example of problem solving of momentum

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Momentum Practice Problems with Answers for High School

These problems about momentum are designed to enhance your understanding of momentum and its definition, as well as to provide practical applications of its formula in various situations.

Momentum is defined as the product of an object's mass and its velocity, represented by the formula \[\vec{P}=m\vec{v}\] Like velocity, momentum is a vector quantity in the same direction as the object's velocity. 

The SI units of momentum are kilogram-meters per second ${\rm kg.m/s}$. 

An object can possess a large momentum if it has either a large mass or a high velocity.

When solving momentum problems, the first step is to assign a positive direction. Then, compare the velocities in this direction. Velocities in the same direction as the positive direction are considered to have positive momentum, while those in the opposite direction have negative momentum.

These problems are straightforward and beneficial for students in classes 10 and 11. For more complex problems appropriate for class 12 or college level, please refer to the sections on momentum and impulse problems .

Momentum Problems   

Problem (1): What is the definition of momentum in physics? 

Solution: the product of a particle's mass and velocity in physics is called the particle's momentum, $\vec{p}=m\vec{v}$. Momentum is a vector quantity like velocity, acceleration , and force. The units of momentum are kg.m/s. 

Practice Problem (2): If a truck has a mass of 1000 kilograms and is traveling with a speed of v = 50 m/s, what is its momentum?

Solution: By definition, momentum is the product of mass and velocity, represented by the formula $P=m\,v$. Thus, substituting the given numerical values into this, we have \begin{align*} P&=mv\\&=1000\times 50\\&=50,000\quad {\rm kg.m/s}\end{align*} In all momentum problems, the question typically asks the magnitude of the momentum, regardless of its direction.

Problem (3): If a car has a mass of 2000 kilograms and travels with a velocity of v = 72 km/h, what is its momentum?

Solution:  Momentum is mass times velocity, $p=mv$ and its units in the SI system are $\rm kg.m/s$. In this problem, we first convert the units of velocity to SI units (i.e., from $\rm km/h$ to $\rm m/s$ by multiplying it by $\frac {10}{36}$). Therefore, the momentum of the car is calculated as follows: \begin{align*} P&=mv\\&=2000\times \Big(72\times \frac{10}{36}\Big)\\&=40,000\quad {\rm kg.m/s}\end{align*} This means the car has a momentum of 40,000 kg.m/s.

Problem (4): An 8-kilogram bowling ball is rolling in a straight line toward you. If its momentum is 16 kg·m/sec, how fast is it traveling?

Solution: using momentum formula $p=mv$ and solving for the velocity , we have \begin{align*} v&=\frac pm\\&=\frac{16}{8}\\&=2\quad {\rm m/s} \end{align*} 

Problem (5): A toy dart gun generates a dart with a momentum of 140 kg.m/s and a velocity of 4 m/s. What is the mass of the dart in grams? 

Solution: by applying momentum formula $p=mv$ and solving for $m$, we get \begin{align*} m&=\frac pv\\&=\frac {140}{4}\\&=35\quad {\rm kg} \end{align*} to convert it to the grams, multiply it by 1000 so the dart's mass is $35,000\,{\rm g}$. 

Problem (6): A beach ball is rolling in a straight line toward you at a speed of 0.5 m/sec. Its momentum is 0.25 kg·m/sec. What is the mass of the beach ball?

Solution:  Given that the momentum is $0.25\,\rm kg\cdot m/s$ and the velocity is $0.5\,\rm m/s$, we can rearrange the formula $p=mv$ to solve for the mass of the ball: \begin{align*} m&=\frac pv\\&=\frac {0.25}{0.5}\\&=0.5\quad {\rm kg} \end{align*}

Problem (7): A 5,000-kilogram truck travels in a straight line with a speed of 54 km/h. What is its momentum?

Solution:  The SI units of momentum are $\rm kg\cdot m/s$. Therefore, we first convert the speed units into SI units by multiplying them by $\frac {10}{36}$. As a result, the truck's speed becomes $v=54\times \frac {10}{36}=15\,{\rm m/s}$. Subsequently, we calculate the momentum as follows: \begin{align*} p&=mv\\&=5000\times 15\\&=75,000\quad {\rm kg.m/s}\end{align*}

Problem (8): A 1,400-kilogram car is traveling in a straight line. Its momentum is equal to that of the truck in the previous question. What is the velocity of the car?

Solution : the car's velocity is \begin{align*} v&=\frac pm\\&=\frac{75,000}{1400}\\&=53.6\quad {\rm m/s}\end{align*}

Problem (9): A bus traveling at a speed of 50 km/h has a momentum of 180,345 kg.m/s. What is the mass of the bus?

Solution : first convert the speed's units to the SI units of velocity ($\rm \frac ms$) by multiplying it by $\frac {10}{36}$. Next, using the formula of momentum $p=mv$ and solving for the mass, we get \begin{align*} m&=\frac pv\\&=\frac{180,345}{50\times \frac{10}{36}}\\&=12984.84\quad {\rm kg}\end{align*}

Problem (10): Which would take more force to stop in 10 seconds: an 8.0-kilogram ball rolling in a straight line at a speed of 0.2 m/sec or a 4.0-kilogram ball rolling along the same path at a speed of 1.0 m/sec?

Solution : According to  Newton's second law of motion , the change in an object's momentum is directly proportional to the force applied. As a result, an object with higher momentum requires a greater force to bring it to a stop.

Consider a $8\,{\rm kg}$-ball with a momentum calculated as $p=mv=(8)(0.2)=1.6\,{\rm kg.m/s}$. On the other hand, a $4\,{\rm kg}$-ball has a momentum of $p=mv=(4)(1)=4\,{\rm kg.m/s}$. Therefore, a larger force is required to stop the 4 kg ball due to its higher momentum.

Problem (11): The momentum of a truck traveling in a straight line at 15 m/sec is 41,500 kg·m/sec. What is the mass of the truck?

Solution : using the momentum formula, we have \begin{align*} m&=\frac pv\\&=\frac {41,500}{15}\\&=2766.6\quad {\rm kg}\end{align*}

Problem (12): The parking brake on a 1500 kg automobile has broken, and the car has reached a momentum of 7500 kg.m/s. What is the velocity of the vehicle?

Solution : using the definition of momentum and solving for the velocity, we have \begin{align*} v&=\frac pm\\&=\frac{7500}{1500}\\&=5\quad{\rm m/s}\end{align*}

Problem (13): A proton with mass of $1.67\times 10^{-27}\,{\rm kg}$ moving at the speed of $5\times 10^{5}\,{\rm m/s}$. What is its momentum?

Solution : momentum is mass multiply by velocity so \begin{align*} p=&mv\\&=(1.67\times 10^{-27})(5\times 10^{5})\\&=8.35\times 10^{-23}\quad {\rm kg.m/s}\end{align*}

Problem (14): A 0.14-kilogram baseball is thrown in a straight line at a velocity of 30 m/sec. What is the momentum of baseball?

Solution : baseball's momentum, $p=mv$, is determined as \begin{align*} p&=mv\\&=(0.14)(30)\\&=4.2\quad {\rm kg.m/s}\end{align*}

Problem (15): A pitcher can throw a 0.14-kg baseball with the same momentum as a 3-g bullet moving at a speed of 3000 m/s. What is baseball's speed? 

