Call Now to Set Up Tutoring:

(888) 888-0446

AP Physics 1 : Newtonian Mechanics

Study concepts, example questions & explanations for AP Physics 1

Create An Account

All AP Physics 1 Resources

7 Diagnostic Tests 170 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #37 : Impulse And Momentum

Two cars of mass 1450kg hit head on, each traveling at $30\frac{m}{s}$ . They then bounce back at $2.5\frac{m}{s}$ If the collision took .25 s , determine the force experienced by a driver of mass 95kg in one of the cars.

Possible Answers:

9.85*10^4N

2.30*10^5N

Impossible to determine

None of these

1.24*10^5N

Correct answer:

1.24*10^5N

Explanation:

Using momentum principle:

$P_i+F\Delta t=P_f$

Since the cars have identical masses and experience the same change in velocity, the forces on them will be identical.

$m_iv_i+F\Delta t=m_fv_f$

Plugging in values:

1450*30+F*.25=1450*-2.5

Solving for :

F=-188500N

|F|=188500N

Now, determine the acceleration of the car:

|F|=188500N=1450*a

$a=130\frac{m}{s^2}$

Now, determine the force experienced by the driver who was in the car, and thus accelerated with the car.

$F=m*a=95*130\frac{m}{s^2}$

F=1.24*10^5N

Report an Error

Example Question #33 : Impulse And Momentum

When firefighters rescue a person from a burning building several stories high, they're prepared with a padded landing for when the person jumps down. As this person falls, they build up greater and greater momentum. How does the padded landing bring this falling person safely to rest?

Possible Answers:

It decreases the force of the collision by decreasing the time of the collision

It increases the force of the collision by decreasing the time of the collision

It increases the force of the collision by increasing the time of the collision

It decreases the force of the collision by increasing the time of the collision

Correct answer:

It decreases the force of the collision by increasing the time of the collision

Explanation:

For this question, we're asked how a padded landing helps a person fall safely from a large height. To do so, we need to consider how the landing interacts with the falling person in changing their momentum.

Let's first remember that a change in an object's momentum is called impulse, which can be written with the following expression.

$I=m\Delta v$

What's more, the units can be rearranged in such a way as to express impulse in terms of average force and time of the collision.

$I=m\Delta v=F_{avg}\Delta t$

Next, we can rearrange the above expression to make it easier to see how force and time of collision are related.

$F_{avg}=\frac{m \Delta v}{\Delta t}$

As we can see in the above expression, the average force of collision is inversely proportional to the time of the collision. In other words, the longer the collision lasts, the less the average force will be.

Relating this to the padded landing example, it's clear that the landing helps to reduce a person's fall by reducing the average force of the collision between the person and the landing. Moreover, the amount of time the collision takes to happen is increased.

Report an Error

Example Question #41 : Impulse And Momentum

Deep in space Object has mass 350kg and is initially traveling with velocity $<30,50>\frac{m}{s}$ . At t=0 , it collides with Object , which has mass 500kg and is initially motionless. The two objects stick together.

Determine the initial momentum of the system.

Possible Answers:

$<17500,10500>\frac{kg*m}{s}$

$<17500,17500>\frac{kg*m}{s}$

Impossible to determine

$<10500,17500>\frac{kg*m}{s}$

None of these

Correct answer:

$<10500,17500>\frac{kg*m}{s}$

Explanation:

Using

$\overrightarrow{p}=m\overrightarrow{v}$

Plugging in values:

$\overrightarrow{p}=350kg*<30,50>\frac{m}{s}$

$\overrightarrow{p}=<10500,17500>\frac{kg*m}{s}$

Report an Error

Example Question #41 : Impulse And Momentum

Determine the impulse experienced by Object

Possible Answers:

None of these

$<6200,10300>\frac{kg*m}{s}$

$<4449,39080>\frac{kg*m}{s}$

$<10300,6200>\frac{kg*m}{s}$

$<4340,7210>\frac{kg*m}{s}$

Correct answer:

