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High School Physics : Momentum

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Example Question #21 : Momentum

A 950kg sports car collides into the rear end of 2200kg SUV stopped at a red light. The bumper lock, the brakes lock, and the two cars skid forward 3.0m before coming to a stop. The police officer, knowing the coefficient of kinetic friction between the tires and the road is about 0.8. He calculates the speed of the sports car at impact. What was that speed?

Possible Answers:

6.9m/s

46.7m/s

22.9m/s

20.5m/s

17.4m/s

Correct answer:

22.9m/s

Explanation:

Knowns

$m_{A}=950kg$

$m_{B}=2200kg$

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

d=3.0m

$\mu =0.8$

Unknowns

$v_{0A}=?$

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.

$-W=\Delta KE$

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

W=Fd

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

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

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

$F_{G=}F_{N}$

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

$F_{G}=mg$

When we put all these equations together we get

$W=\mu mgd$

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

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

Therefore our final equation should look like

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

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.

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

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

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

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

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

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

$v_{0}=6.9m/s$

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

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

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.

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

Now we can solve for our missing variable.

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

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

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Example Question #22 : Momentum

Two bumper cars at an amusement park collide elastically as one approaches the other directly from the rear. The car in front (CarA) has a mass of 550kg and the car behind it (CarB) has a mass of 450kg. The car in front was traveling at 3.70m/s while the car behind hit him with a velocity of 4.50m/s. What are their final velocities after the collision?

Possible Answers:

$v_{Af}=4.5m/s$

$v_{Bf}=3.7m/s$

$v_{Af}=5.4 m/s$

$v_{Bf}=2.3m/s$

$v_{Af}=3.6m/s$

$v_{Bf}=4.4m/s$

$v_{Af}=4.4m/s$

$v_{Bf}=3.6m/s$

$v_{Af}=8 m/s$

$v_{Bf}=0m/s$

Correct answer:

$v_{Af}=4.4m/s$

$v_{Bf}=3.6m/s$

Explanation:

Knowns

$m_{A}=550kg$

$m_{B}=450kg$

$v_{A0}=3.7m/s$

$v_{B0}=4.5m/s$

Unknowns

$v_{Af}=?$

$v_{Bf}=?$

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

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

We also know that the momentum is conserved meaning that

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

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

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

We can now substitute this equation into our momentum equation.

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

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

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

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

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

$3620=1000v_{Bf}$

$v_{Bf}=3.6m/s$

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

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

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

$v_{Af}=4.4m/s$

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Example Question #291 : Motion And Mechanics

A 50kg crate slides along the floor for 10s before stopping. If it was initially moving with a velocity of $12\frac{m}{s}$ , what is the force of friction?

Possible Answers:

-6N

-60N

-100N

-600N

-10N

Correct answer:

-60N

Explanation:

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

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

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

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

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.

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

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

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

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

-60N=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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Example Question #291 : Motion And Mechanics

A 2.5kg hammer moving with a velocity of $20\frac{m}{s}$ strikes a nail. The two are in contact for 0.2s , after which the hammer has a velocity of $0\frac{m}{s}$ . With how much force did the hammer strike the nail?

Possible Answers:

-125N

75N

250N

300N

112N

Correct answer:

250N

Explanation:

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

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

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

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

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.

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

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

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

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

-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, $F_{hammer}=-F_{nail}$ . This means that if the nail exerts -250N of force on the hammer, then the hammer must exert 250N of force on the nail.

Our answer will be 250N .

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Example Question #25 : Momentum

A 100kg crate slides along a floor with a starting velocity of $21\frac{m}{s}$ . If the force due to friction is -8N , how long will it take for the box to come to rest?

Possible Answers:

108.7s

262.5s

504.2s

121.3s

284.5s

Correct answer:

262.5s

Explanation:

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

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

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

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

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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Example Question #26 : Momentum

A 0.04kg 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 -6N on the ball and the two are in contact for 0.01s , what is the final momentum of the ball?

Possible Answers:

$1.76\frac{kg\cdot m}{s}$

$1.12\frac{kg\cdot m}{s}$

$1.34\frac{kg\cdot m}{s}$

$0.98\frac{kg\cdot m}{s}$

$1.4\frac{kg\cdot m}{s}$

Correct answer:

$1.34\frac{kg\cdot m}{s}$

Explanation:

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

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

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

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

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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Example Question #1 : Calculating Momentum

An 800kg car strikes a 1200kg 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?

Possible Answers:

$37.5\frac{m}{s}$

$51.2\frac{m}{s}$

$38.5\frac{m}{s}$

$41\frac{m}{s}$

$72.5\frac{m}{s}$

Correct answer:

$37.5\frac{m}{s}$

Explanation:

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, p=mv , 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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Example Question #291 : Motion And Mechanics

Susan pushes a car 1100kg with 50N 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?

Possible Answers:

123s

87s

145s

132s

212s

Correct answer:

132s

Explanation:

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

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

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

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

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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Example Question #29 : Momentum

A 2.2kg 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 180N of force, how long were the two in contact?

Possible Answers:

1.2 s

0.09s

0.12s

1.02s

2.4s

Correct answer:

0.12s

Explanation:

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

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

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

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

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 180N of force on the nail, then the nail must exert -180N 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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Example Question #291 : Motion And Mechanics

A 3.1kg 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 200N of force, how long were the two in contact?

Possible Answers:

0.64s

3.5s

0.01s

0.31s

0.03s

Correct answer:

0.31s

Explanation:

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

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

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

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

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 200N of force on the nail, then the nail must exert -200N 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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High School Physics : Momentum

Study concepts, example questions & explanations for High School Physics

All High School Physics Resources

Example Questions

Example Question #21 : Momentum

Example Question #22 : Momentum

Example Question #291 : Motion And Mechanics

Example Question #291 : Motion And Mechanics

Example Question #25 : Momentum

Example Question #26 : Momentum

Example Question #1 : Calculating Momentum

Example Question #291 : Motion And Mechanics

Example Question #29 : Momentum

Example Question #291 : Motion And Mechanics

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