Lesson 3 of this chapter of our Tutorial has focused on situations involving two-dimensional collisions. We have learned that for collisions occurring in an isolated system the total system momentum is conserved. The unique aspect of two-dimensional collisions is that we can analyze the x-dimension and the y-dimension independently of one another and observe that the total momentum in the x-direction is the same before the collision as it is after the collision. Similarly, the total momentum in the y-direction is the same before the collision as it is after the collision. Total system momentum is conserved in each dimension.
On the previous page of Lesson 3 we analyzed several hit-and-stick, right angle collisions. 
On this page of Lesson 3 we will analyze hit-and-bounce, glancing collisions. We will start with one of the simpler cases with one object initially in motion and the other object initially at rest. The diagram at the right depicts a typical before-and-after representation of this type of collision. Object B is at rest and Object A is in motion before the collision. After the collision, the two objects bounce away from each other and are in motion, likely at different speeds. Before the collision, only Object A possesses the momentum. And that momentum is directed to the right (or east). So the total momentum of the system is directed rightward. Since momentum is conserved, then we would expect that the total momentum of the system after the collision is also directed rightward (and only rightward). The system should not have any post-collision momentum in the vertical direction. But how can this be when both Object A and Object B are moving vertically after the collision? The only way for this to be the case is if the vertical component of momentum of Object A is equal to the vertical component of momentum of Object B. That is, the y-momentum of the two individual objects are equal and opposite such that they sum or add to 0.
Developing Momentum Equations
Now let's develop a symbolic representation of the verbal description given in the previous paragraph. We will use the following symbols in our representation; their definition is given.
mA: mass of object A
mB: mass of object B
vA: velocity of object A before the collision
vB: velocity of object B before the collision (0 m/s in our scenario)
vA': velocity of object A after the collision
vB': velocity of object B after the collision
θA: angle between the horizontal and the post-collision direction of object A
θB: angle between the horizontal and the post-collision direction of object B
We can re-draw our diagram with the symbols included. For clarity sake, we have included the x-y axis convention in the diagram.
The total system momentum before the collision in each dimension is represented by the following expressions:
x-dimension: mA•vA + mB•vB = mA•vA (vB is 0 m/s; the 2nd term cancels)
y-dimension: 0 kg•m/s
We must consider vector components for the after-collision analysis. Each object is moving at an angle to the axes. As such, each object has both an x- and a y-component of momentum. The values can be determined by using the sine and cosine functions. The expressions for the momentum values are mA•vA' and mB•vB'. By multiplying these values by the cosine and sine of the given angles, the x- and y-components can be calculated. The application of the sine and cosine functions to this situation is shown below.
We can summarize the above in table form:
| |
Before Collision Momentum |
After-Collision Momentum |
| x-dimension |
mA•vA |
mA•vA'•cosθA + mB•vB'•cosθB |
| y-dimension |
0 |
mB•vB'•cosθB - mA•vA'•cosθA |
Note in the last row of the table that the momentum of Object A after the collision is subtracted from the momentum of Object B. Object A is moving downward or in the negative direction after the collision. Because momentum is a vector, we have to account for this negative-direction of motion by subtracting the value for Object A's momentum from that of Object B.
Now we can conduct our after-collision momentum analysis using our symbolic representations in order to generate some equations for this situation. The equations are generated using the understanding the total momentum of the system before the collision is equal to the total momentum of the system after the collision ... for both dimensions.
x-dimension: mA•vA = mA•vA'•cosθA + mB•vB'•cosθB
y-dimension: 0 = mB•vB'•cosθB - mA•vA'•cosθA
The equation for the y-dimension can be rearranged, leading to the final two equations:
Next we will see how to implement this thinking and these equations with a couple of problems.
Example 1 - Puck Collision on an Air Table
A 0.50-kg puck (Puck A) moving rightward at 82 cm/s collides with a 0.25-kg stationary puck (Puck B) on an air table. After the collision, the 0.50-kg puck travels in a direction that is 23.6° from the original line of motion and the 0.25-kg puck travels in a direction that is 51.7° from the original line of motion. Determine their post-collision velocities.
