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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  

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Impulse and Momentum – Physics Example Problem

Impulse and momentum are physical concepts that are easily seen from Newton’s Laws of Motion.

Start with this equation of motion for constant acceleration.

v = v 0 + at

where v = velocity v 0 = initial velocity a = acceleration t = time

If you rearrange the equation, you get

v – v 0 = at

Newton’s second law deals with force.

where F = force m = mass a = acceleration

solve this for a and get

Stick this into the velocity equation and get

v – v 0 = (F/m)t

Multiply both sides by m

mv – mv 0 = Ft

The left side of the equation deals with momentum (often denoted by a lower-case p) and the right side is impulse (often denoted by an upper-case letter J).

Mass times velocity is known as momentum and force applied over time is called impulse.

Impulse and Momentum Example Problem

Question: A 50 kg mass is sitting on a frictionless surface. An unknown constant force pushes the mass for 2 seconds until the mass reaches a velocity of 3 m/s.

a) What is the initial momentum of the mass? b) What is the final momentum of the mass? c) What was the force acting on the mass? d) What was the impulse acting on the mass?

Part a) What is the initial momentum?

Momentum is mass times velocity. Since the mass is at rest, the initial velocity is 0 m/s.

momentum = m⋅v = (50 kg)⋅(0 m/s) = 0 kg⋅m/s

Part b) What is the final momentum?

After the force is finished acting on the mass, the velocity is 3 m/s.

momentum = m⋅v = (50 kg)⋅(3 m/s) = 150 kg⋅m/s

Part c) What was the force acting on the mass?

From parts a and b, we know mv 0 = 0 kg⋅m/s and mv = 150 kg⋅m/s.

150 kg⋅m/s – 0 kg⋅m/s = Ft 150 kg⋅m/s = Ft

Since the force was in effect over 2 seconds, t = 2 s.

150 kg⋅m/s = F ⋅ 2s F = (15 kg⋅m/s) / 2 s F = 75 kg⋅m/s 2

Unit Fact: kg⋅m/s 2 can be denoted by the derived SI unit Newton (symbol N )

Part d) What was the impulse acting on the mass?

The impulse is the force multiplied by the time passed. It is also equal to the change in momentum over the same time period.

Ft = 75 N ⋅ 2 s Ft = 150 Ns or 150 kg⋅m/s

The impulse was 150 kg⋅m/s.

These problems are relatively simple as long as you keep your units straight. Impulse and momentum should have the same units: mass⋅velocity or force⋅time. Check your units when you check your answer.

Another possible way to cause errors is to confuse your vector directions. Velocity and Force are both vector quantities. In this example, the mass was pushed in the direction of the final velocity. If another force pushed in the opposite direction to slow down the mass, the force would have a negative value compared to the velocity vector.

If you found this helpful, check out other Physics Example Problems .

Related Posts

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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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

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Momentum Practice Problems

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Momentum Practice Problems – Tutorial and Examples

The formula for calculating momentum is =

Momentum = mass X velocity

Where P = momentum, V = velocity and M = mass

Based on the above definition, it is clear that the momentum of a car and a bicycle both travelling at 20 m/s will not be the same, because although the velocity of the two objects are the same, their mass is different. The car would have greater momentum, due to its larger mass.

The SI unit for velocity = m/s

SI unit for Mass = kg

So therefore momentum = kg x m/s and SI unit for momentum is kg x m/s

Momentum must always have a direction and so the final answer must reflect the direction of the momentum or velocity.

1. Find the momentum of a round stone weighing 12.05kg rolling down a hill at 8m/s.

Formula – P= kg x m/s

= 12.05kg x 8m/s

= 96.4 kg x m/s down hill

Note that the final answer has the proper SI unit of momentum (kg x m/s) after it and it also mentions the direction of the movement.

2. A cannon ball weighing 35kg is shot from a cannon towards the east at 220mls, calculate the momentum of the cannon ball.

