Calculating Momentum - Physics

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Question

A crate slides along the floor for before stopping. If it was initially moving with a velocity of $3\frac{m}{s}$ , what is the force of friction?

Answer

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

Remember that momentum is equal to mass times velocity: . We can rewrite this equation in terms of force.

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

$\frac{kgm}{s^2}s=Ns=Ft$

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

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

Expand this equation to include our given values.

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

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

$(100kg* 0\frac{m}s{})-(100kg3\frac{m}{s})=F(3s)$

$(-100kg3\frac{m}{s})=F(3s)$

$(-300\frac{kgm}{s})=F(3s)$

$\frac{(-300\frac{kgm}{s})}{3s}}=F$

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Question

A crate slides along the floor for before stopping. If it was initially moving with a velocity of $12\frac{m}{s}$ , what is the force of friction?

Answer

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

Remember that momentum is equal to mass times velocity: . We can rewrite this equation in terms of force.

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

$\frac{kgm}{s^2}s=Ns=Ft$

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

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

Expand this equation to include our given values.

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

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

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

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

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

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

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

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Question

A crate slides along a floor with a starting velocity of $11\frac{m}{s}$ . If the force due to friction is , how long will it take before stopping?

Answer

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

Remember that momentum is equal to mass times velocity: . We can rewrite this equation in terms of force.

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

$\frac{kgm}{s^2}s=Ns=Ft$

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

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

Expand this equation to include our given values.

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

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

$(300kg0\frac{m}{s})-(300kg11\frac{m}{s})=-35N\Delta t$

$(-300kg11\frac{m}{s})=-35N\Delta t$

$(-3300\frac{kgm}{s})=-35N\Delta t$

$\frac{(-3300\frac{kgm}{s})}{-35N}= \Delta t$

$94.29s= \Delta t$

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Question

A ball hits a brick wall with a velocity of $10\frac{m}{s}$ and bounces back at the same speed. If the ball is in contact with the wall for , what is the force exerted by the wall on the ball?

Answer

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

Remember that momentum is equal to mass times velocity: . We can rewrite this equation in terms of force.

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

$\frac{kgm}{s^2}s=Ns=Ft$

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

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

Expand this equation to include our given values.

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

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 these given values, we can solve for the force that acts on the ball.

$(0.5kg-10\frac{m}{s})-(0.5kg10\frac{m}{s})=F(0.1s)$

$(-5\frac{kgm}{s})-(5\frac{kgm}{s})=F(0.1s)$

$(-10\frac{kgm}{s})=F(0.1s)$

$\frac{(-10\frac{kgm}{s})}{0.1s}=F$

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

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Question

A ball moving at $30\frac{m}{s}$ strikes a piece of paper. It breaks through the paper and continues moving in the same direction. If the paper exerted a force of on the ball and the two are in contact for , what is the final momentum of the ball?

Answer

Remember that momentum is equal to mass times velocity: . We can rewrite this equation in terms of force.

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

$\frac{kgm}{s^2}s=Ns=Ft$

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

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

Expand this equation to include our given values.

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

We are given the mass and initial velocity, allowing us to solve for the initial momentum, and the force and time, allowing us to evaluate the right side of the equation. Using these values we can solve for the final momentum of the ball.

$p_{final}-(0.02kg30\frac{m}{s})=(-10N)(0.01s)$

$p_{final}-(0.6\frac{kgm}{s})=(-0.1Ns)$

$p_{final}=-0.1\frac{kgm}{s}+0.6\frac{kgm}{s}$

$p_{final}=0.5\frac{kgm}{s}$

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Question

Susan pushes a car with of force on a frictionless surface. How long does she need to push the car to get it to a velocity of $4\frac{m}{s}$ if it starts at rest?

Answer

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

Remember that momentum is equal to mass times velocity: . We can rewrite this equation in terms of force.

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

$\frac{kgm}{s^2}s=Ns=Ft$

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

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

Expand this equation to include our given values.

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

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

$(1000kg4\frac{m}{s})-(1000kg0\frac{m}{s})=(35N)(\Delta t)$

$4000\frac{kgm}{s}=35N\Delta t$

$\frac{4000\frac{kg*m}{s}}{35N}=\Delta t$

$114.3s=\Delta t$

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Question

A hammer moving with a velocity of $12\frac{m}{s}$ strikes a nail. The two are in contact for , after which the hammer has a velocity of $0\frac{m}{s}$ . What is the force of the hammer on the nail?

Answer

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

Remember that momentum is equal to mass times velocity: . We can rewrite this equation in terms of force.

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

$\frac{kgm}{s^2}s=Ns=Ft$

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

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

Expand this equation to include our given values.

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

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

$(2kg0\frac{m}{s})-(2kg12\frac{m}{s})=F(0.2s)$

$(-2kg12\frac{m}{s})=F(0.2s)$

$(-24\frac{kg m}{s})=F(0.2s)$

$\frac{-24\frac{kg m}{s}}{0.2s}=F$

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, $F_{hammer}=-F_{nail}$ . This means that if the nail exerts of force on the hammer, then the hammer must exert of force on the nail.

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Question

A hammer moving with a velocity of $11\frac{m}{s}$ strikes a nail, after which the hammer has a velocity of $0\frac{m}{s}$ . If the hammer strikes the nail with of force, how long were the two in contact?

Answer

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

Remember that momentum is equal to mass times velocity: . We can rewrite this equation in terms of force.

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

$\frac{kgm}{s^2}s=Ns=Ft$

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

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

Expand this equation to include our given values.

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

Since the hammer is not moving at the end, its final velocity is zero. This allows us to set up the left side of our equation.

$(1.8kg0\frac{m}{s})-(1.8kg11\frac{m}{s})=F\Delta t$

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 our equation since we are looking purely at the momentum of the hammer.

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

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

$(0)-(1.8kg11\frac{m}{s})=(-160N)(\Delta t)$

$(-1.8kg11\frac{m}{s})=(-160N)(\Delta t)$

$-19.8\frac{kgm}{s}=-160N\Delta t$

$\frac{-19.8\frac{kgm}{s}}{-160N}=\Delta t$

$0.12s=\Delta t$

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Question

A ball hits a brick wall with a velocity of $30\frac{m}{s}$ and bounces back at the same speed. If the ball is in contact with the wall for , what is the value of the force exerted by the wall on the ball?

Answer

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

Remember that momentum is equal to mass times velocity: . We can rewrite this equation in terms of force.

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

$\frac{kgm}{s^2}s=Ns=Ft$

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

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

Expand this equation to include our given values.

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

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 this understanding we can solve for the force in our equation.

$(0.25kg* -30\frac{m}{s})-(0.25kg* 30\frac{m}{s})=F(0.1s)$

$(-7.5\frac{kg\cdot m}{s})-(7.5\frac{kg\cdot m}{s})=F(0.1s)$

$(-15\frac{kg\cdot m}{s})=F(0.1s)$

$\frac{(-15\frac{kg\cdot m}{s})}{0.1s}=F$

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

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Question

A hammer moving with a velocity of $20\frac{m}{s}$ strikes a nail. The two are in contact for , after which the hammer has a velocity of $0\frac{m}{s}$ . With how much force did the hammer strike the nail?

Answer

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

Remember that momentum is equal to mass times velocity: . We can rewrite this equation in terms of force.

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

$\frac{kgm}{s^2}s=Ns=Ft$

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

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

Expand this equation to include our given values.

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

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

$(2.5kg* 0\frac{m}{s})-(2.5kg20\frac{m}{s})=(F)(0.2s)$

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

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

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

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, $F_{hammer}=-F_{nail}$ . This means that if the nail exerts of force on the hammer, then the hammer must exert of force on the nail.

Our answer will be .

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Question

A crate slides along a floor with a starting velocity of $21\frac{m}{s}$ . If the force due to friction is , how long will it take for the box to come to rest?

Answer

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

Remember that momentum is equal to mass times velocity: . We can rewrite this equation in terms of force.

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

$\frac{kgm}{s^2}s=Ns=Ft$

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

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

Expand this equation to include our given values.

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

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

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

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

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

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

$262.5s= \Delta t$

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Question

A ball strikes a piece of paper moving at $35\frac{m}{s}$ . It breaks through the paper and continues on in the same direction. If the paper exerted a force of on the ball and the two are in contact for , what is the final momentum of the ball?

Answer

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

Remember that momentum is equal to mass times velocity: . We can rewrite this equation in terms of force.

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

$\frac{kgm}{s^2}s=Ns=Ft$

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

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

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

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

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

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

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

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

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

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

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Question

An car strikes a car from behind. The bumpers lock and they move forward together. If their new final velocity is equal to $15\frac{m}{s}$ , what was the initial velocity of the first car?

Answer

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, , we can set up the following equation.

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

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.

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

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

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

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

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

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Question

Susan pushes a car with of force. How long does she need to push it to get it to a velocity of $6\frac{m}{s}$ if it starts at rest and there is no friction?

Answer

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

Remember that momentum is equal to mass times velocity: . We can rewrite this equation in terms of force.

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

$\frac{kgm}{s^2}s=Ns=Ft$

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

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

Expand this equation to include our given values.

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

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

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

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

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

$132s=\Delta t$

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Question

A crate slides along a floor with a starting velocity of $21\frac{m}{s}$ . If the force due to friction is , how long will it take before the crate slides to a stop?

Answer

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

Remember that momentum is equal to mass times velocity: . We can rewrite this equation in terms of force.

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

$\frac{kgm}{s^2}s=Ns=Ft$

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

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

Expand this equation to include our given values.

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

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

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

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

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

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

$7.5s= \Delta t$

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Question

A hammer moving with a velocity of $10\frac{m}{s}$ strikes a nail, after which the hammer has a velocity of $0\frac{m}{s}$ . If the hammer strikes the nail with of force, how long were the two in contact?

Answer

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

Remember that momentum is equal to mass times velocity: . We can rewrite this equation in terms of force.

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

$\frac{kgm}{s^2}s=Ns=Ft$

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

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

Expand this equation to include our given values.

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

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

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

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 $F_{hammer}=-F_{nail}$ . This means that if the hammer exerts of force on the nail, then the nail must exert of force on the hammer.

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

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

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

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

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

$0.12s=\Delta t$

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Question

A hammer moving with a velocity of $\small 20\frac{m}{s}$ strikes a nail, after which the hammer has a velocity of $\small 0\frac{m}{s}$ . If the hammer strikes the nail with of force, how long were the two in contact?

Answer

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

Remember that momentum is equal to mass times velocity: . We can rewrite this equation in terms of force.

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

$\frac{kgm}{s^2}s=Ns=Ft$

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

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

Expand this equation to include our given values.

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

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

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

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 $F_{hammer}=-F_{nail}$ . This means that if the hammer exerts of force on the nail, then the nail must exert of force on the hammer.

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

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

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

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

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

$0.31s=\Delta t$

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Question

A crate slides along the floor for before stopping. If it was initially moving with a velocity of $7.7\frac{m}{s}$ , what is the force of friction?

Answer

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

Remember that momentum is equal to mass times velocity: . We can rewrite this equation in terms of force.

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

$\frac{kgm}{s^2}s=Ns=Ft$

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

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

Expand this equation to include our given values.

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

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

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

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

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

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

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Question

A hammer moving with a velocity of $17\frac{m}{s}$ strikes a nail. The two are in contact for , after which the hammer has a velocity of $0\frac{m}{s}$ . How much force went into the nail?

Answer

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

Remember that momentum is equal to mass times velocity: . We can rewrite this equation in terms of force.

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

$\frac{kgm}{s^2}s=Ns=Ft$

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

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

Expand this equation to include our given values.

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

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

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

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

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

$\frac{-54.4\frac{kg \cdot m}{s}}{0.02s}=F$

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 $F_{hammer}=-F_{nail}$ .

This means that if the nail exerts of force on the hammer, then the hammer must exert of force on the nail; therefore, our answer will be .

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Question

Courtney pushes a car with of force along a frictionless surface. If the car is initially at rest, how long does she need to push for the car to reach a velocity of $10\frac{m}{s}$ ?

Answer

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

Remember that momentum is equal to mass times velocity: . We can rewrite this equation in terms of force.

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

$\frac{kgm}{s^2}s=Ns=Ft$

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

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

Expand this equation to include our given values.

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

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

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

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

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

$300s=\Delta t$

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