F\Delta t$

Since the box is not moving at the end, its final velocity is zero. Plug in the given values and solve for the time.

$\displaystyle (100kg* 0\frac{m}{s})-(100kg* 21\frac{m}{s})=(-8N) \Delta t$

$\displaystyle (-100kg* 21\frac{m}{s})=(-8N) \Delta t$

$\displaystyle (-2100\frac{kg\cdot m}{s})=(-8N) \Delta t$

$\displaystyle \frac{(-2100\frac{kg\cdot m}{s})}{-8N}= \Delta t$

$\displaystyle 262.5s= \Delta t$

The fastest way to solve a problem like this is with momentum.

Remember that momentum is equal to mass times velocity: $\displaystyle p=mv$. We can rewrite this equation in terms of force.

$\displaystyle mv=(kg)(\frac{m}{s})=(kg)(\frac{m}{s})*(\frac{s}{s})$

$\displaystyle \frac{kg*m}{s^2}*s=N*s=Ft$

Using this transformation, we can see that momentum is also equal to force times time.

$\displaystyle \Delta p=F\Delta t$ can also be thought of as $\displaystyle (p_{final}-p_{initial})=F\Delta t$.

Since we're looking for $\displaystyle p_{final}$, we're going to leave that part alone in the problem, but we can expand the rest.

$\displaystyle p_{final}-mv_{initial}=F\Delta t$

From here, plug in the given values and solve for the final momentum.

$\displaystyle p_{final}-(0.04kg* 35\frac{m}{s})=(-6N)(0.01s)$

$\displaystyle p_{final}-(1.4\frac{kg\cdot m}{s})=(-0.06N\cdot s)$

At this point, remember that $\displaystyle 1N=1\frac{kg\cdot m}{s^2}$, so sides are now working in the same units. 

$\displaystyle p_{final}=-0.06\frac{kg\cdot m}{s}+1.4\frac{kg\cdot m}{s}$

$\displaystyle p_{final}=1.34\frac{kg\cdot m}{s}$

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 initial velocity of the first car.

$\displaystyle (800kg* v_{1initial})+(1200kg* 0\frac{m}{s})=(800kg+1200kg)* 15\frac{m}{s}$

$\displaystyle 800kg* v_{1initial}=(2000kg)*15\frac{m}{s}$

$\displaystyle 800kg* v_{1initial}=30000\frac{kg\cdot m}{s}$

$\displaystyle v_{1initial}=\frac{30,000\frac{kg\cdot m}{s}}{800kg}$

$\displaystyle v_{1initial}=37.5\frac{m}{s}$

The fastest way to solve a problem like this is with momentum.

Remember that momentum is equal to mass times velocity: $\displaystyle p=mv$. We can rewrite this equation in terms of force.

$\displaystyle mv=(kg)(\frac{m}{s})=(kg)(\frac{m}{s})*(\frac{s}{s})$

$\displaystyle \frac{kg*m}{s^2}*s=N*s=Ft$

Using this transformation, we can see that momentum is also equal to force times time.

$\displaystyle \Delta p=F\Delta t$ can also be thought of as $\displaystyle (p_{final}-p_{initial})=F\Delta t$.

Expand this equation to include our given values.

$\displaystyle (mv_{final}-mv_{initial})=F\Delta t$

Since the car starts from rest, its initial velocity is zero. Plug in the given values and solve for the time.

$\displaystyle (1100kg* 6\frac{m}{s})-(1100kg* 0\frac{m}{s})=(50N)(\Delta t)$

$\displaystyle 6600\frac{kg\cdot m}{s}=(50N) \Delta t$

$\displaystyle \frac{6600\frac{kg\cdot m}{s}}{50N}=\Delta t$

$\displaystyle 132s=\Delta t$

The fastest way to solve a problem like this is with momentum.

Remember that momentum is equal to mass times velocity: $\displaystyle p=mv$. We can rewrite this equation in terms of force.

$\displaystyle mv=(kg)(\frac{m}{s})=(kg)(\frac{m}{s})*(\frac{s}{s})$

$\displaystyle \frac{kg*m}{s^2}*s=N*s=Ft$

Using this transformation, we can see that momentum is also equal to force times time.

$\displaystyle \Delta p=F\Delta t$ can also be thought of as $\displaystyle (p_{final}-p_{initial})=F\Delta t$.

Expand this equation to include our given values.

$\displaystyle (mv_{final}-mv_{initial})=F\Delta t$

Since the hammer is not moving at the end, its final velocity is zero. 

$\displaystyle (2.2kg* 0\frac{m}{s})-(2.2kg* 10\frac{m}{s})=F\Delta t$

The problem gave us the force of the hammer on the nail, but not the force of the nail on the hammer, which is what we need for the equation as we are looking purely at the momentum of the hammer.  Fortunately, Newton's third law can help us. It states that $\displaystyle F_{hammer}=-F_{nail}$. This means that if the hammer exerts $\displaystyle 180N$ of force on the nail, then the nail must exert $\displaystyle -180N$ of force on the hammer.

We can plug that value in for the force and solve for the time.

$\displaystyle 0-(2.2kg* 10\frac{m}{s})=(-180N)(\Delta t)$

$\displaystyle -(22\frac{kg\cdot m}{s})=(-180N)(\Delta t)$

$\displaystyle -22\frac{kg\cdot m}{s}=(-180N) \Delta t$

$\displaystyle \frac{-22\frac{kg\cdot m}{s}}{-180N}=\Delta t$

$\displaystyle 0.12s=\Delta t$

The fastest way to solve a problem like this is with momentum.

Remember that momentum is equal to mass times velocity: $\displaystyle p=mv$. We can rewrite this equation in terms of force.

$\displaystyle mv=(kg)(\frac{m}{s})=(kg)(\frac{m}{s})*(\frac{s}{s})$

$\displaystyle \frac{kg*m}{s^2}*s=N*s=Ft$

Using this transformation, we can see that momentum is also equal to force times time.

$\displaystyle \Delta p=F\Delta t$ can also be thought of as $\displaystyle (p_{final}-p_{initial})=F\Delta t$.

Expand this equation to include our given values.

$\displaystyle (mv_{final}-mv_{initial})=F\Delta t$

Since the hammer is not moving at the end, its final velocity is zero.

$\displaystyle (3.1kg* 0\frac{m}{s})-(3.1kg* 20\frac{m}{s})=F\Delta t$

The problem gave us the force of the hammer on the nail, but not the force of the nail on the hammer, which is what we need as we are looking purely at the momentum of the hammer.

Fortunately, Newton's third law can help us. It states that $\displaystyle F_{hammer}=-F_{nail}$. This means that if the hammer exerts $\displaystyle 200N$ of force on the nail, then the nail must exert $\displaystyle -200N$ of force on the hammer.

We can plug that value in for the force and solve.

$\displaystyle 0-(3.1kg* 20\frac{m}{s})=(-200N)(\Delta t)$

$\displaystyle -(3.1kg* 20\frac{m}{s})=(-200N)(\Delta t)$

$\displaystyle -62\frac{kg\cdot m}{s}=-200N* \Delta t$

$\displaystyle \frac{-62\frac{kg\cdot m}{s}}{-200N}=\Delta t$

$\displaystyle 0.31s=\Delta t$