Solution : Since the momentum of the bullet, $p_2=m_2v_2$, is the same as the baseball's one $p_1=m_1v_1$, so equating those and solving for the unknown speed of the baseball, we get \begin{align*} p_1&=p_2\\m_1v_1&=m_2v_2\\\Rightarrow v_2&=\frac{m_1 v_1}{m_2}\\&=\frac{0.003\times 3000}{0.14}\\&=64.28\quad {\rm m/s}\end{align*} in above we converter the bullet's mass to the SI unit of mass $kg$ by dividing it by $1000$. 

Problem (16): Another pitcher throws the same baseball in a straight line. Its momentum is 2.1 kg·m/sec. What is the velocity of the ball?

Solution : using momentum formula, $p=mv$ and solving for the velocity, we have \begin{align*} v&=\frac pm\\&=\frac {2.1}{0.14}\\&=15\quad{\rm m/s} \end{align*}

Problem (17):  A 1-kilogram turtle crawls in a straight line at a speed of 0.01 m/sec. What is the turtle’s momentum?

Solution : momentum is defined as the product of mass and velocity, so \begin{align*} p&=mv\\&=(1)(0.01)\\&=0.01\quad {\rm kg.m/s} \end{align*}

Practice Problem (18): A bicycle and its rider have a mass of 100 kg. At what speed do they have the same momentum as an 1800-kg car traveling at 2 m/s?

Solution : the momenta of bicycle and car are equal so we have \begin{align*} p_c&=p_b\\m_b\,v_b &= m_c\, v_c \\(100)v_b &= (1800)(10) \\ \Rightarrow v_b &= \frac{1800\times 2}{100}\\&=36\quad {\rm m/s}\end{align*}

To solve introductory momentum problems and questions, you must first learn the definition of momentum as the product of mass and velocity and then apply it to find the unknown. All the above problems are easy and could be solved without any difficulty. 

In this article, we solved some simple problems on momentum that are helpful for high school students. 

We learned that momentum is defined as the product of an object's mass and its velocity. \[\vec{p}=m\vec{v}\]

Momentum is a vector quantity in physics, like velocity and displacement. It has both a magnitude and a direction. 

Date Created: 10/26/2020

Author : Dr. Ali Nemati  

© 2015 All rights reserved. by Physexams.com

8.1 Linear Momentum, Force, and Impulse

Section learning objectives.

By the end of this section, you will be able to do the following:

• Describe momentum, what can change momentum, impulse, and the impulse-momentum theorem
• Describe Newton’s second law in terms of momentum
• Solve problems using the impulse-momentum theorem

Teacher Support

The learning objectives in this section will help your students master the following standards:

• (C) calculate the mechanical energy of, power generated within, impulse applied to, and momentum of a physical system.

Section Key Terms

[BL] [OL] Review inertia and Newton’s laws of motion.

[AL] Start a discussion about movement and collision. Using the example of football players, point out that both the mass and the velocity of an object are important considerations in determining the impact of collisions. The direction as well as the magnitude of velocity is very important.

Momentum, Impulse, and the Impulse-Momentum Theorem

Linear momentum is the product of a system’s mass and its velocity . In equation form, linear momentum p is

You can see from the equation that momentum is directly proportional to the object’s mass ( m ) and velocity ( v ). Therefore, the greater an object’s mass or the greater its velocity, the greater its momentum. A large, fast-moving object has greater momentum than a smaller, slower object.

Momentum is a vector and has the same direction as velocity v . Since mass is a scalar , when velocity is in a negative direction (i.e., opposite the direction of motion), the momentum will also be in a negative direction; and when velocity is in a positive direction, momentum will likewise be in a positive direction. The SI unit for momentum is kg m/s.

Momentum is so important for understanding motion that it was called the quantity of motion by physicists such as Newton. Force influences momentum, and we can rearrange Newton’s second law of motion to show the relationship between force and momentum.

Recall our study of Newton’s second law of motion ( F net = m a ). Newton actually stated his second law of motion in terms of momentum: The net external force equals the change in momentum of a system divided by the time over which it changes. The change in momentum is the difference between the final and initial values of momentum.

In equation form, this law is

where F net is the net external force, Δ p Δ p is the change in momentum, and Δ t Δ t is the change in time.

We can solve for Δ p Δ p by rearranging the equation

F net Δ t F net Δ t is known as impulse and this equation is known as the impulse-momentum theorem . From the equation, we see that the impulse equals the average net external force multiplied by the time this force acts. It is equal to the change in momentum. The effect of a force on an object depends on how long it acts, as well as the strength of the force. Impulse is a useful concept because it quantifies the effect of a force. A very large force acting for a short time can have a great effect on the momentum of an object, such as the force of a racket hitting a tennis ball. A small force could cause the same change in momentum, but it would have to act for a much longer time.

[OL] [AL] Explain that a large, fast-moving object has greater momentum than a smaller, slower object. This quality is called momentum.

[BL] [OL] Review the equation of Newton’s second law of motion. Point out the two different equations for the law.

Newton’s Second Law in Terms of Momentum

When Newton’s second law is expressed in terms of momentum, it can be used for solving problems where mass varies, since Δ p = Δ ( m v ) Δ p = Δ ( m v ) . In the more traditional form of the law that you are used to working with, mass is assumed to be constant. In fact, this traditional form is a special case of the law, where mass is constant. F net = m a F net = m a is actually derived from the equation:

For the sake of understanding the relationship between Newton’s second law in its two forms, let’s recreate the derivation of F net = m a F net = m a from

by substituting the definitions of acceleration and momentum.

The change in momentum Δ p Δ p is given by

If the mass of the system is constant, then

By substituting m Δ v m Δ v for Δ p Δ p , Newton’s second law of motion becomes

for a constant mass.

we can substitute to get the familiar equation

when the mass of the system is constant.

[BL] [OL] [AL] Show the two different forms of Newton’s second law and how one can be derived from the other.

Tips For Success

We just showed how F net = m a F net = m a applies only when the mass of the system is constant. An example of when this formula would not apply would be a moving rocket that burns enough fuel to significantly change the mass of the rocket. In this case, you can use Newton’s second law expressed in terms of momentum to account for the changing mass without having to know anything about the interaction force by the fuel on the rocket.

Hand Movement and Impulse

In this activity you will experiment with different types of hand motions to gain an intuitive understanding of the relationship between force, time, and impulse.

• one tub filled with water
• Try catching a ball while giving with the ball, pulling your hands toward your body.
• Next, try catching a ball while keeping your hands still.
• Hit water in a tub with your full palm. Your full palm represents a swimmer doing a belly flop.
• After the water has settled, hit the water again by diving your hand with your fingers first into the water. Your diving hand represents a swimmer doing a dive.
• Explain what happens in each case and why.
• a football player colliding with another, or a car moving at a constant velocity
• a car moving at a constant velocity, or an object moving in the projectile motion
• a car moving at a constant velocity, or a racket hitting a ball
• a football player colliding with another, or a racket hitting a ball

[OL] [AL] Discuss the impact one feels when one falls or jumps. List the factors that affect this impact.

Links To Physics

Engineering: saving lives using the concept of impulse.

Cars during the past several decades have gotten much safer. Seat belts play a major role in automobile safety by preventing people from flying into the windshield in the event of a crash. Other safety features, such as airbags, are less visible or obvious, but are also effective at making auto crashes less deadly (see Figure 8.2 ). Many of these safety features make use of the concept of impulse from physics. Recall that impulse is the net force multiplied by the duration of time of the impact. This was expressed mathematically as Δ p = F net Δ t Δ p = F net Δ t .