$<6200,10300>\frac{kg*m}{s}$

Explanation:

Using

$\overrightarrow{p}=m\overrightarrow{v}$

Plugging in values:

$\overrightarrow{p}=350kg*<30,50>\frac{m}{s}$

$\overrightarrow{p}=<10500,17500>\frac{kg*m}{s}$

The momentum will be the same in the final state, so again using

$\overrightarrow{p}=m\overrightarrow{v}$

Solving for velocity:

$\frac{\overrightarrow{p}}{m}=\overrightarrow{v}$

Plugging in values (the total mass is equal to the combined masses:

$\frac{<10500,17500>}{850}=\overrightarrow{v}$

$\overrightarrow{v_f}=<12.4,20.6>\frac{m}{s}$

Definition of impulse:

$J=\Delta \overrightarrow{p}$

$J=m\Delta\overrightarrow{v}$

J=500(<12.4,20.6>-<0,0>)

$J=<6200,10300>\frac{kg*m}{s}$

Report an Error

Example Question #43 : Impulse And Momentum

A ball of mass 30g is thrown at a target. The ball strikes with a velocity of $20\frac{m}{s}$ and bounces back with equal magnitude. Determine the magnitude of impulse experienced by the ball.

Possible Answers:

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

$6.8\frac{kg\cdot m}{s}$

$3.6\frac{kg\cdot m}{s}$

$5.5\frac{kg\cdot m}{s}$

$7.7\frac{kg\cdot m}{s}$

Correct answer:

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

Explanation:

Impulse is defined as change in momentum:

$\Delta P= P_f-P_i$

Using

P=mv

Combining equations:

$\Delta P= mv_f-mv_i$

Plugging in values:

$\Delta P= .030*-20-.030*20$

$|\Delta P| =1.2\frac{kg*m}{s}$

Report an Error

Example Question #221 : Linear Motion And Momentum

Frictionless cart is traveling at $<10,15>\frac{m}{s}$ when it hits identical frictionless cart which was previously motionless. After the collision, cart is traveling at $<1,1.5>\frac{m}{s}$ . Determine the final velocity of cart .

Possible Answers:

$<9,13.5>\frac{m}{s}$

$<0,0>\frac{m}{s^2}$

$<19,3.5>\frac{m}{s}$

None of these

$<18,27>\frac{m}{s}$

Correct answer:

$<9,13.5>\frac{m}{s}$

Explanation:

Using conservation of momentum:

Pi=P_f

$m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}$

Solving for $v_{2f}$

$\frac{m_1v_{1i}+m_2v_{2i}-m_1v_{1f}}{m_1}=m_2v_{2f}$

Using m_1=m_2 (since the carts are identical):

$v_{1i}+v_{2i}-v_{1f}=v_{2f}$

Plugging in values:

<10,15>+<0,0>-<1,1.5>=<9,13.5>

$v_{2f}=<9,13.5>\frac{m}{s}$

Report an Error

Example Question #42 : Impulse And Momentum

During time period , a rocket ship deep in space of mass 2.55*10^5kg travels from <0,10>km to <5,5>km . During time period , the rocket fires. During time period , the rocket travels from <-15,0>km to <-30,-5>km .

Time periods , , and all took 10seconds.

Determine the impulse during time period .

Possible Answers:

None of these

$<5.1*10^8,0>\frac{\textup{kg}\cdot \textup{m}}{\textup{s}}$

$\Delta\overrightarrow{P}=<0,5.1*10^8>\frac{\textup{kg}\cdot \textup{m}}{\textup{s}}$

$\Delta\overrightarrow{P}=<0,0>\frac{\textup{kg}\cdot \textup{m}}{\textup{s}}$

$\Delta\overrightarrow{P}=<5.1*10^8,5.1*10^8>\frac{\textup{kg}\cdot \textup{m}}{\textup{s}}$

Correct answer:

$<5.1*10^8,0>\frac{\textup{kg}\cdot \textup{m}}{\textup{s}}$

Explanation:

Finding initial momentum:

$\overrightarrow{P}=m\overrightarrow{v}$

$\overrightarrow{v}=\frac{<x_f-x_i,y_f-y_i>}{t}$

Combining equations:

$\overrightarrow{P}=m\frac{<x_f-x_i,y_f-y_i>}{t}$

Plugging in values:

$\overrightarrow{P}=2.55*10^5*\frac{<5000-0,5000-10000>}{10}$

$\overrightarrow{P_i}=<1.275*10^8,-1.275*10^8>\frac{kg*m}{s}$

$\overrightarrow{P}=m\overrightarrow{v}$

$\overrightarrow{v}=\frac{<x_f-x_i,y_f-y_i>}{t}$

Combining equations:

$\overrightarrow{P}=m\frac{<x_f-x_i,y_f-y_i>}{t}$

Converting to and plugging in values:

$\overrightarrow{P}=2.55*10^5*\frac{<-30000+15000,-5000-0>}{10}$

$\overrightarrow{P}_f=<-3.825*10^8,-1.275*10^8>\frac{kg*m}{s}$

Using

$Impulse=\Delta\overrightarrow{P}=\overrightarrow{P_f}-\overrightarrow{P_i}$ Plugging in values:

$\overrightarrow{\Delta P} =<-3.825*10^8,-1.275*10^8>-<1.275*10^8,-1.275*10^8>$

$\Delta\overrightarrow{P}=<5.1*10^8,0>\frac{\textup{kg}\cdot \textup{m}}{\textup{s}}$

Report an Error

Example Question #46 : Impulse And Momentum

Ty throws a 0.10kg snowball directly at his brother who also throws a snowball directly at Ty with a mass of 0.20kg , because he dipped it in a bucket of water. Ty throws his snowball at $20.0\frac{m}{s}$ and his brother throws his at $-15\frac{m}{s}$ and the two snowballs make perfect contact in mid flight, causing a perfectly inelastic collision. Neglecting air resistance, what is the combined final speed of the snowballs and their direction following the collision?

Possible Answers:

$-6.6\frac{m}{s}$ ; negative x-direction

$3.3\frac{m}{s}$ ; positive x-direction

$-3.3\frac{m}{s}$ ; negative x-direction

$6.6\frac{m}{s}$ ; positive x-direction

Correct answer:

$-3.3\frac{m}{s}$ ; negative x-direction

Explanation:

Use the conservation of momentum for inelastic collisions (objects that stick together).

$m_{Ty}(v_{Ty})+m_{brother}(v_{brother})=(m_{Ty}+m_{brother})v_{final}$

Plug in and solve for the final velocity.

$0.1Kg(20\frac{m}{s})+0.2Kg(-15\frac{m}{s})=(0.1Kg+0.2Kg)v_{final}$

$2j+(-3j)=(0.3Kg)v_{final}$

$v_{final}=-3.33\frac{m}{s}$

The final velocity of the combined snowballs is $-3.3\frac{m}{s}$ in the negative x-direction.

Report an Error

Example Question #47 : Impulse And Momentum

On a toy car set, two children apply a piece of double sided tape to the front of two toy cars so that when they collide they will stick together causing a perfectly inelastic collision. The two identical cars have a combined mass of 0.075kg and their final combined speed was $2.60\frac{m}{s}$ in the rightward direction. If car "A" was shot to the right with a speed measured at $17.0\frac{m}{s}$ right before the collision . At what speed was the other car traveling right before the collision?

Possible Answers:

$-6.1\frac{m}{s}$

$-11.8\frac{m}{s}$

$-18.4\frac{m}{s}$

$-24.5\frac{m}{s}$

Correct answer:

$-11.8\frac{m}{s}$

Explanation:

Use the conservation of momentum, perfectly inelastic collision.

$m_{A}(v_{A})+m_{B}(v_{B})=(m_{A}+m_{B})v_{final}$

Plug in and solve for the speed of car B.

$0.0375Kg(17\frac{m}{s})+0.0375Kg(v_{B})=(0.075Kg)2.6\frac{m}{s}$

$0.6375j+0.0375Kg(v_{B})=0.19j$

$v_{B}=-11.8\frac{m}{s}$

The velocity of car B right before the collision was $-11.8\frac{m}{s}$ .

Report an Error

Example Question #48 : Impulse And Momentum

A spaceship is stationary deep in space with it’s rocket broken. The crew decides to propel it to earth by throwing tennis balls out the back window. If they throw one tennis ball every second at $30 \frac{m}{s}$ , how many balls will it take the ship to reach $1.1*10^4 \frac{m}{s}$ ? Each tennis ball has a mass of 50 g and the ship has a mass of 7.0*10^6kg .

Possible Answers:

None of these

$5.13*10^{10}balls$

$6.26*10^{14}balls$

$8.44*10^{7}balls$

$3.95*10^{18}balls$

Correct answer:

$5.13*10^{10}balls$

Explanation:

The net momentum of the system needs to stay constant, that is, any momentum given to the tennis balls, the space ship must gain in the opposite direction.

m_sv_s+n_bm_bv_b=0

Where

m_s is the mass of the ship.

v_s is the velocity of the ship

n_b is the number of balls thrown

m_b is the mass of a ball

v_b is the velocity of a thrown ball

Solving for

$n_b=\frac{m_sv_s}{m_b(-v_b)}$

Since the velocities are given as magnitudes, the negative sign can be ignored

$n_b=\frac{m_sv_s}{m_b(v_b)}$

Plugging in values

$n_b=\frac{7*10^6*1.1*10^4}{.050(30)}$

$n_b=5.13*10^{10}$

Report an Error

View AP Physics 1 Tutors

Hailu Getachew
Certified Tutor

Michigan State University, Doctor of Engineering, Mechanical Engineering.

View AP Physics 1 Tutors

Aaron
Certified Tutor

Tulsa Community College, Associate in Science, Mechanical Engineering.

View AP Physics 1 Tutors

Cameo
Certified Tutor

University of Oklahoma Norman Campus, Bachelor of Science, Environmental Engineering.

All AP Physics 1 Resources

7 Diagnostic Tests 170 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

LSAT Tutors in Houston, SAT Tutors in Miami, Physics Tutors in New York City, SAT Tutors in Chicago, SSAT Tutors in Atlanta, English Tutors in Phoenix, Reading Tutors in Seattle, Algebra Tutors in Philadelphia, SAT Tutors in San Francisco-Bay Area, Algebra Tutors in Houston

Popular Courses & Classes

ISEE Courses & Classes in Los Angeles, SSAT Courses & Classes in Chicago, SSAT Courses & Classes in Atlanta, Spanish Courses & Classes in Seattle, GMAT Courses & Classes in Houston, GRE Courses & Classes in Dallas Fort Worth, ACT Courses & Classes in Denver, GMAT Courses & Classes in Los Angeles, GRE Courses & Classes in Los Angeles, GRE Courses & Classes in Boston

Popular Test Prep

GRE Test Prep in Seattle, SAT Test Prep in Seattle, ACT Test Prep in Denver, LSAT Test Prep in San Diego, GMAT Test Prep in Houston, GRE Test Prep in Dallas Fort Worth, LSAT Test Prep in Seattle, ISEE Test Prep in Seattle, ACT Test Prep in Atlanta, SAT Test Prep in Chicago

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

AP Physics 1 : Newtonian Mechanics

Study concepts, example questions & explanations for AP Physics 1

All AP Physics 1 Resources

Example Questions

Example Question #37 : Impulse And Momentum

Example Question #33 : Impulse And Momentum

Example Question #41 : Impulse And Momentum

Example Question #41 : Impulse And Momentum

Example Question #43 : Impulse And Momentum

Example Question #221 : Linear Motion And Momentum

Example Question #42 : Impulse And Momentum

Example Question #46 : Impulse And Momentum

Example Question #47 : Impulse And Momentum

Example Question #48 : Impulse And Momentum

All AP Physics 1 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: