Call Now to Set Up Tutoring:

(888) 888-0446

High School Physics : Momentum

Study concepts, example questions & explanations for High School Physics

Create An Account

All High School Physics Resources

6 Diagnostic Tests 233 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #17 : Calculating Momentum

An 11.2kg ball moving at $39\frac{m}{s}$ strikes a 15kg ball at rest. After the collision the 11.2kg ball is moving with a velocity of $30\frac{m}{s}$ . What is the velocity of the second ball?

Possible Answers:

$6.7\frac{m}{s}$

$67.1\frac{m}{s}$

$3.4\frac{m}{s}$

$7.6\frac{m}{s}$

$5.9\frac{m}{s}$

Correct answer:

$6.7\frac{m}{s}$

Explanation:

This is an example of an elastic collision. We start with two masses and end with two masses with no loss of energy.

We can use the law of conservation of momentum to equate the initial and final terms.

$m_1v_{1initial}+m_2v_{2initial}=m_1v_{1final}+m_2v_{2final}$

Plug in the given values and solve for $\small v_{2final}$ .

$(11.2kg*39\frac{m}{s})+(15kg*0\frac{m}{s})=(11.2kg*30\frac{m}{s})+(15kg*v_{2final})$

$436.8\frac{kg\cdot m}{s}+0=(336\frac{kg\cdot m}{s})+15kg( v_{2final})$

${436.8\frac{kg\cdot m}{s}}-{336\frac{kg\cdot m}{s}}=15kg( v_{2final})$

${100.8\frac{kg\cdot m}{s}}=15kg( v_{2final})$

$\frac{100.8\frac{kg\cdot m}{s}}{15kg}=v_{2final}$

$6.7\frac{m}{s}=v_{2final}$

Report an Error

Example Question #18 : Calculating Momentum

A cat is playing with a 0.2kg toy mouse. He bats it with of force and is in contact with the toy for 0.1s . What is the new velocity of the mouse?

Possible Answers:

$0.5\frac{m}{s}$

$4\frac{m}{s}$

$2\frac{m}{s}$

$1\frac{m}{s}$

$0.2\frac{m}{s}$

Correct answer:

$1\frac{m}{s}$

Explanation:

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

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

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

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

$\Delta p=F\Delta t$ can also be thought of as $(p_{2}-p_{1})=F\Delta t$ .

Expand this equation to include our given values.

$(mv_{2}-mv_{1})=F\Delta t$

Since the toy starts at rest, we know the initial velocity.

$v_1=0\frac{m}{s}$

We are also given the mass, force, and time. Using these given values allows us to solve for the final velocity.

$0.2kg(v_2)-0.2kg(0\frac{m}{s})=2N(0.1s)$

$0.2kg(v_2)=2\frac{kg\cdot m}{s^2}( 0.1s)$

$0.2kg(v_2)=0.2\frac{kg\cdot m}{s}$

$v_2=\frac{0.2\frac{kg\cdot m}{s}}{0.2kg}$

$v_2=1\frac{m}{s}$

Report an Error

Example Question #41 : Momentum

A 0.8kg hammer moving with a velocity of $12\frac{m}{s}$ strikes a nail, after which the hammer has a velocity of $0\frac{m}{s}$ . If the hammer strikes the nail with 190N of force, how long were the two in contact?

Possible Answers:

1824s

0.1s

0.05s

0.01s

Correct answer:

0.05s

Explanation:

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

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

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

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

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

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

Expand this equation to include our given values.

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

Since the hammer is not moving at the end, its final velocity is zero. We can plug in the given values on the left side of the equation.

$(0.8kg* 0\frac{m}{s})-(0.8kg*12\frac{m}{s})=F \Delta t$

The problem gave us the force of the hammer on the nail, but not the force of the nail on the hammer, which is what we need as we are looking purely at the momentum of the hammer.

Fortunately, Newton's third law can help us. It states that $F_{hammer}=-F_{nail}$ . This means that if the hammer exerts 190N of force on the nail, then the nail must exert -190N of force on the hammer.

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

$(0)-(0.8kg* 12\frac{m}{s})=(-190N)(\Delta t)$

$(-0.8kg* 12\frac{m}{s})=(-190N)(\Delta t)$

$-9.6\frac{kg\cdot m}{s}=(-190N) \Delta t$

$\frac{-9.6\frac{kg\cdot m}{s}}{-190N}=\Delta t$

$0.05s=\Delta t$

Report an Error

Example Question #22 : Calculating Momentum

Susan pushes a car 1412kg with 18N of force along a frictionless surface. How long does she need to push the car to get it to a velocity of $2\frac{m}{s}$ , if it starts at rest?

Possible Answers:

33.45s

111.23s

156.89s

96.23s

301.22s

Correct answer:

156.89s

Explanation:

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

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

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

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

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

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

Expand this equation to include our given values.