Airbags allow the net force on the occupants in the car to act over a much longer time when there is a sudden stop. The momentum change is the same for an occupant whether an airbag is deployed or not. But the force that brings the occupant to a stop will be much less if it acts over a larger time. By rearranging the equation for impulse to solve for force F net = Δ p Δ t , F net = Δ p Δ t , you can see how increasing Δ t Δ t while Δ p Δ p stays the same will decrease F net . This is another example of an inverse relationship. Similarly, a padded dashboard increases the time over which the force of impact acts, thereby reducing the force of impact.

Cars today have many plastic components. One advantage of plastics is their lighter weight, which results in better gas mileage. Another advantage is that a car will crumple in a collision , especially in the event of a head-on collision. A longer collision time means the force on the occupants of the car will be less. Deaths during car races decreased dramatically when the rigid frames of racing cars were replaced with parts that could crumple or collapse in the event of an accident.

Grasp Check

You may have heard the advice to bend your knees when jumping. In this example, a friend dares you to jump off of a park bench onto the ground without bending your knees. You, of course, refuse. Explain to your friend why this would be a foolish thing. Show it using the impulse-momentum theorem.

• Bending your knees increases the time of the impact, thus decreasing the force.
• Bending your knees decreases the time of the impact, thus decreasing the force.
• Bending your knees increases the time of the impact, thus increasing the force.
• Bending your knees decreases the time of the impact, thus increasing the force.

Solving Problems Using the Impulse-Momentum Theorem

Talk about the different strategies to be used while solving problems. Make sure that students know the assumptions made in each equation regarding certain quantities being constant or some quantities being negligible.

Worked Example

Calculating momentum: a football player and a football.

(a) Calculate the momentum of a 110 kg football player running at 8 m/s. (b) Compare the player’s momentum with the momentum of a 0.410 kg football thrown hard at a speed of 25 m/s.

No information is given about the direction of the football player or the football, so we can calculate only the magnitude of the momentum, p . (A symbol in italics represents magnitude.) In both parts of this example, the magnitude of momentum can be calculated directly from the definition of momentum:

To find the player’s momentum, substitute the known values for the player’s mass and speed into the equation.

To find the ball’s momentum, substitute the known values for the ball’s mass and speed into the equation.

The ratio of the player’s momentum to the ball’s momentum is

Although the ball has greater velocity, the player has a much greater mass. Therefore, the momentum of the player is about 86 times greater than the momentum of the football.

Calculating Force: Venus Williams’ Racquet

During the 2007 French Open, Venus Williams ( Figure 8.3 ) hit the fastest recorded serve in a premier women’s match, reaching a speed of 58 m/s (209 km/h). What was the average force exerted on the 0.057 kg tennis ball by Williams’ racquet? Assume that the ball’s speed just after impact was 58 m/s, the horizontal velocity before impact is negligible, and that the ball remained in contact with the racquet for 5 ms (milliseconds).

Recall that Newton’s second law stated in terms of momentum is

As noted above, when mass is constant, the change in momentum is given by

where v f is the final velocity and v i is the initial velocity. In this example, the velocity just after impact and the change in time are given, so after we solve for Δ p Δ p , we can use F net = Δ p Δ t F net = Δ p Δ t to find the force.

To determine the change in momentum, substitute the values for mass and the initial and final velocities into the equation above.

Now we can find the magnitude of the net external force using F net = Δ p Δ t F net = Δ p Δ t

This quantity was the average force exerted by Venus Williams’ racquet on the tennis ball during its brief impact. This problem could also be solved by first finding the acceleration and then using F net = m a , but we would have had to do one more step. In this case, using momentum was a shortcut.

Practice Problems

• 0.5 kg ⋅ m/s
• 15 kg ⋅ m/s
• 50 kg ⋅ m/s

A 155-g baseball is incoming at a velocity of 25 m/s. The batter hits the ball as shown in the image. The outgoing baseball has a velocity of 20 m/s at the angle shown.

What is the magnitudde of the impulse acting on the ball during the hit?

• 2.68 kg⋅m/s.
• 5.42 kg⋅m/s.
• 6.05 kg⋅m/s.
• 8.11 kg⋅m/s.

Check Your Understanding

What is linear momentum?

• the sum of a system’s mass and its velocity
• the ratio of a system’s mass to its velocity
• the product of a system’s mass and its velocity
• the product of a system’s moment of inertia and its velocity

If an object’s mass is constant, what is its momentum proportional to?

• Its velocity
• Its displacement
• Its moment of inertia

What is the equation for Newton’s second law of motion, in terms of mass, velocity, and time, when the mass of the system is constant?

• F net = Δ v Δ m Δ t F net = Δ v Δ m Δ t
• F net = m Δ t Δ v F net = m Δ t Δ v
• F net = m Δ v Δ t F net = m Δ v Δ t
• F net = Δ m Δ v Δ t F net = Δ m Δ v Δ t

Give an example of a system whose mass is not constant.

• A spinning top
• A baseball flying through the air
• A rocket launched from Earth
• A block sliding on a frictionless inclined plane

Use the Check Your Understanding questions to assess whether students master the learning objectives of this section. If students are struggling with a specific objective, the assessment will help identify which objective is causing the problem and direct students to the relevant content.

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Impulse and Momentum

Now, the bears I live with average, the males, eight to twelve hundred pounds [360 to 540 kg]. They're the largest bears in the world…. They've been clocked at 41 [mph] and they've run a hundred meter dash in 5.85 seconds, which a human on steroids doesn't even approach. Timothy Treadwell, 2001

• Compute the speed of a grizzly bear using Mr. Treadwell's hundred meter statement.
• Compute the momentum of a grizzly bear using the speed you calculated in part a. and the average mass stated by Mr. Treadwell.
• How fast would a 250 lb man have to run to have the same momentum you calculated in part b? (Do not use a calculator to compute your answer.)
• How fast would a 4000 lb car have to drive to have the same momentum you calculated in part b? (Do not use a calculator to compute your answer.)

For the first month of its journey, the Mars Orbiter Spacecraft actually orbited the Earth. After a series of orbit raising maneuvers, the tiny main engine gave the spacecraft enough speed in the right direction to escape. For the next nine months the main engine was only used twice and only very briefly (less than a minute of total burn time) to correct the spacecraft's trajectory.

Nine months after leaving Earth orbit, the spacecraft arrived at Mars. To enter orbit around Mars, the spacecraft needed to slow down — a maneuver called orbit insertion. Since the main engine hadn't been used much in nine months, a quick little test was needed.

Press Release: September 22, 2014 Mars Orbiter Spacecraft's Main Liquid Engine Successfully Test Fired The 440 newton Liquid Apogee Motor (LAM) of India's Mars Orbiter Spacecraft, last fired on December 01, 2013, was successfully fired for a duration of 3.968 seconds at 1430 hrs IST today (September 22, 2014). This operation of the spacecraft's main liquid engine was also used for the spacecraft's trajectory correction and changed its velocity by 2.18 metre/second. With this successful test firing, Mars Orbiter Insertion (MOI) operation of the spacecraft is scheduled to be performed on the morning of September 24, 2014 at 07:17:32 hrs IST by firing the LAM along with eight smaller liquid engines for a duration of about 24 minutes. ISRO, 2014

• Using the press release, determine the mass of the Mars Orbiter Spacecraft at the time of this test firing.
• Why was the qualifying phrase "at the time of this test firing" added to the end of the previous question? Why does the mass of the spacecraft vary with time?

After all, the faster we go, the more difficult it is to avoid collisions with small objects and the more damage such a collision will wreak. Even if we are fortunate enough to miss all sizable objects, we can scarcely expect to miss the dust and individual atoms that are scattered throughout space. At two-tenths of the speed of light, dust and atoms might not do significant damage even in a voyage of 40 years, but the faster you go, the worse it is — space begins to become abrasive. When you begin to approach the speed of light, each hydrogen atom becomes a cosmic ray particle and will fry the crew. (A hydrogen atom or its nucleus striking the ship at nearly the speed of light is a cosmic ray particle, and there is no difference if the ship strikes the hydrogen atom or nucleus at nearly the speed of light. As Sancho Panza: "Whether the stone strikes the pitcher, or the pitcher strikes the stone, it is bad for the pitcher.") So 60,000 kilometers per second may be the practical speed limit for space travel. Isaac Asimov, 1987

The density of the interstellar medium is about one hydrogen atom per cubic centimeter. Imagine a 1,000 tonne, 4 by 6 meter, classroom-sized interstellar spacecraft traveling at 60,000 km/s on its way to Proxima Centauri (the nearest solar system to our own) 4.243 light years away.

Kinematics:

• How long would it take our hypothetical spacecraft to complete its hypothetical journey?

Impulse-Momentum:

• Determine the momentum of our spacecraft.
• What mass of interstellar medium is swept up during the journey?
• What impulse does the interstellar medium deliver to the spacecraft?
• How does this impulse compare to the momentum of the spacecraft?

Work-Energy:

• Determine the kinetic energy of our spacecraft.
• What is the effective drag force of the interstellar medium during the journey?
• How much work does the interstellar medium do on the spacecraft?
• How does this work compare to the kinetic energy of the spacecraft?
• Write something.
• In older passenger cars, body panels were attached to a single frame around the perimeter, making them very rigid. This is known as body-over-frame construction. In newer cars, different body parts have stress-bearing elements within them and these parts are then welded to each other. This is known as unitized body construction. Repairing "unibody" cars after collision is comparatively difficult as stress (and thus damage) are distributed throughout the different parts. Why then are cars now built this way
• To escape from a horrible fire, two people are forced to jump from the third story of a burning building on to solid concrete. Which person is more likely to sustain serious injuries: the jumper who comes to an abrupt halt when he lands or the jumper who bounces after impact?
• What is the basic idea behind crash safety features in cars like seatbelts, airbags, crumple zones, etc. What quantities in the impulse-momentum theorem ( F ∆ t  =  m ∆ v ) change as a result of these features, how are they changed, and how does this result in increased safety during a crash?
• The phrase "roll with the punches" has its origin in boxing. What does it mean to roll with a punch (or ride a punch)? How does rolling reduce the severity of a punch?
• Is it possible for a motorcycle to have more momentum (that is, a momentum with a larger magnitude) than a train?
• When hit, the velocity of a 0.145 kg baseball changes from +20 m/s to −20 m/s. What is the magnitude of the impulse delivered by the bat to the ball?
• the ground is soft and the ball stops dead
• the ground is hard and the ball bounces straight up at 2.0 m/s
• Draw a free body diagram showing all the forces acting on the model rocket.
• the weight of the rocket
• the net force on the rocket while the engine was running
• the net impulse on the rocket while the engine was running
• the speed of the rocket when the engine stopped
• the height of the rocket above the ground when the engine stopped
• What is the acceleration of the rocket after the engine shut down?
• What maximum height above the ground did the rocket reach?
• What does the area under this curve represent?
• Calculate its cumulative value at 2 s intervals. Compile your results in a table like the one below.
• Sketch a graph of this quantity with respect to time.

statistical

• Use the given data to create a force-time graph.
• Determine the impulse acting on the projectile as a function of time.
• Compute the launch speed of the projectile.

Data adapted from Kampen, Kaczmarczik, and Rath; 2006 .

investigative

• Which plate have you chosen to work with?
• What is the speed of this plate ?
• What is the area of this plate?
• Is the plate you've chosen continental or oceanic?
• What is a typical density of this kind of crust?
• What is a typical thickness of this type of crust?
• Compute the momentum of the tectonic plate you've chosen from the data you've found. State your answer to the nearest order of magnitude (the nearest power of ten). Don't forget the unit.
• the thrust-time data used to generate the graph on the data sheet
• the propellant mass
• the mass after firing (i.e., the mass of the empty rocket)
• impulse provided by the motor
• fractional propellant mass loss (You need to determine the initial and final mass of the rocket. Assume that the loss of mass is directly proportional to the cumulative impulse. When the impulse is zero at the beginning, the mass loss is zero. When the impulse reaches its final value, the mass loss is 100%.)
• mass of the rocket
• speed of the rocket (Don't forget to include the force of gravity in your calculations.)
• acceleration of the rocket
• altitude of the rocket

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Conservation of Momentum Example Problem

Momentum is a measurement of inertia in motion. When a mass has velocity, it has momentum. Momentum is calculated by the equation

momentum = mass x velocity momentum = mv

This conservation of momentum example problem illustrates the principle of conservation of momentum after a collision between two objects.

Consider a 42,000 kg train car travelling at 10 m/s toward another train car. After the two cars collide, they couple together and move along at 6 m/s. What is the mass of the second train car?

In a closed system, momentum is conserved. This means the total momentum of the system remains unchanged before and after the collision.

Before the collision, the total momentum was the sum of the momentums of both train cars.

The first car’s (blue freight car) momentum is

momentum Blue = mv momentum Blue = (42,000 kg)(10 m/s) momentum Blue = 420,000 kg·m/s

The second car’s (tanker car) momentum is

momentum tanker = mv momentum tanker = m(0 m/s) momentum tanker = 0 kg·m/s

Add these two together to get the total momentum of the system prior to collision.

Total momentum = momentum Blue + momentum tanker Total momentum = 420,000 kg·m/s + 0 kg·m/s Total momentum = 420,000 kg·m/s

After the collision, the two cars couple together into one mass. This new mass is

mass Blue + mass tanker 42,000 kg + mass tanker

The velocity the new pair of cars is traveling is 6 m/s. Because momentum is conserved, we know the total momentum of the cars after the collision is equal to the momentum prior to the collision.

Total Momentum = 420,000 kg·m/s Total Momentum = mv Total momentum = (42,000 kg + mass tanker )·(6 m/s)

420,000 kg·m/s = (42,000 kg + mass tanker )·(6 m/s)

Divide both sides by 6 m/s

70,000 kg = 42,000 kg + mass tanker

Subtract 42,000 kg from both sides

70,000 kg – 42,000 kg = mass tanker 28,000 kg = mass tanker

The mass of the second car is equal to 28,000 kg.

Remember, the momentum of a system is conserved. The momentum of the individual masses may change, but the net momentum of the system does not change.

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Home » Science » Physics » How to Solve Momentum Problems

How to Solve Momentum Problems

Here, we will look at how to solve momentum problems in both one and two dimensions using the law of conservation of linear momentum . According to this law, the total momentum of a system of particles remains constant as long as no external forces act on them. Therefore, solving momentum problems involve calculating the total momentum of a system before and after an interaction, and equating the two.

1D Momentum Problems

A ball with a mass of 0.75 kg travelling at a speed of 5.8 m s -1 collides with another ball of mass 0.90 kg, also travelling in the same distance at a speed of 2.5 m s -1 . After the collision, the lighter ball travels at a speed of 3.0 m s -1 in the same direction. Find the velocity of the larger ball.

How to Solve Momentum Problems – Example 1

  Example 2

An object of mass 0.32 kg traveling at a speed of  5 m s -1  collides with a stationary object having a mass of 0.90 kg. After the collision, the two particles stick and travel together. Find at which speed they travel.

A bullet having a mass of 0.015 kg is fired off a 2 kg gun. Immediately after firing, the bullet is travelling at a speed of 300 m s -1 . Find the recoil speed of the gun, assuming the gun was stationary before firing the bullet.