= 35kg x 220m/s

= 7700 kg x m/s east

Momentum questions generally appear with general science and/or basic physics.   Tests with momentum questions:

GED   —  HESI   —  NLN PAX   —  TEAS

Practice Problems

1.  Three cars are travelling down an even road at a velocity of 110 m/s, calculate the car with the highest momentum if they are all moving at the same speed, but the first car weighs 2500kg, second car weighs 2650kg and third car weighs 2009kg?

a. First car b. Second car c. Third car d. All have same momentum

2. What is the momentum of a log of wood that weighs 700kg rolling down a hill at 4.6m/s.

a. 3320 kg x m/s down hill b. 3320 kg x m/s c. 3320 down hill d. 3320 M

3. An object that weighs 500g is rolling along the road at 3.5m/s, what is the momentum of the object?

a. 124.9 kg x m/s along road b. 17. 50 kg x m/s along road c. 1750 kg x m/s along  road d. 1.75 kg x m/s along road

4. A javelin is thrown into a field at 18m/s. if the Javelin weighs 1.5kg, what is the momentum?

a. 1.2 kg x m/s into the field b. 12 kg x m/s into the field c. 27 kg x m/s into the field d. 2.7 kg x m/s into the field

5. Which of these object has greater momentum, a 2kg truck moving east at 3.5m/s or a 4.3kg truck moving south at 1.5m/s?

a. First truck at 7 kg x m/s moving east b. Second truck at 7.45 kg x m/s due south c. First truck at 6.45 kg x m/s due east d. Second truck at 7 kg x m/s due south

6. A bullet weighing 350g is shot towards a target at a velocity of 250m/s. Calculate the momentum of the bullet

a. 1.4 kg x m/s towards target b. 87.5 kg x m/s towards target c. 87500 kg x m/s towards target d. 8.75 kg x m/s towards target

1. B Momentum is a product of velocity and mass.  If they are all traveling at the same speed, the car that weighs the most would have the highest momentum.

2. A  4.6 X 700 = 3220 down the hill

3. D First convert 500g to kg = 500/1000 = 0.5kg, momentum = 0.5 x 3.5 = 1.75 kg x m/s along the road.

4. C p = 1.5 x 18 = 27 kg x m/s into the field.

5. A The momentum of first object = 2 x 3.5 = 7; momentum of second truck = 4.3 x 1.5 = 6.45. First truck has more momentum at 7 kg x m/s moving east.

6. B First convert 350g to kg = 350/1000 = 0.35kg. Momentum of bullet = 0.35 x 250 = 87.5 kg x m/s towards target.

Most Common Momentum Questions on a Test

Here is what type of questions to expect:  

• Force and momentum – Explain how force and momentum are related and how analyze motion given the force and momentum.
• Momentum in two-dimensional motion: Analyze and calculate momentum in two-dimensional motion, for example, projectiles.

Most Common Mistakes Answering Momentum Questions

• Confusing mass and weight: Mass and weight are similar not the same thing, and can’t be used interchangeably.
• Confusing velocity and speed: They sound the same but are different = velocity has direction and speed doesn’t.
• Watch your units!

Got a Question? Email me anytime - [email protected]

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Mechanics: Momentum and Collisions

Momentum and collisions: problem set overview.

There are 15 ready-to-use problem sets on the topic of Momentum and Collisions. The problems target your ability to use the momentum equation and the impulse-momentum change theorem in order to analyze physical situations involving collisions and impulses, to use momentum conservation principles to analyze a collision or an explosion, to combine a momentum analysis with other forms of analyzes (Newton's laws, kinematics, etc.) to solve for an unknown quantity, and to analyze two-dimensional collisions.   

An object which is moving has momentum. The amount of momentum ( p ) possessed by the moving object is the product of mass ( m ) and velocity ( v ). In equation form:

p = m • v

An equation such as the one above can be treated as a sort of recipe for problem-solving. Knowing the numerical values of all but one of the quantities in the equations allows one to calculate the final quantity in the equation. An equation can also be treated as a statement which describes qualitatively how one variable depends upon another. Two quantities in an equation could be thought of as being either directly proportional or inversely proportional. Momentum is directly proportional to both mass and velocity. A two-fold or three-fold increase in the mass (with the velocity held constant) will result in a two-fold or a three-fold increase in the amount of momentum possessed by the object. Similarly, a two-fold or three-fold increase in the velocity (with the mass held constant) will result in a two-fold or a three-fold increase in the amount of momentum possessed by the object. Thinking and reasoning proportionally about quantities allows you to predict how an alteration in one variable would effect another variable.  

• Impulse-Momentum Change Equation

In a collision, a force acts upon an object for a given amount of time to change the object's velocity. The product of force and time is known as impulse . The product of mass and velocity change is known as momentum change . In a collision the impulse encountered by an object is equal to the momentum change it experiences.

Impulse = Momentum Change

F • t = mass • Delta v

Several problems in this unit test your understanding of the above relationship. In many of these problems, a piece of extraneous information is provided. Without an understanding of the above relationships, you will be tempted to force such information into your calculations. Physics is about conceptual ideas and relationships; and problems test your mathematical understanding of these relationships. If you treat this problem set as a mere exercise in the algebraic manipulation of physics equations, then you are likely to become frustrated quickly. As you proceed through this problem set, be concepts-minded. Do not strip physics of its conceptual meaning.  #WordsForTheWise Our 8-minute video on the Impulse-Momentum Change Relationship will expand on the use of this equation.​  

Several of the problems in this unit demand that you be able to calculate the velocity change of an object. This calculation becomes particularly challenging when the collision involves a rebounding effect - that is, the object is moving in one direction before the collision and in the opposite direction after the collision. Velocity is a vector and is distinguished from speed in that it has a direction associated with it. This direction is often expressed in mathematics as a + or - sign. In a collision, the velocity change is always computed by subtracting the initial velocity value from the final velocity value. If an object is moving in one direction before a collision and rebounds or somehow changes direction, then its velocity after the collision has the opposite direction as before. Mathematically, the before-collision velocity would be + and the after-collision velocity would be - in sign. Ignoring this principle will result in great difficulty when analyzing any collision involving the rebounding of an object.  

The Momentum Conservation Principle

In a collision between two objects, each object is interacting with the other object. The interaction involves a force acting between the objects for some amount of time. This force and time constitutes an impulse and the impulse changes the momentum of each object. Such a collision is governed by Newton's laws of motion; and as such, the laws of motion can be applied to the analysis of the collision (or explosion) situation. So with confidence it can be stated that ...

In a collision between object 1 and object 2, the force exerted on object 1 (F 1 ) is equal in magnitude and opposite in direction to the force exerted on object 2 (F 2 ). In equation form:

F 1 = - F 2

The above statement is simply an application of Newton's third law of motion to the collision between objects 1 and 2. Now in any given interaction, the forces that are exerted upon an object act for the same amount of time. You can't contact another object and not be contacted yourself (by that object). And the duration of time during which you contact the object is the same as the duration of time during which that object contacts you. Touch a wall for 2.0 seconds, and the wall touches you for 2.0 seconds. Such a contact interaction is mutual; you touch the wall and the wall touches you. It's a two-way interaction - a mutual interaction; not a one-way interaction. Thus, it is simply logical to state that in a collision between object 1 and object 2, the time during which the force acts upon object 1 (t 1 ) is equal to the time during which the force acts upon object 2 (t 2 ). In equation form:

The basis for the above statement is simply logic. Now we have two equations which relate the forces exerted upon individual objects involved in a collision and the times over which these forces occur. It is accepted mathematical logic to state the following:

If A = - B and C = D then A • C = - B • D

The above logic is fundamental to mathematics and can be used here to analyze our collision.

If F 1 = - F 2 and t 1 = t 2 then F 1 • t 1 = - F 2 • t 2

The above equation states that in a collision between object 1 and object 2, the impulse experienced by object 1 (F 1 • t 1 ) is equal in magnitude and opposite in direction to the impulse experienced by object 2 (F 2 • t 2 ). Objects encountering impulses in collisions will experience a momentum change. The momentum change is equal to the impulse. Thus, if the impulse encountered by object 1 is equal in magnitude and opposite in direction to the impulse experienced by object 2, then the same can be said of the two objects' momentum changes. The momentum change experienced by object 1 (m 1 • Delta v 1 ) is equal in magnitude and opposite in direction to the momentum change experienced by object 2 (m 2 • Delta v 2 ). This statement could be written in equation form as

m 1 • Delta v 1 = - m 2 • Delta v 2

This equation claims that in a collision, one object gains momentum and the other object loses momentum. The amount of momentum gained by one object is equal to the amount of momentum lost by the other object. The total amount of momentum possessed by the two objects does not change. Momentum is simply transferred from one object to the other object.

Put another way, it could be said that when a collision occurs between two objects in an isolated system, the sum of the momentum of the two objects before the collision is equal to the sum of the momentum of the two objects after the collision.  If the system is indeed isolated from external forces, then the only forces contributing to the momentum change of the objects are the interaction forces between the objects.  As such, the momentum lost by one object is gained by the other object and the total system momentum is conserved. And so the sum of the momentum of object 1 and the momentum of object 2 before the collision is equal to the sum of the momentum of object 1 and the momentum of object 2 after the collision. The following mathematical equation is often used to express the above principle.

m 1 • v 1 + m 2 • v 2 = m 1 • v 1 ' + m 2 • v 2 '

The symbols m 1 and m 2 in the above equation represent the mass of objects 1 and 2. The symbols v 1 and v 2 in the above equation represent the velocities of objects 1 and 2 before the collision. And the symbols v 1 ' and v 2 ' in the above equation represent the velocities of objects 1 and 2 after the collision. (Note that a ' symbol is used to indicate after the collision.) Our video titled Action-Reaction and the Law of Momentum Conservation provides a thorough discussion of the above principles.   

Direction Matters

Momentum is a vector quantity; it is fully described by both a magnitude (numerical value) and a direction. The direction of the momentum vector is always in the same direction as the velocity vector. Because momentum is a vector, the addition of two momentum vectors is conducted in the same manner by which any two vectors are added. For situations in which the two vectors are in opposite directions, one vector is considered negative and the other positive. Successful solutions to many of the problems in this unit demands that attention be given to the vector nature of momentum.  

Two-Dimensional Collision Problems

A two-dimensional collision is a collision in which the two objects are not originally moving along the same line of motion. They could be initially moving at right angles to one another or at least at some angle (other than 0 degrees and 180 degrees) relative to one another. In such cases, vector principles must be combined with momentum conservation principles in order to analyze the collision. The underlying principle of such collisions is that both the "x" and the "y" momentum are conserved in the collision. The analysis involves determining pre-collision momentum for both the x- and the y- directions. If it is a hit-and-stick collision, then the total amount of system momentum before the collision (and after) can be determined by using the Pythagorean theorem. Since the two colliding objects travel together in the same direction after the collision, the total momentum is simply the total mass of the objects multiplied by their velocity.  

Momentum Plus Problems

A momentum plus problem is a problem type in which the analysis and solution includes a combination of momentum conservation principles and other principles of mechanics. Such a problem typically involves two analyses which must be conducted separately. One of the analyses is a collision analysis to determine the speed of one of the colliding objects before or after the collision. The second analyses typically involves Newton's laws and/or kinematics. These two models (Newton's laws and kinematics) allows a student to make a prediction about how far an object will slide or how high it will roll or slide or swing after the collision with the other object.

When solving momentum plus problems, it is important to take the time to identify the known and the unknown quantities. It is helpful to organize such known quantities in two columns - a column for information pertaining to the collision analysis and a column for information pertaining to the Newton's law and/or kinematic analysis.  

Habits of an Effective Problem-Solver

An effective problem solver by habit approaches a physics problem in a manner that reflects a collection of disciplined habits. While not every effective problem solver employs the same approach, they all have habits which they share in common. These habits are described briefly here. An effective problem-solver...

• ...reads the problem carefully and develops a mental picture of the physical situation. If needed, they sketch a simple diagram of the physical situation to help visualize it.
• ...identifies the known and unknown quantities in an organized manner, often times recording them on the diagram itself. They equate given values to the symbols used to represent the corresponding quantity (e.g., m = 1.50 kg, v i = 2.68 m/s,  F = 4.98 N, t = 0.133 s, v f = ???).
• ...plots a strategy for solving for the unknown quantity; the strategy will typically center around the use of physics equations and be heavily dependent upon an understanding of physics principles.
• ...identifies the appropriate formula(s) to use, often times writing them down. Where needed, they perform the needed conversion of quantities into the proper unit.
• ...performs substitutions and algebraic manipulations in order to solve for the unknown quantity.

Read more...  

Additional Readings/Study Aids:

The following pages from The Physics Classroom tutorial may serve to be useful in assisting you in the understanding of the concepts and mathematics associated with these problems.

• Real World Applications
• Momentum Conservation Principle
• Isolated Systems
• Collision Analysis
• Explosion Analysis
• Kinematic Equations

Watch a Video

We have developed and continue to develop Video Tutorials on introductory physics topics. You can find these videos on our YouTube channel . We have an entire Playlist on the topic of Momentum and Collisions .  

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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?