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

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

$(1412kg* 2\frac{m}{s})-(1412kg* 0\frac{m}{s})=(18N)(\Delta t)$

$2824\frac{kg\cdot m}{s}=18N* \Delta t$

$\frac{2824\frac{kg\cdot m}{s}}{18N}=\Delta t$

$156.89s=\Delta t$

Report an Error

Example Question #23 : Calculating Momentum

A 0.12kg tennis ball strikes a racket, moving at $35\frac{m}{s}$ . After striking the racket, it bounces back at a speed of $45\frac{m}{s}$ . What is the change in momentum?

Possible Answers:

$1.2\frac{kg*m}{s}$

$-19.2\frac{kg*m}{s}$

$-0.6\frac{kg*m}{s}$

$-9.6\frac{kg*m}{s}$

$4.8\frac{kg*m}{s}$

Correct answer:

$-9.6\frac{kg*m}{s}$

Explanation:

The change in momentum is the final momentum minus the initial momentum, or $\Delta p =p_{final}-p_{initial}$ .

Notice that the problem gives us the final SPEED of the ball but not the final VELOCITY. Since the ball "bounced back," it begins to move in the opposite direction, so its velocity at this point will be negative.

Plug in our values to solve:

$\Delta p =p_{final}-p_{initial}$

$\Delta p =(0.12kg*-45\frac{m}{s})-(0.12kg*35\frac{m}{s})$

$\Delta p=-5.4\frac{kg*m}{s}-4.2\frac{kg*m}{s}$

$\Delta p = -9.6\frac{kg*m}{s}$

Report an Error

Example Question #24 : Calculating Momentum

A 0.12kg tennis ball strikes a racket, moving at $35\frac{m}{s}$ . After striking the racket, it bounces back at a speed of $45\frac{m}{s}$ . If the ball is in contact with the racket for 0.12s , how much force did the racket exert on the ball?

Possible Answers:

u -80N

40N

1.152N

810N

Correct answer:

u -80N

Explanation:

The change in momentum is equal to the force exerted on the ball times time, or $\Delta p = F*\Delta t$ .

It is also equal to the final momentum minus the initial momentum, or $\Delta p =p_{final}-p_{initial}$ .

Notice that the problem gives us the final SPEED of the ball but not the final VELOCITY. Since the ball "bounces back," it is moving in the opposite direction, so its velocity at this point will be negative.

Plug in our values to solve:

$\Delta p =p_{final}-p_{initial}$

$\Delta p =(0.12kg*-45\frac{m}{s})-(0.12kg*35\frac{m}{s})$

$\Delta p=-5.4\frac{kg*m}{s}-4.2\frac{kg*m}{s}$

$\Delta p = -9.6\frac{kg*m}{s}$

Plug that into our first equation to solve for the force the racket exerted on the ball:

$\Delta p = F*\Delta t$

$-9.6\frac{kg*m}{s} = F*0.12s$

$\frac{-9.6\frac{kg*m}{s}}{0.12s}{} = F$

$-80\frac{kg*m}{s^{2}}=F$

Since $N = \frac{kg*m}{s^{2}}$ , we can rewrite our answer as:

-80N=F

Report an Error

Example Question #311 : Motion And Mechanics

A 0.12kg tennis ball moving at $35\frac{m}{s}$ strikes a racket. After striking the racket, it bounces back at a speed of $45\frac{m}{s}$ . If the racket exerts 160N of force on the ball, how long were they in contact?

Possible Answers:

16.67s

0.12s

9.6s

0.06s

150.4s

Correct answer:

0.06s

Explanation:

The change in the ball's momentum is equal to the force exerted on the ball times time, or $\Delta p = F*\Delta t$ .

It is also equal to the final momentum minus the initial momentum, or $\Delta p =p_{final}-p_{initial}$

Notice that the problem gives us the final SPEED of the ball but not the final VELOCITY. Since the ball "bounced back," it is moving in the opposite direction, and its velocity at this point will be negative.

Plug in our values to solve:

$\Delta p =p_{final}-p_{initial}$

$\Delta p =(0.12kg*-45\frac{m}{s})-(0.12kg*35\frac{m}{s})$

$\Delta p=-5.4\frac{kg*m}{s}-4.2\frac{kg*m}{s}$

$\Delta p = -9.6\frac{kg*m}{s}$

Plug that into our first equation:

$\Delta p = F*\Delta t$

$-9.6\frac{kg*m}{s}=-160N*\Delta t$

Since our force is going in the same direction as our change in momentum, make both negative.

$\frac{-9.6\frac{kg*m}{s}}{-160N}=\Delta t$

Since $N = \frac{kg*m}{s^{2}}$ , we can rewrite our equation's units as:

$\frac{-9.6 \frac{kg*m}{s}}{-160\frac{kg*m}{s^{2}}}=\Delta t$

Then, we can simplify the equation's units to get our final answer:

$0.06s=\Delta t$

Report an Error

Example Question #42 : Momentum

Possible Answers:

$-9.6\frac{m}{s^2}$

$-1.2\frac{m}{s^2}$

$-80\frac{m}{s^2}$

$-666.67\frac{m}{s^2}$

$-1200\frac{m}{s^2}$

Correct answer:

$-666.67\frac{m}{s^2}$

Explanation:

First, we need to find the force applied to the ball since F=ma . We know the mass, but now we need the force.

