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 $\displaystyle 2720N$.

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 time.

$\displaystyle (117kg* 0\frac{m}{s})-(117kg* 21\frac{m}{s})=-330N* \Delta t$

$\displaystyle -(117kg* 21\frac{m}{s})=-330N* \Delta t$

$\displaystyle (-2457\frac{kg\cdot m}{s})=-330N* \Delta t$

$\displaystyle \frac{(-2457\frac{kg\cdot m}{s})}{-330N}= \Delta t$

$\displaystyle 7.5s= \Delta t$

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

Expand this equation to include our given values.

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

Since the car is not moving at the beginning, its initial velocity is zero. Plug in the given values and solve for the time.

$\displaystyle (1200kg* 10\frac{m}{s})-(1200kg* 0\frac{m}{s})=(40N)\Delta t$

$\displaystyle 12000\frac{kg\cdot m}{s}=(40N) \Delta t$

$\displaystyle \frac{12000\frac{kg\cdot m}{s}}{40N}=\Delta t$

$\displaystyle 300s=\Delta t$

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.

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

Even though the ball is bouncing back at the same "speed", its velocity will now be negative as it is moving in the opposite direction. Using what we know from the question, we can solve for the force.

$\displaystyle (0.05kg*-9.7\frac{m}{s})-(0.05kg* 9.7\frac{m}{s})=F(0.2s)$

$\displaystyle (-0.485\frac{kg\cdot m}{s})-(0.485\frac{kg\cdot m}{s})=F(0.2s)$

$\displaystyle (-0.97\frac{kg\cdot m}{s})=F(0.2s)$

$\displaystyle \frac{(-0.97\frac{kg\cdot m}{s})}{0.2s}=F$

$\displaystyle -4.85N=F$

Our answer is negative because the force is acting on the ball in the OPPOSITE direction from the way it was originally heading.

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 $\displaystyle \small v_{1initial}$.

$\displaystyle (0.75kg*v_{1initial})+(2.5kg* 0\frac{m}{s})=(0.75kg*20\frac{m}{s})+(2.5kg* 22\frac{m}{s})$

$\displaystyle (0.75kg* v_{1initial})+(0)=(15\frac{kg\cdot m}{s} )+(55\frac{kg\cdot m}{s})$

$\displaystyle (0.75kg* v_{1initial})=(70\frac{kg\cdot m}{s} )$

$\displaystyle v_{1initial}=\frac{70\frac{kg\cdot m}{s}}{0.75kg}$

$\displaystyle v_{1initial}=93.33\frac{m}{s}$

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

Since we are solving for $\displaystyle p_{final}$, we can leave that part of the equation alone and expand the rest to include our given values.

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

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

$\displaystyle p_{final}-(0.05kg*40\frac{m}{s})=(-18N)(0.01s)$

$\displaystyle p_{final}-(2\frac{kg\cdot m}{s})=(-0.18N\cdot s)$

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

$\displaystyle p_{final}=-0.18\frac{kg\cdot m}{s}+2\frac{kg\cdot m}{s}$

$\displaystyle p_{final}=1.82\frac{kg\cdot m}{s}$

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 $\displaystyle \small m_2$.

$\displaystyle (12kg*76.4\frac{m}{s})+(m_2*0\frac{m}{s})=(12kg* 22.2\frac{m}{s})+(m_2* 13.7\frac{m}{s})$

$\displaystyle (916.8\frac{kg\cdot m}{s})+(0)=(266.4\frac{kg\cdot m}{s})+m_2(22.2\frac{m}{s})$

$\displaystyle (916.8\frac{kg\cdot m}{s})-(266.4\frac{kg\cdot m}{s})=m_2(22.2\frac{m}{s})$

$\displaystyle (650.4\frac{kg\cdot m}{s})=m_2( 22.2\frac{m}{s})$

$\displaystyle \frac{650.4\frac{kg\cdot m}{s}}{22.2\frac{m}{s}}=m_2$

$\displaystyle 29.3kg=m_2$

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.

$\displaystyle (3400kg* 28.7\frac{m}{s})+(3400kg*0\frac{m}{s})=(3400kg+3400kg)v_{final}$

$\displaystyle 97580\frac{kg\cdot m}{s}=(6800kg)v_{final}$

$\displaystyle \frac{97580\frac{kg\cdot m}{s}}{6800kg}=v_{final}$

$\displaystyle 14.4\frac{m}{s}=v_{final}$

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 of the joined cars is given. Plug in these values and solve for the initial velocity of the first car.

$\displaystyle (950kg*v_{1initial})+(2200kg* 0\frac{m}{s})=(950kg+2200kg)* 15\frac{m}{s}$

$\displaystyle 950kg* v_{1initial}=(3150kg)*15\frac{m}{s}$

$\displaystyle 950kg* v_{1initial}=47250\frac{kg\cdot m}{s}$

$\displaystyle v_{1initial}=\frac{47250\frac{kg\cdot m}{s}}{950kg}$

$\displaystyle v_{1initial}=49.7\frac{m}{s}$