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By the end of this section, you will be able to:
  • Express momentum as a two-dimensional vector
  • Write equations for momentum conservation in component form
  • Calculate momentum in two dimensions, as a vector quantity

It is far more common for collisions to occur in two dimensions; that is, the angle between the initial velocity vectors is neither zero nor 180 ° . Let’s see what complications arise from this.

The first idea we need is that momentum is a vector; like all vectors, it can be expressed as a sum of perpendicular components (usually, though not always, an x -component and a y -component, and a z -component if necessary). Thus, when we write down the statement of conservation of momentum for a problem, our momentum vectors can be, and usually will be, expressed in component form.

The second idea we need comes from the fact that momentum is related to force:

F = d p d t .

Expressing both the force and the momentum in component form,

F x = d p x d t , F y = d p y d t , F z = d p z d t .

Remember, these equations are simply Newton’s second law, in vector form and in component form. We know that Newton’s second law is true in each direction, independently of the others. It follows therefore (via Newton’s third law) that conservation of momentum is also true in each direction independently.

These two ideas motivate the solution to two-dimensional problems: We write down the expression for conservation of momentum twice: once in the x -direction and once in the y -direction.

p f , x = p 1,i , x + p 2,i , x p f , y = p 1,i , y + p 2,i , y

This procedure is shown graphically in [link] .

Figure a, titled break initial momentum into x and y components shows vector p 1 i as a solid arrow pointing to the right and down. Its components are shown as dashed arrows: p 1 i y points down from the tail of p 1 i and p 1 i x points to the right from the head of p 1 i y to the head of p 1 i. Vector p 2 i is shown as a solid arrow with its tail at the head of vector p 1 i, and is shorter than p 1 i. Vector p 2 i points to the right and up. Its components are shown as dashed arrows: p 2 i x points to the right from the tail of p 2 i and p 2 i y points up from the head of p 2 i x to the head of p 2 i. Vector p f points from the tail of p 1 i to the head of p 2 i, pointing to the right and slightly down. Figure b titled add x and y components to obtain x and y components of final momentum shows the vector sums of the components. P 1 i y is a downward arrow. P 2 i y is a shorter upward arrow, aligned with its tail at the head of P 1 i y. P f y is a short downward arrow that starts at the tail of P 1 i y and ends at the head of P 2 i y. P 1 i x is a rightward arrow. P 2 i x is a shorter rightward arrow, aligned with its tail at the head of P 1 i x. P f x is a long rightward arrow that starts at the tail of P 1 i x and ends at the head of P 2 i x. Figure c, titled add x and y components of final momentum shows the right triangle formed by sides p f x and p f y and hypotenuse p f. Arrows from figure b indicate that p f x and p f y are the same in figures b and c.
(a) For two-dimensional momentum problems, break the initial momentum vectors into their x - and y -components. (b) Add the x - and y -components together separately. This gives you the x - and y -components of the final momentum, which are shown as red dashed vectors. (c) Adding these components together gives the final momentum.

We solve each of these two component equations independently to obtain the x - and y -components of the desired velocity vector:

v f , x = m 1 v 1,i , x + m 2 v 2,i , x m v f , y = m 1 v 1,i , y + m 2 v 2,i , y m .

(Here, m represents the total mass of the system.) Finally, combine these components using the Pythagorean theorem,

v f = | v f | = v f , x 2 + v f , y 2 .

Problem-solving strategy: conservation of momentum in two dimensions

The method for solving a two-dimensional (or even three-dimensional) conservation of momentum problem is generally the same as the method for solving a one-dimensional problem, except that you have to conserve momentum in both (or all three) dimensions simultaneously:

  1. Identify a closed system.
  2. Write down the equation that represents conservation of momentum in the x -direction, and solve it for the desired quantity. If you are calculating a vector quantity (velocity, usually), this will give you the x -component of the vector.
  3. Write down the equation that represents conservation of momentum in the y -direction, and solve. This will give you the y -component of your vector quantity.
  4. Assuming you are calculating a vector quantity, use the Pythagorean theorem to calculate its magnitude, using the results of steps 3 and 4.

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
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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
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David Reply
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David
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emma Reply
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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
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Adjanou
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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
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Maurice Reply
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Maurice
answer
Magreth
progressive wave
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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
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Source:  OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5
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