The change in momentum is equal to the force exerted on the ball times time, or $\Delta p = F*\Delta t$ .

The change in momentum is also equal to the final momentum minus the initial momentum, or $\Delta p =p_{final}-p_{initial}$ .

Notice that the problem gives us the final SPEED of the ball but not the final VELOCITY. Since the ball "bounces back," is moving in the opposite direction, and its velocity will be negative at this point.

Plug in our values to solve:

$\Delta p =p_{final}-p_{initial}$

$\Delta p =(0.12kg*-45\frac{m}{s})-(0.12kg*35\frac{m}{s})$

$\Delta p=-5.4\frac{kg*m}{s}-4.2\frac{kg*m}{s}$

$\Delta p = -9.6\frac{kg*m}{s}$

Plug that into our other equation:

$\Delta p = F*\Delta t$

$-9.6\frac{kg*m}{s} = F*0.12s$

$\frac{-9.6\frac{kg*m}{s}}{0.12s}{} = F$

$-80\frac{kg*m}{s^{2}}=F$

Since $N=\frac{kg*m}{s^{2}}$ , we can rewrite our answer as:

-80N=F

Plug that into our first equation:

F=ma

-80N=0.12kg*a

$\frac{-80N}{0.12kg}=a$

$-666.67\frac{N}{kg}=a$

Since $N=\frac{kg*m}{s^{2}}$ , we can rewrite our answer as:

$-666.67\frac{\frac{kg*m}{s^{2}}}{kg}=a$

Then, we can reduce our answer's units to:

$-666.67\frac{m}{s^2}=a$

Report an Error

Example Question #312 : Motion And Mechanics

A 100kg crate slides along the floor for before stopping. If it was initially moving with a velocity of $3\frac{m}{s}$ , what is the force of friction?

Possible Answers:

-30N

-100N

-300N

10N

-10N

Correct answer:

-100N

Explanation:

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

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

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

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

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

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

Expand this equation to include our given values.

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

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

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

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

$(-300\frac{kg*m}{s})=F(3s)$

$\frac{(-300\frac{kg*m}{s})}{3s}}=F$

-100N=F

Report an Error

Example Question #28 : Calculating Momentum

A 300kg crate slides along a floor with a starting velocity of $11\frac{m}{s}$ . If the force due to friction is -35N , how long will it take before stopping?

Possible Answers:

-90s

94.29s

35s

942.9s

35.5s

Correct answer:

94.29s

Explanation:

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

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

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

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

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

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

Expand this equation to include our given values.

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

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

$(300kg*0\frac{m}{s})-(300kg*11\frac{m}{s})=-35N*\Delta t$

$(-300kg*11\frac{m}{s})=-35N*\Delta t$

$(-3300\frac{kg*m}{s})=-35N*\Delta t$

$\frac{(-3300\frac{kg*m}{s})}{-35N}= \Delta t$

$94.29s= \Delta t$

Report an Error

View Tutors

Tracy
Certified Tutor

Clarion University of Pennsylvania, Bachelor of Science, Physics.

View Tutors

Nicole
Certified Tutor

Western Governors University, Bachelor of Education, Special Education.

View Tutors

Sharif
Certified Tutor

NCCU, Bachelors, Physics and Mathematics.

All High School Physics Resources

6 Diagnostic Tests 233 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

Reading Tutors in Boston, ISEE Tutors in San Diego, MCAT Tutors in Los Angeles, Math Tutors in Houston, ACT Tutors in San Francisco-Bay Area, Physics Tutors in Houston, English Tutors in Chicago, ACT Tutors in Phoenix, MCAT Tutors in Chicago, English Tutors in Miami

Popular Courses & Classes

GMAT Courses & Classes in San Diego, SAT Courses & Classes in Atlanta, GMAT Courses & Classes in Boston, LSAT Courses & Classes in Philadelphia, ACT Courses & Classes in San Francisco-Bay Area, SAT Courses & Classes in San Diego, ISEE Courses & Classes in Los Angeles, Spanish Courses & Classes in Philadelphia, GMAT Courses & Classes in New York City, ACT Courses & Classes in Boston

Popular Test Prep

SAT Test Prep in Atlanta, SSAT Test Prep in Boston, MCAT Test Prep in Phoenix, SSAT Test Prep in New York City, GMAT Test Prep in Boston, GMAT Test Prep in San Diego, ACT Test Prep in Seattle, MCAT Test Prep in San Francisco-Bay Area, ACT Test Prep in Miami, GMAT Test Prep in San Francisco-Bay Area

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

* Signature (your digital signature is legally binding)

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

High School Physics : Momentum

Study concepts, example questions & explanations for High School Physics

All High School Physics Resources

Example Questions

Example Question #17 : Calculating Momentum

Example Question #18 : Calculating Momentum

Example Question #41 : Momentum

Example Question #22 : Calculating Momentum

Example Question #23 : Calculating Momentum

Example Question #24 : Calculating Momentum

Example Question #311 : Motion And Mechanics

Example Question #42 : Momentum

Example Question #312 : Motion And Mechanics

Example Question #28 : Calculating Momentum

All High School Physics Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: