5.3 Non-Linear Examples

Applying the law of conservation of momentum in two-dimensional collisions is more complex than that of a one-dimensional collision.  Essentially, the law of conservation of momentum is applied twice-once in the x and once in the y directions.  View the demonstration of adding vectors for a 2-D collision then read the following example of how to apply the law of conservation of momentum correctly to solve two-dimensional collision problems.

Example Problem

An 8.0-kg mass collides elastically with a 5.0-kg mass that is at rest.  Initially, the 8.0-kg mass was travelling to the right at 4.5 m/s.  After the collision, the mass moves with a speed of 3.65 m/s and at an angle of 27° to its original direction.

What is the final velocity for the 5.0-kg mass?

Example Solution
 

«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«msub»«mi»m«/mi»«mn»1«/mn»«/msub»«mo»=«/mo»«mn»8«/mn»«mo».«/mo»«mn»0«/mn»«mo»§#160;«/mo»«mi»k«/mi»«mi»g«/mi»«mspace linebreak=¨newline¨»«/mspace»«msub»«mover»«mi»v«/mi»«mo»§#8594;«/mo»«/mover»«mrow»«mn»1«/mn»«mi»i«/mi»«/mrow»«/msub»«mo»=«/mo»«mn»4«/mn»«mo».«/mo»«mn»5«/mn»«mo»§#160;«/mo»«mfrac bevelled=¨true¨»«mi»m«/mi»«mi»s«/mi»«/mfrac»«mo»§#160;«/mo»«mo»[«/mo»«msup»«mn»0«/mn»«mo»§#8728;«/mo»«/msup»«mo»]«/mo»«mspace linebreak=¨newline¨»«/mspace»«msub»«mover»«mi»v«/mi»«mo»§#8594;«/mo»«/mover»«mrow»«mn»1«/mn»«mi»f«/mi»«/mrow»«/msub»«mo»=«/mo»«mn»3«/mn»«mo».«/mo»«mn»65«/mn»«mo»§#160;«/mo»«mfrac bevelled=¨true¨»«mi»m«/mi»«mi»s«/mi»«/mfrac»«mo»§#160;«/mo»«mo»[«/mo»«msup»«mn»27«/mn»«mo»§#8728;«/mo»«/msup»«mo»]«/mo»«mspace linebreak=¨newline¨»«/mspace»«msub»«mover»«mi»v«/mi»«mo»§#8594;«/mo»«/mover»«mrow»«mn»2«/mn»«mi»i«/mi»«/mrow»«/msub»«mo»=«/mo»«mn»0«/mn»«/math»

Required: The final velocity of mass 2, «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«msub»«mover»«mi»v«/mi»«mo»§#8594;«/mo»«/mover»«mrow»«mn»2«/mn»«mi»f«/mi»«/mrow»«/msub»«/math».

Analysis and Solution: 

Initially, mass 1 moves to the right and mass 2 is at rest.  Thus, the total momentum is in the x-direction (to the right).  After the collision, mass 1 travels at an angle of 27° to its original direction of motion; it now has momentum in the y-direction as well as the x-direction.  For momentum to be conserved, mass 2 must have momentum in both the x- and y- directions.

Define motion to the right as positive, which follows the usual convention.  As well, measure angles counterclockwise from the horizontal.

Use the law of conservation of momentum to solve this question.  Because momentum is conserved in both the x- and y-directions, we can set up two sets of equations-one each to solve for the x- and y- components of the final velocity of mass 2.

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 «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle indentalign=¨left¨»«msub»«mi»v«/mi»«mrow»«mn»2«/mn»«mi»y«/mi»«mi»f«/mi»«/mrow»«/msub»«mo»=«/mo»«mo»-«/mo»«mfrac»«mrow»«mo»(«/mo»«mn»8«/mn»«mo».«/mo»«mn»0«/mn»«mo»§#160;«/mo»«mi»k«/mi»«mi»g«/mi»«mo»)«/mo»«mo»(«/mo»«mn»3«/mn»«mo».«/mo»«mn»65«/mn»«mo»§#160;«/mo»«mi»m«/mi»«mo»/«/mo»«mi»s«/mi»«mo»)«/mo»«mi»sin«/mi»«mo»(«/mo»«msup»«mn»27«/mn»«mo»§#8728;«/mo»«/msup»«mo»)«/mo»«/mrow»«mrow»«mn»5«/mn»«mo».«/mo»«mn»0«/mn»«mo»§#160;«/mo»«mi»k«/mi»«mi»g«/mi»«/mrow»«/mfrac»«mspace linebreak=¨newline¨»«/mspace»«msub»«mi»v«/mi»«mrow»«mn»2«/mn»«mi»y«/mi»«mi»f«/mi»«/mrow»«/msub»«mo»=«/mo»«mo»-«/mo»«mn»2«/mn»«mo».«/mo»«mn»651«/mn»«mo»§#160;«/mo»«mn»304«/mn»«mo»§#160;«/mo»«mn»508«/mn»«mo»§#160;«/mo»«mi»m«/mi»«mo»/«/mo»«mi»s«/mi»«/mstyle»«/math»

Once you know the final velocity in the x and y directions, you can find the final velocity and direction of motion for mass 2. Remember, do not use the rounded-off values in your calculations-use rounded numbers for the final paraphrase.

Determine the magnitude of the velocity for mass 2 by using Pythagorean Theorem

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Determine the direction of the velocity using the inverse of tangent

«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»tan«/mi»«mi»§#945;«/mi»«mo»=«/mo»«mfrac»«msub»«mi»v«/mi»«mrow»«mn»2«/mn»«mi»f«/mi»«mi»y«/mi»«/mrow»«/msub»«msub»«mi»v«/mi»«mrow»«mn»2«/mn»«mi»f«/mi»«mi»x«/mi»«/mrow»«/msub»«/mfrac»«mspace linebreak=¨newline¨/»«mi»§#945;«/mi»«mo»=«/mo»«msup»«mi»tan«/mi»«mrow»«mo»-«/mo»«mn»1«/mn»«/mrow»«/msup»«mfenced open=¨[¨ close=¨]¨»«mfrac»«mrow»«mo»-«/mo»«mn»2«/mn»«mo».«/mo»«mn»651«/mn»«mo»§#160;«/mo»«mn»304«/mn»«mo»§#160;«/mo»«mn»508«/mn»«mo»§#160;«/mo»«mi»m«/mi»«mo»/«/mo»«mi»s«/mi»«/mrow»«mrow»«mn»1«/mn»«mo».«/mo»«mn»996«/mn»«mo»§#160;«/mo»«mn»521«/mn»«mo»§#160;«/mo»«mn»899«/mn»«mo»§#160;«/mo»«mi»m«/mi»«mo»/«/mo»«mi»s«/mi»«/mrow»«/mfrac»«/mfenced»«mspace linebreak=¨newline¨/»«mi»§#945;«/mi»«mo»=«/mo»«mo»-«/mo»«mn»53«/mn»«mo».«/mo»«mn»019«/mn»«mo»§#160;«/mo»«mn»017«/mn»«mo»§#160;«/mo»«msup»«mn»07«/mn»«mo»§#8728;«/mo»«/msup»«mspace linebreak=¨newline¨/»«mi»§#945;«/mi»«mo»=«/mo»«mo»-«/mo»«msup»«mn»53«/mn»«mrow»«mo»§#8728;«/mo»«mo»§#160;«/mo»«/mrow»«/msup»«mi»c«/mi»«mi»o«/mi»«mi»r«/mi»«mi»r«/mi»«mi»e«/mi»«mi»c«/mi»«mi»t«/mi»«mo»§#160;«/mo»«mi»t«/mi»«mi»o«/mi»«mo»§#160;«/mo»«mn»2«/mn»«mo»§#160;«/mo»«mi»s«/mi»«mi»i«/mi»«mi»g«/mi»«mi»n«/mi»«mi»i«/mi»«mi»f«/mi»«mi»i«/mi»«mi»c«/mi»«mi»a«/mi»«mi»n«/mi»«mi»t«/mi»«mo»§#160;«/mo»«mi»d«/mi»«mi»i«/mi»«mi»g«/mi»«mi»i«/mi»«mi»t«/mi»«mi»s«/mi»«/math»

This answer seems plausible-mass 2 has to move to the right and down to conserve momentum.