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.

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

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

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

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

\(\displaystyle 0.31s=\Delta t\)

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

\(\displaystyle (175kg* 0\frac{m}{s})-(175kg* 7.7\frac{m}{s})=F(4s)\)

\(\displaystyle -(175kg* 7.7\frac{m}{s})=F(4s)\)

\(\displaystyle (-1347.5\frac{kg\cdot m}{s})=F(4s)\)

\(\displaystyle \frac{(-1347.5\frac{kg\cdot m}{s})}{4s}}=F\)

\(\displaystyle -336.88N=F\)

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.

\(\displaystyle (3.2kg* 0\frac{m}{s)}-(3.2kg* 17\frac{m}{s})=(F)(0.02s)\)

\(\displaystyle -(3.2kg* 17\frac{m}{s})=(F)(0.02s)\)

\(\displaystyle (-54.4\frac{kg \cdot m}{s})=(F)(0.02s)\)

\(\displaystyle \frac{-54.4\frac{kg \cdot m}{s}}{0.02s}=F\)

\(\displaystyle -2720N=F\)

This equation solves for the force of the nail on the hammer, as we were looking purely at the momentum of the hammer; however, we need to find the force of the hammer on the nail. Newton's third law states that \(\displaystyle F_{hammer}=-F_{nail}\).

This means that if the nail exerts \(\displaystyle -2720N\) of force on the hammer, then the hammer must exert \(\displaystyle 2720N\) of force on the nail; therefore, our answer will be \(\displaystyle 2720N\).