Before the collision, the system momentum is possessed by Puck A. It's value is
psystem-before = pA = mA•vA = (0.50 kg)•(82 cm/s) = 41 kg•cm/s, Rightward
This momentum is x-momentum. There is no y-momentum. After the collision, there should be 41 kg•cm/s of rightward momentum and 0 kg•cm/s of vertical (y) momentum. And so we can state ...
x-dimension: 41 = (0.50)•vA'•cos(23.6°) + (0.25)•vB'•cos(51.7°)
y-dimension: 0 = (0.25)•vB'•sin(51.7°) - (0.50)•vA'•sin(23.6°)
(NOTE: units have been dropped to simplify the presentation. We will add them back in at the end of the solution.)
There are two unknowns in the above two equations. With as many equations as unknowns we know that at least in theory we can solve for the values of the unknowns. There are numerous methods of solving a system of two equations and two unknowns. We will take the approach of using the y-dimension equation to write an expression for vA' as a function of vB' and then taking that expression for vA' and substituting it into the x-dimension equation in order to solve for vB'. Once we determine vB', we can use our original expression to solve for vA'. Here it goes ... step-by-step:
From the y-dimension equation:
(0.25)•vB'•sin(51.7°) = (0.50)•vA'•sin(23.6°)
vA' = (0.25)•vB'•sin(51.7°) / [(0.50)•sin(23.6°)]
vA' = (0.980115232 ...)•vB'
(NOTE: we will use unrounded numbers through the entirety of our calculations and round to two significant digits once we are done.)
Now we will substitute this expression for vA' into the x-dimension equation, reducing that equation to an equation with a single unknown.
41 = (0.50)•[(0.980115232 ...)•vB']•cos(23.6°) + (0.25)•vB'•cos(51.7°)
Now we will perform careful algebraic steps to solve for vB'.
41 = (0.449070535 ...)•vB' + (0.154944757 ...)•vB'
41 = (0.604015293 ...)•vB'
vB' = 67.87907604 ... cm/s
Now that we have determined the value of vB', we will substitute it back into our original equation for vA' in order to solve for vA'.
vA' = (0.980115232 ...)•vB'
vA' = (0.980115232 ...)•(67.87907604 ...)
vA' = 66.52931636 ... cm/s
Our solution yields post-collision velocities of 67 cm/s for Puck A and 68 cm/s for Puck B.
vA' = 67 cm/s
vB' = 68 cm/s
Example 2 - Billiards Table Collision
A 161-gram billiard ball (Ball A) moving rightward at 1.50 m/s collides with a 169-gram stationary ball (Ball B). After the collision, Ball A moves in a direction that is 65.3° from the original line of motion and Ball B moves in a direction that is 28.0° from the original line of motion. Determine their post-collision velocities.
Before the collision, the system momentum is possessed by Ball A. It's value is
psystem-before = pA = mA•vA = (161 g)•(1.5 m/s) = 241.5 g•m/s, Rightward
This momentum is x-momentum. There is no y-momentum in the system. After the collision, there should be 241.5 g•m/s of rightward momentum and 0 g•m/s of vertical (y) momentum. And so we can state ...
x-dimension: 241.5 = (161)•vA'•cos(65.3°) + (169)•vB'•cos(28.0°)
y-dimension: 0 = (169)•vB'•sin(28.0°) - (161)•vA'•sin(65.3°)
(NOTE: units have been dropped to simplify the presentation. We will add them back in at the end of the solution.)
There are two unknowns in the above two equations. With as many equations as unknowns we know that at least in theory we can solve for the values of the unknowns. There are numerous methods of solving a system of two equations and two unknowns. We will take the approach of using the y-dimension equation to write an expression for vA' as a function of vB' and then taking that expression for vA' and substituting it into the x-dimension equation in order to solve for vB'. Once we determine vB', we can use our original expression to solve for vA'. Here it goes ... step-by-step:
From the y-dimension equation:
(161)•vA'•sin(65.3°) = (169)•vB'•sin(28.0°)
vA' = (169)•vB'•sin(28.0°) / [(161)•vB'•sin(65.3°)]
vA' = (0.542426974 ...)•vB'
(NOTE: we will use unrounded numbers through the entirety of our calculations and round velocity values to the second decimal place once we are done.)
Now we will substitute this expression for vA' into the x-dimension equation, reducing that equation to an equation with a single unknown.
241.5 = (161)•[(0.542426974 ...)•vB']•cos(65.3°) + (169)•vB'•cos(28.0°)
Now we will perform careful algebraic steps to solve for vB'.
