<< Chapter < Page Chapter >> Page >
Illustration of collision of two cars with masses m 1 and m 2. The system of interest is the two cars before and after the collision. Before the collision, car m 2 is in front and moving forward with velocity v 2, and car m 1 is behind it, moving forward with velocity v 1. Net vector F = 0 and vectors p 1 plus p 2 equal p tot. After the collision, car m 2 is in front and moving forward with velocity v 2 prime which is larger than v 2 before the collision, and car m 1 is behind it, moving forward with velocity v 1 prime that is less than v 1 before the collision. Vectors p 1 prime plus p 2 prime equal p tot prime.
The two cars together form the system that is to be analyzed. It is important to remember that the contents (the mass) of the system do not change before, during, or after the objects in the system interact.

Problem-solving strategy: conservation of momentum

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

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

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 ( [link] ). 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?

An illustration of two lab carts on a track, stuck together.
Two lab carts collide and stick together after the collision.

Strategy

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.

Solution

Conservation of momentum is

p f = p i .

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

p i = m 1 v 1 i ^ + m 2 v 2 i ^ .

The final momentum of the now-linked carts is

p f = ( m 1 + m 2 ) v f .

Equating:

( m 1 + m 2 ) v f = m 1 v 1 i ^ + m 2 v 2 i ^ v f = ( m 1 v 1 + m 2 v 2 m 1 + m 2 ) i ^ .

Substituting the given numbers:

v f = [ ( 0.675 kg ) ( 0.75 m/s ) + ( 0.5 kg ) ( 1.33 m/s ) 1.175 kg ] i ^ = ( 0.997 m/s ) i ^ .

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.)

Check Your Understanding 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?

If the smaller cart were rolling at 1.33 m/s to the left, then conservation of momentum gives
( m 1 + m 2 ) v f = m 1 v 1 i ^ m 2 v 2 i ^ v f = ( m 1 v 1 m 2 v 2 m 1 + m 2 ) i ^ = [ ( 0.675 kg ) ( 0.75 m/s ) ( 0.500 kg ) ( 1.33 m/s ) 1.175 kg ] i ^ = ( 0.135 m/s ) i ^
Thus, the final velocity is 0.135 m/s to the left.

Got questions? Get instant answers now!

Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
Aislinn Reply
cm
tijani
what is titration
John Reply
what is physics
Siyaka Reply
A mouse of mass 200 g falls 100 m down a vertical mine shaft and lands at the bottom with a speed of 8.0 m/s. During its fall, how much work is done on the mouse by air resistance
Jude Reply
Can you compute that for me. Ty
Jude
what is the dimension formula of energy?
David Reply
what is viscosity?
David
what is inorganic
emma Reply
what is chemistry
Youesf Reply
what is inorganic
emma
Chemistry is a branch of science that deals with the study of matter,it composition,it structure and the changes it undergoes
Adjei
please, I'm a physics student and I need help in physics
Adjanou
chemistry could also be understood like the sexual attraction/repulsion of the male and female elements. the reaction varies depending on the energy differences of each given gender. + masculine -female.
Pedro
A ball is thrown straight up.it passes a 2.0m high window 7.50 m off the ground on it path up and takes 1.30 s to go past the window.what was the ball initial velocity
Krampah Reply
2. A sled plus passenger with total mass 50 kg is pulled 20 m across the snow (0.20) at constant velocity by a force directed 25° above the horizontal. Calculate (a) the work of the applied force, (b) the work of friction, and (c) the total work.
Sahid Reply
you have been hired as an espert witness in a court case involving an automobile accident. the accident involved car A of mass 1500kg which crashed into stationary car B of mass 1100kg. the driver of car A applied his brakes 15 m before he skidded and crashed into car B. after the collision, car A s
Samuel Reply
can someone explain to me, an ignorant high school student, why the trend of the graph doesn't follow the fact that the higher frequency a sound wave is, the more power it is, hence, making me think the phons output would follow this general trend?
Joseph Reply
Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
Joseph
Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing
Joseph
"Generation of electrical energy from sound energy | IEEE Conference Publication | IEEE Xplore" ***ieeexplore.ieee.org/document/7150687?reload=true
Ryan
what's motion
Maurice Reply
what are the types of wave
Maurice
answer
Magreth
progressive wave
Magreth
hello friend how are you
Muhammad Reply
fine, how about you?
Mohammed
hi
Mujahid
A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string?
yasuo Reply
Who can show me the full solution in this problem?
Reofrir Reply
Practice Key Terms 3

Get Jobilize Job Search Mobile App in your pocket Now!

Get it on Google Play Download on the App Store Now




Source:  OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'University physics volume 1' conversation and receive update notifications?

Ask