We took the bullet’s direction to be positive. So, the negative sign indicates that the gun is travelling in the answer indicates that the gun is travelling in the opposite direction.

Example 4: The ballistic Pendulum

From conservation of momentum, we have:

From conservation of energy, we have:

2D Momentum Problems

How to Solve Momentum Problems – Example 5

Show that for an oblique collision (a “glancing blow”) when a body collides elastically  with another body having the same mass at rest , the two bodies would move off at an angle of 90 o between them.

How to Solve Momentum Problems – Example 6

How to Solve Momentum Problems – Example 6 Velocity vector Triangle

We know the collision is elastic. Then,

Canceling out the common factors, we get:

About the Author: Nipun

​you may also like these.

Impulse and Momentum

by Gary Gende and Owen "O-Dog" Zahorcak, January 1998

Table of Contents

New quantities.

• General Problem Solving Strategy
• Example Problem #1
• Example Problem #2

Sample Problems

Solutions to sample problems.

• Go back to Tutorial Page

Here's the quantities you can know:

• /\ Delta (aka the change)
• /\ p Impulse

These quantities are defined and explained on other pages execept Impulse and momentum. They are defined below.

Momentum is defined as the product of an object's mass and velocity This means p(momentum)=m(mass)*V(velocity) by the definition of momentum Momentum is measured in N(Newton)m(meters)/s(per second)

Impulse is the product of force and time. As this suggests the unit for measuring impulse is N(Newton)s(seconds) A formula for finding impulse is shown and explained below.

From other chapters we know that

We can use these formulas to show that /\ p(Impulse)=Ft

• F = ma....it just does
• F = m( /\ V/t) ....a= /\ a/t
• F = (m(V-V o ))/t = (mV-mV o )/t.... /\ V = V-V o and The Distributive Property
• F = (p-p o )/ /\ t = /\ p/t....p=mv and /\ p = p-p o
• /\ p = Ft....multiply both sides by t

*Several of the formula in the work above become important*

F= /\ p/t is Newton's Second Law of Motion.

So now we have all the formulas we need for solving impulse and momentum problems:

• F = m( /\ V/t)....(From above)

Go back to: Table of Contents

General Problem Solving Strategy:

• Read the problem.
• Go through the problem and figure out what is given or implied Make a list, and identify the quantities you know.
• Find any formula that will allow you to calculate anything that you don't know, and apply it.
• Add what you just found in the last step to your list of knowns.
• Check to see if you have found the answer. If not, repeat the previous two steps until you are done.

Mother of all Example Problems

Joe the alien fires a 20 Kg photon torpedo at 200 m/s from his stationary UFO in deep space. The torpedo has special no fuel engines that can give 650 N of force. What is the momentum of the torpedo? How long must Joe fire the engines to get 3000 Ns of impulse. What speed is Joe attempting to get the torpedo to go? If Joe is firing at you and your UFO can accelerate at a rate of 30 m/s/s, will you be able to out accellerate the torpedo?

a. Momentum is just p=mv so plug in the known m (20) and V (200) 20 * 200 = 4000 Kgm/s

b. The needed formula is /\ p = Ft. F (650) and /\ p (1500) are given. divide both sides of the equation by F to get /\ p/F = t then plug in the known values to get 3000/650 = t = 4.62 Ns

c. The needed formula now is F = m( /\ V/t). F is still 650 N, the torpedo still has a mass of 20 Kg and time can be found from part b. When you plug them in you get 650 = 20( /\ V/2.31). Multiplying both sides by 2.31/20 gives you (4.62 * 650)/20 = /\ V = 150 m/s. Since this is the change in speed (hence the /\ ), 150 needs to be added to the original 200. The final speed is 350 m/s.

d. F = ma is the importanr equation now. You know F (650 N) and m (20 Kg) so just divide both sides of the equation by m to solve it for a and plug in the numbers to get 650/20 = a = 32.5 m/s/s. So you better hope you have a big enough head start.

The answers to each problem follow it in parentheses. They also link to a solution to the problem. Try the problem, check your answer, and go to the solution if you do not understand.

What is the momentum of a 23 Kg cannon shell going 530 m/s?

Momentum: Explanation, Review, and Examples

• The Albert Team
• Last Updated On: September 21, 2023

Momentum is a term you’ve likely heard used in everyday language that also has a special meaning in physics. When a sports team is described as having “momentum”, you know that the team is on a winning streak and hard to stop. Momentum in physics has a similar but more precise meaning. This post will explain what momentum is, the equation for momentum, and how to calculate momentum.  

What We Review

What Is Momentum In Physics?

Definition of momentum.

In Physics, momentum is a quantity of motion that applies to moving objects. Every object that is in motion has momentum. As we’ll explain in further detail below, how much momentum a moving object has depends on its mass and velocity. 

How Momentum Relates to Velocity and Mass

Momentum is the product of an object’s mass and velocity. This means that momentum is directly proportional to both mass and velocity. The larger the mass of the object, the more momentum it has. Similarly, for velocity, objects that are moving faster also have more momentum. 

For example, a heavy truck traveling on the highway has more momentum than a smaller car traveling at the same speed because it has a greater mass. Having more momentum also makes it harder for the truck to stop. 

An object’s momentum can also change as its motion changes. For example, a child sledding down a steep hill gets faster and faster as they slide down. As the child’s velocity increases, they also gain momentum.  

Check out this video from Conceptual Academy for more examples of momentum. 

Momentum Units

The units for momentum come from the equation for momentum and the fact that momentum is the product of mass and velocity. The standard units for mass are kilograms (kg) and the units for velocity are meters per second (m/s). Therefore, momentum is measured in units of \text{kg}\cdot\text{m/s} . 

Is Momentum a Vector?

As a reminder, vectors are quantities that have both a magnitude and direction. As you’ll recall from your study of kinematics , velocity is a vector quantity because it tells you both how fast an object is moving and the direction it is moving in, such as 25\text{m/s} East. Since momentum depends on velocity, momentum is also a vector.  

How To Calculate Momentum

Equation for momentum.

The equation for momentum is:

…where:

• p is momentum
• v is velocity

Step-By-Step Guide for Using the Equation for Momentum

When using this equation to calculate momentum, you’ll want to take the following steps:

• Identify the mass, in kilograms.
• Identify the velocity, in meters per second. 
• Substitute these values into the equation p=mv .
• Solve by multiplying the mass and the velocity. 

For example, say a problem asks you to calculate the momentum of a 15\text{ kg} object moving at 6\text{ m/s}\text{ North} . Applying the steps above produces:

• The mass, m , of the object is 15\text{ kg} .
• The velocity, v , of the object is 6\text{ m/s}\text{ North} .
• Substituting these values into the momentum equation gives: p=(15\text{ kg})(6\text{ m/s}\text{ North}) .
• Multiplying mass and velocity gives: p=90\text{kg}\cdot\text{ m/s}\text{ North} .

How To Find Momentum Using Examples

Applying the equation for momentum to real-world situations.

We can find momentum in a variety of real-world situations whenever we know mass and velocity. 

For example, if you are a baseball catcher, you might be curious about the momentum of the fastballs you catch. The average fastball pitch is around 90\text{ mph} or 40\text{ m/s} . The mass of a baseball is about 5\text{ ounces} or 0.145\text{ kg} . Substituting these values into the momentum equation produces:

Another real-world example where momentum has important applications is with cars. Cars often have a large amount of momentum due to their large mass and the ability to move at high velocities. And when a car has a lot of momentum, it can be harder for it to come to a stop in an emergency or accident. For example, an average-sized car can have a mass of about 1{,}800\text{ kg} . If the car is traveling at 32\text{ m/s} , (about 60\text{ mph} ), its momentum is:

Momentum Experiments

You can see momentum in action, and calculate momentum, in a variety of experiments. To calculate the momentum of an object in an experiment, you will need a scale to measure the object’s mass, a ruler to measure distance, and a stopwatch to measure the time. By measuring the distance and time an object travels, you can calculate its velocity using the formula: v=\frac{d}{t} . 

For example, you can do an experiment at home by rolling a ball down a ramp and having it hit a stationary object. You’ll notice that the steeper the ramp is, the more velocity and momentum the ball has, and the bigger impact it has on the stationary object it hits. If you compare two balls with different masses rolling down the same ramp, they have the same velocity but the heavier ball has more momentum and a bigger impact on the object it hits. 

Momentum is a quantity of motion that depends on an object’s mass and velocity. You now know how to calculate the momentum of any object and can apply this to find momentum in many everyday situations. As you’ll learn in future posts, knowing how to calculate momentum can provide important insights into collisions and forces.

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Ex 1)   A 5.0 kg mass has its velocity change from 8.0 m/s east to 2.0 m/s east.  Find the objects change in momentum .

m = 5.0 kg V i = 8.0 m/s East V f = 2.0 m/s East Δp = ?

Δp = mΔV

= (5.0 kg)(2.0 m/s - 8.0 m/s)

= -30. kgm/s East

or +30. kg m/s West

Ex 2) A 5.0 kg mass moving with a vector of 8.0 m/s east has an impulse applied to it which causes its velocity to change to 20. m/s East .

Find Impulse :

m = 5.0 kg V i = 8.0 m/s East V f = 20. m/s East J = ?

J = mΔv = (5.0 kg)(12. m/s East)

= 60. kg m/s east

= 60. Ns East

Find the force if the impulse was applied for 3.0 sec.

t = 3 seconds

V i = 8.0 m/s East

V f = 20. m/s East

J = 60. kg m/s east

J = Ft = 60. Ns East

F(3.0 sec) = 60. Ns East

F = 20. N East

Golf Ball hitting steel super slo mo

School Blocks YouTube? Click below.

GolfBallhittingsuperslomo

Ex 3) How long would it take for a net upward force of 100. N , to increase the speed of a 50. kg object from 100. m/s to 150. m/s .

V i = 100. m/s

V f = 150. m/s

FΔt  = mΔv

(100. N)t  =  50. kg(50. m/s)

t  =  25. secs

  Racquetball Slow-Mo - HiViz

Ex 4)   A 1.0 kg ball traveling @ 4.0  m/s strikes a wall and bounces straight back @ 2.0 m/s . 

Find Δp  

V i = 4.0 m/s

V f = -2.0 m/s

(opposite direction)

Δp = ?

(a) Δp = mΔv

= (1.0 kg)(-2.0 m/s - 4.0 m/s)

= - 6.0 kgm/s

(b) What is impulse applied to the ball?   

J =  Δp = -6.0 kgm/s

(c) What is impulse applied to the wall ?

J = +6.0 kgm/s

3rd Law, Action Reaction

Computer animation of a Newtons' cradle

DemonDeLuxe (Dominique Toussaint)

World's Lightest Material 200 times lighter than styrofoam

HRL Laboratories - See the video

Using an innovative fabrication process developed at HRL, researchers created a “micro-lattice” structure of interconnected hollow tubes with a wall thickness of 100 nanometers, 1,000 times thinner than a human hair. Download   high resolution image. High resolution video is available upon request by emailing   [email protected] . Photo credit: Photo by Dan Little © HRL Laboratories,  

D p = m D v

Momentum only changes in the x direction

toward wall +

V i = Vsin q

V f = -Vsin q

= 2.1 kgm/s left

©Tony Mangiacapre., - All Rights Reserved [ Home] Established 1995 Use any material on this site (w/ attribution)

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9.6: Conservation of Linear Momentum (Part 2)

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Problem-Solving Strategy: Conservation of Momentum

Using conservation of momentum requires four basic steps. The first step is crucial:

• Identify a closed system (total mass is constant, no net external force acts on the system).
• Write down an expression representing the total momentum of the system before the “event” (explosion or collision).
• Write down an expression representing the total momentum of the system after the “event.”
• Set these two expressions equal to each other, and solve this equation for the desired quantity

Example \(\PageIndex{1}\): Colliding Carts

Two carts in a physics lab roll on a level track, with negligible friction. These carts have small magnets at their ends, so that when they collide, they stick together (Figure \(\PageIndex{1}\)). The first cart has a mass of 675 grams and is rolling at 0.75 m/s to the right; the second has a mass of 500 grams and is rolling at 1.33 m/s, also to the right. After the collision, what is the velocity of the two joined carts?

We have a collision. We’re given masses and initial velocities; we’re asked for the final velocity. This all suggests using conservation of momentum as a method of solution. However, we can only use it if we have a closed system. So we need to be sure that the system we choose has no net external force on it, and that its mass is not changed by the collision.

Defining the system to be the two carts meets the requirements for a closed system: The combined mass of the two carts certainly doesn’t change, and while the carts definitely exert forces on each other, those forces are internal to the system, so they do not change the momentum of the system as a whole. In the vertical direction, the weights of the carts are canceled by the normal forces on the carts from the track.

Conservation of momentum is

\[\vec{p}_{f} = \vec{p}_{i} \ldotp \nonumber\]

Define the direction of their initial velocity vectors to be the +x-direction. The initial momentum is then

\[\vec{p}_{i} = m_{1} v_{1}\; \hat{i} + m_{2} v_{2}\; \hat{i} \ldotp \nonumber\]

The final momentum of the now-linked carts is

\[\vec{p}_{f} = (m_{1} + m_{2}) \vec{v}_{f} \ldotp \nonumber\]

\[\begin{align*} (m_{1} + m_{2}) \vec{v}_{f} & = m_{1} v_{1}\; \hat{i} + m_{2} v_{2}\; \hat{i} \\[4pt] \vec{v}_{f} & = \left(\dfrac{m_{1} v_{1} + m_{2} v_{2}}{m_{1} + m_{2}}\right) \hat{i} \ldotp \end{align*}\]

Substituting the given numbers:

\[\begin{align*} \vec{v}_{f} & = \Bigg[ \frac{(0.675\; kg)(0.75\; m/s) + (0.5\; kg)(1.33\; m/s)}{1.175\; kg} \Bigg] \hat{i} \\[4pt] & = (0.997\; m/s) \hat{i} \ldotp \end{align*}\]

Significance

The principles that apply here to two laboratory carts apply identically to all objects of whatever type or size. Even for photons, the concepts of momentum and conservation of momentum are still crucially important even at that scale. (Since they are massless, the momentum of a photon is defined very differently from the momentum of ordinary objects. You will learn about this when you study quantum physics.)

Exercise\(\PageIndex{1}\)

Suppose the second, smaller cart had been initially moving to the left. What would the sign of the final velocity have been in this case?

Example \(\PageIndex{2}\): A Bouncing Superball

A superball of mass 0.25 kg is dropped from rest from a height of h = 1.50 m above the floor. It bounces with no loss of energy and returns to its initial height (Figure \(\PageIndex{2}\)).

• What is the superball’s change of momentum during its bounce on the floor?
• What was Earth’s change of momentum due to the ball colliding with the floor?
• What was Earth’s change of velocity as a result of this collision?

(This example shows that you have to be careful about defining your system.)

Since we are asked only about the ball’s change of momentum, we define our system to be the ball. But this is clearly not a closed system; gravity applies a downward force on the ball while it is falling, and the normal force from the floor applies a force during the bounce. Thus, we cannot use conservation of momentum as a strategy. Instead, we simply determine the ball’s momentum just before it collides with the floor and just after, and calculate the difference. We have the ball’s mass, so we need its velocities.

• p 0 = the magnitude of the ball’s momentum at time t 0 , the moment it was released; since it was dropped from rest, this is zero.
• p 1 = the magnitude of the ball’s momentum at time t 1 , the instant just before it hits the floor.
• p 2 = the magnitude of the ball’s momentum at time t 2 , just after it loses contact with the floor after the bounce.

The ball’s change of momentum is

\[\begin{align*} \Delta \vec{p} & = \vec{p}_{2} - \vec{p}_{1} \\[4pt] & = p_{2}\; \hat{j} - (-p_{1}\; \hat{j}) \\[4pt] & = (p_{2} + p_{1}) \hat{j} \ldotp \end{align*}\]

Its velocity just before it hits the floor can be determined from either conservation of energy or kinematics. We use kinematics here; you should re-solve it using conservation of energy and confirm you get the same result.

We want the velocity just before it hits the ground (at time t 1 ). We know its initial velocity v 0 = 0 (at time t 0 ), the height it falls, and its acceleration; we don’t know the fall time. We could calculate that, but instead we use

\[\vec{v}_{1} = - \hat{j} \sqrt{2gy} = -5.4\; m/s\; \hat{j} \ldotp \nonumber\]

Thus the ball has a momentum of

\[\begin{align*} \vec{p}_{1} & = - (0.25\; kg)(-5.4\; m/s\; \hat{j}) \\[4pt] & = - (1.4\; kg\; \cdotp m/s) \hat{j} \ldotp \end{align*}\]

We don’t have an easy way to calculate the momentum after the bounce. Instead, we reason from the symmetry of the situation.

Before the bounce, the ball starts with zero velocity and falls 1.50 m under the influence of gravity, achieving some amount of momentum just before it hits the ground. On the return trip (after the bounce), it starts with some amount of momentum, rises the same 1.50 m it fell, and ends with zero velocity. Thus, the motion after the bounce was the mirror image of the motion before the bounce. From this symmetry, it must be true that the ball’s momentum after the bounce must be equal and opposite to its momentum before the bounce. (This is a subtle but crucial argument; make sure you understand it before you go on.) Therefore,

\[\vec{p}_{2} = - \vec{p}_{1} = + (1.4\; kg\; \cdotp m/s) \hat{j} \ldotp \nonumber\]

Thus, the ball’s change of momentum during the bounce is

\[\begin{align*} \Delta \vec{p} & = \vec{p}_{2} - \vec{p}_{1} \\ & = (1.4\; kg\; \cdotp m/s) \hat{j} - (-1.4\; kg\; \cdotp m/s) \hat{j} \\ & = + (2.8\; kg\; \cdotp m/s) \hat{j} \ldotp \end{align*}\]

• What was Earth’s change of momentum due to the ball colliding with the floor? Your instinctive response may well have been either “zero; the Earth is just too massive for that tiny ball to have affected it” or possibly, “more than zero, but utterly negligible.” But no—if we re-define our system to be the Superball + Earth, then this system is closed (neglecting the gravitational pulls of the Sun, the Moon, and the other planets in the solar system), and therefore the total change of momentum of this new system must be zero. Therefore, Earth’s change of momentum is exactly the same magnitude: $$\Delta \vec{p}_{Earth} = -2.8\; kg\; \cdotp m/s\; \hat{j} \ldotp$$
• What was Earth’s change of velocity as a result of this collision? This is where your instinctive feeling is probably correct: \[\begin{align*} \Delta \vec{v}_{Earth} & = \frac{\Delta \vec{p}_{Earth}}{M_{Earth}} \\[4pt] & = - \frac{2.8\; kg\; \cdotp m/s}{5.97 \times 10^{24}\; kg}\; \hat{j} \\[4pt] & = - (4.7 \times 10^{-25}\; m/s) \hat{j} \ldotp \end{align*}\] This change of Earth’s velocity is utterly negligible

It is important to realize that the answer to part (c) is not a velocity; it is a change of velocity, which is a very different thing. Nevertheless, to give you a feel for just how small that change of velocity is, suppose you were moving with a velocity of 4.7 x 10 −25 m/s. At this speed, it would take you about 7 million years to travel a distance equal to the diameter of a hydrogen atom.

Exercise \(\PageIndex{2}\)

Would the ball’s change of momentum have been larger, smaller, or the same, if it had collided with the floor and stopped (without bouncing)? Would the ball’s change of momentum have been larger, smaller, or the same, if it had collided with the floor and stopped (without bouncing)?

Example \(\PageIndex{3}\): Ice hockey 1

Two hockey pucks of identical mass are on a flat, horizontal ice hockey rink. The red puck is motionless; the blue puck is moving at 2.5 m/s to the left (Figure \(\PageIndex{3}\)). It collides with the motionless red puck. The pucks have a mass of 15 g. After the collision, the red puck is moving at 2.5 m/s, to the left. What is the final velocity of the blue puck?

We’re told that we have two colliding objects, we’re told the masses and initial velocities, and one final velocity; we’re asked for both final velocities. Conservation of momentum seems like a good strategy. Define the system to be the two pucks; there’s no friction, so we have a closed system.

Before you look at the solution, what do you think the answer will be?

The blue puck final velocity will be:

• 2.5 m/s to the left
• 2.5 m/s to the right
• 1.25 m/s to the left
• 1.25 m/s to the right
• something else

Define the +x-direction to point to the right. Conservation of momentum then reads

\[\begin{align*} \vec{p_{f}} & = \vec{p_{i}} \\ mv_{r_{f}}\; \hat{i} + mv_{b_{f}}\; \hat{i} & = mv_{r_{i}}\; \hat{i} - mv_{b_{i}}\; \hat{i} \ldotp \end{align*}\]

Before the collision, the momentum of the system is entirely and only in the blue puck. Thus,

\[mv_{r_{f}}\; \hat{i} + mv_{b_{f}}\; \hat{i} = - mv_{b_{i}}\; \hat{i} \nonumber\]

\[v_{r_{f}}\; \hat{i} + v_{b_{f}}\; \hat{i} = - v_{b_{i}}\; \hat{i} \ldotp \nonumber\]

(Remember that the masses of the pucks are equal.) Substituting numbers:

\[\begin{align*} - (2.5\; m/s) \hat{i} + \vec{v}_{b_{f}} & = - (2.5\; m/s) \hat{i} \\ \vec{v}_{b_{f}} & = 0 \ldotp \end{align*}\]

Evidently, the two pucks simply exchanged momentum. The blue puck transferred all of its momentum to the red puck. In fact, this is what happens in similar collision where m 1 = m 2 .

Exercise \(\PageIndex{3}\)

Even if there were some friction on the ice, it is still possible to use conservation of momentum to solve this problem, but you would need to impose an additional condition on the problem. What is that additional condition?

On November 12, 2014, the European Space Agency successfully landed a probe named Philae on Comet 67P/ Churyumov/Gerasimenko (Figure \(\PageIndex{4}\)). During the landing, however, the probe actually landed three times, because it bounced twice. Let’s calculate how much the comet’s speed changed as a result of the first bounce.

Let’s define upward to be the +y-direction, perpendicular to the surface of the comet, and y = 0 to be at the surface of the comet. Here’s what we know:

• The mass of Comet 67P: M c = 1.0 x 10 13 kg
• The acceleration due to the comet’s gravity: \(\vec{a}\) = −(5.0 x 10 −3 m/s 2 ) \(\hat{j}\)
• Philae ’s mass: M p = 96 kg
• Initial touchdown speed: \(\vec{v}_{1}\) = −(1.0 m/s) \(\hat{j}\)
• Initial upward speed due to first bounce: \(\vec{v}_{2}\) = (0.38 m/s) \(\hat{j}\)
• Landing impact time: \(\Delta\)t = 1.3 s

We’re asked for how much the comet’s speed changed, but we don’t know much about the comet, beyond its mass and the acceleration its gravity causes. However, we are told that the Philae lander collides with (lands on) the comet, and bounces off of it. A collision suggests momentum as a strategy for solving this problem.

If we define a system that consists of both Philae and Comet 67/P, then there is no net external force on this system, and thus the momentum of this system is conserved. (We’ll neglect the gravitational force of the sun.) Thus, if we calculate the change of momentum of the lander, we automatically have the change of momentum of the comet. Also, the comet’s change of velocity is directly related to its change of momentum as a result of the lander “colliding” with it.

Let \(\vec{p}_{1}\) be Philae ’s momentum at the moment just before touchdown, and \(\vec{p}_{2}\) be its momentum just after the first bounce. Then its momentum just before landing was

\[\vec{p}_{1} = M_{p} \vec{v}_{1} = (96\; kg)(-1.0\; m/s\; \hat{j}) = - (96\; kg\; \cdotp m/s) \hat{j} \nonumber\]

and just after was

\[\vec{p}_{2} = M_{p} \vec{v}_{2} = (96\; kg)(+0.38\; m/s\; \hat{j}) = (36.5\; kg\; \cdotp m/s) \hat{j} \ldotp \nonumber\]

Therefore, the lander’s change of momentum during the first bounce is

\[\begin{align*} \Delta \vec{p} & = \vec{p}_{2} \vec{p}_{1} \\ & = (36.5\; kg\; \cdotp m/s) \hat{j} - (-96.0\; kg\; \cdotp m/s\; \hat{j}) \\ & = (133\; kg\; \cdotp m/s) \hat{j} \end{align*}\]

Notice how important it is to include the negative sign of the initial momentum.

Now for the comet. Since momentum of the system must be conserved, the comet’s momentum changed by exactly the negative of this:

\[\Delta \vec{p}_{c} = - \Delta \vec{p} = - (133\; kg\; \cdotp m/s) \hat{j} \ldotp \nonumber\]

Therefore, its change of velocity is

\[\Delta \vec{v}_{c} = \frac{\Delta \vec{p}_{c}}{M_{c}} = \frac{-(133\; kg\; \cdotp m/s) \hat{j}}{1.0 \times 10^{13}\; kg} = - (1.33 \times 10^{-11}\; m/s) \hat{j} \ldotp \nonumber\]

This is a very small change in velocity, about a thousandth of a billionth of a meter per second. Crucially, however, it is not zero.

Exercise \(\PageIndex{4}\)

The changes of momentum for Philae and for Comet 67/P were equal (in magnitude). Were the impulses experienced by Philae and the comet equal? How about the forces? How about the changes of kinetic energies?

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The sports announcer says, "Going into the all-star break, the Chicago White Sox have the momentum ." The headlines declare "Chicago Bulls Gaining Momentum ." The coach pumps up his team at half-time, saying "You have the momentum ; the critical need is that you use that momentum and bury them in this third quarter."

Momentum is a commonly used term in sports. A team that has the momentum is on the move and is going to take some effort to stop. A team that has a lot of momentum is really on the move and is going to be hard to stop . Momentum is a physics term; it refers to the quantity of motion that an object has. A sports team that is on the move has the momentum. If an object is in motion ( on the move ) then it has momentum.

In physics, the symbol for the quantity momentum is the lower case p . Thus, the above equation can be rewritten as

where m is the mass and v is the velocity. The equation illustrates that momentum is directly proportional to an object's mass and directly proportional to the object's velocity.

The units for momentum would be mass units times velocity units. The standard metric unit of momentum is the kg•m/s. While the kg•m/s is the standard metric unit of momentum, there are a variety of other units that are acceptable (though not conventional) units of momentum. Examples include kg•mi/hr, kg•km/hr, and g•cm/s. In each of these examples, a mass unit is multiplied by a velocity unit to provide a momentum unit. This is consistent with the equation for momentum.  

Momentum as a Vector Quantity

The momentum equation as a guide to thinking.

From the definition of momentum, it becomes obvious that an object has a large momentum if both its mass and its velocity are large. Both variables are of equal importance in determining the momentum of an object. Consider a Mack truck and a roller skate moving down the street at the same speed. The considerably greater mass of the Mack truck gives it a considerably greater momentum. Yet if the Mack truck were at rest, then the momentum of the least massive roller skate would be the greatest. The momentum of any object that is at rest is 0. Objects at rest do not have momentum - they do not have any " mass in motion ." Both variables - mass and velocity - are important in comparing the momentum of two objects.

The momentum equation can help us to think about how a change in one of the two variables might affect the momentum of an object. Consider a 0.5-kg physics cart loaded with one 0.5-kg brick and moving with a speed of 2.0 m/s. The total mass of loaded cart is 1.0 kg and its momentum is 2.0 kg•m/s. If the cart was instead loaded with three 0.5-kg bricks, then the total mass of the loaded cart would be 2.0 kg and its momentum would be 4.0 kg•m/s. A doubling of the mass results in a doubling of the momentum.

Check Your Understanding

Express your understanding of the concept and mathematics of momentum by answering the following questions. Click the button to view the answers.

1. Determine the momentum of a ...

a. 60-kg halfback moving eastward at 9 m/s. b. 1000-kg car moving northward at 20 m/s. c. 40-kg freshman moving southward at 2 m/s.

A. p = m*v = 60 kg*9 m/s

p = 540 kg•m/s, east

B. p = m*v = 1000 kg*20 m/s

p = 20 000 kg•m/s, north

C. p = m*v = 40 kg*2 m/s

p = 80 kg•m/s, south

2. A car possesses 20 000 units of momentum. What would be the car's new momentum if ...

a. its velocity was doubled. b. its velocity was tripled. c. its mass was doubled (by adding more passengers and a greater load) d. both its velocity was doubled and its mass was doubled.

A. p = 40 000 units (doubling the velocity will double the momentum)

B. p = 60 000 units (tripling the velocity will triple the momentum)

C. p = 40 000 units (doubling the mass will double the momentum)

D. p = 80 000 units (doubling the velocity will double the momentum and doubling the mass will also double the momentum; the combined result is that the momentum is doubled twice -quadrupled)

3. A halfback (m = 60 kg), a tight end (m = 90 kg), and a lineman (m = 120 kg) are running down the football field. Consider their ticker tape patterns below.

Compare the velocities of these three players. How many times greater are the velocity of the halfback and the velocity of the tight end than the velocity of the lineman?

Which player has the greatest momentum? Explain.

A. The tight end travels twice the distance of the lineman in the same amount of time. Thus, the tight end is twice as fast ( v tight end = 6 m/s ). The halfback travels three times the distance of the lineman in the same amount of time. Thus, the halfback is three times as fast ( v halfback = 9 m/s ).

B. Both the halfback and the tight end have the greatest momentum. The each have the same amount of momentum - 540 kg*m/s . The lineman only has 360 kg*m/s.

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