241.5 = (36.49264201 ...)•vB' + (149.2181432 ...)•vB'
241.5 = (185.7107852 ...)•vB'
vB' = 1.300409127 ... m/s
Now that we have determined the value of vB', we will substitute it back into our original equation for vA' in order to solve for vA'.
vA' = (0.542426974 ...)•vB'
vA' = (0.542426974 ...)•(1.300409127 ...)
vA' = 0.705376987 ... m/s
Our solution yields post-collision velocities of 0.71 m/s for Ball A and 1.30 m/s for Ball B.
vA' = 0.71 m/s
vB' = 1.30 m/s
Taking a Deeper Glance at 2D Collisions
The analyses on this page include one object in motion colliding with another object at rest in a glancing collision. But the law of momentum conservation is powerful enough to handle any set of circumstances. As a book keeping tool, momentum conservation is powerful enough to determine after-collision velocities if we know enough about the before collision parameters (mass, velocity, and direction of both objects) and the post-collision directions of both objects. To illustrate the power of this book keeping tool, consider the final example problem.
A 2.00-kg object (A) moving East at 5.00 m/s collides with a 1.00-kg object (B) moving at 4.00 m/s in a direction that is 45.7° S of E. After the collision, Object A moves in a direction that is 28.3° S of E. Object B moves in a direction that is 8.7° N of E. Determine their post-collision velocities.
The solution begins by determining the total x- and total y-momentum of the system before the collision. In this situation, Object B is moving in both the x- and the y-dimension. No problem! We know vectors. We will begin by using sine and cosine to determine the total x- and total y-momentum of the system:
x-dimension: (2.00)•(5.00) + (1.00)•(4.00)•cos(45.7°) = 10.00 + 2.793661142 ... = 12.793661142 ..., East
y-dimension: (1.00)•(4.00)•sin(45.7°) = 2.862770935 ..., South
(NOTE: units have been dropped to simplify the presentation. We will add them back in at the end of the solution. And we will carry unrounded numbers throughout the solution and round to two decimal places at the very end of the solution.)
After the collision, both Object A and Object B have x- and y-momentum and both contribute to the total system momentum in each direction. We can write the following momentum conservation equations for the two dimensions:
x-dimension: 12.793661142 ...= (2.00)•vA'•cos(28.3°) + (1.00)•vB'•cos(8.7°)
y-dimension: 2.862770935 ...(South) = (2.00)•vA'•sin(28.3°) - (1.00)•vB'•sin(8.7°)
(NOTE: units have been dropped to simplify the presentation. We will add them back in at the end of the solution. And we will carry unrounded numbers throughout the solution and round to two decimal places at the very end of the solution.)
There are two unknowns in the above two equations. With as many equations as unknowns we know that at least in theory we can solve for the values of the unknowns. There are numerous methods of solving a system of two equations and two unknowns. We will take the approach of using the y-dimension equation to write an expression for vA' as a function of vB' and then taking that expression for vA' and substituting it into the x-dimension equation in order to solve for vB'. Once we determine vB', we can use our original expression to solve for vA'. Here it goes ... step-by-step:
From the y-dimension equation:
2.862770935 ...(South) = (2.00)•vA'•sin(28.3°) - (1.00)•vB'•sin(8.7°)
(2.00)•vA'•sin(28.3°) = 2.862770935 ... + (1.00)•vB'•sin(8.7°)
(0.948176418 ...)•vA' = 2.862770935 ... + (0.15126082 ...)•vB'
vA' = 3.019238699 ... + 0.159528139 ... • vB'
Now we will substitute this expression for vA' into the x-dimension equation, reducing that equation to an equation with a single unknown.
12.793661142 ... = (2.00)•[3.019238699 ... + 0.159528139 ... • vB']•cos(28.3°) + (1.00)•vB'•cos(8.7°)
Now we will perform careful algebraic steps to solve for vB'.
12.793661142 ... = 5.316742599 ... + (0.280921827...)•vB' + (0.988493886...)•vB'
7.476918541 ... = (1.269415713 ...)•vB'
vB' = (7.476918541 ...) / (1.269415713 ...)
vB' = 5.890047259 ... m/s
Now that we have determined the value of vB', we will substitute it back into our original equation for vA' in order to solve for vA'.
vA' = 3.019238699 ... + 0.159528139 ... • vB'
vA' = 3.019238699 ... + 0.159528139 ... • (5.890047259 ...)
vA' = 3.019238699 ... + 0.939628277
vA' = 3.958866977 ... m/s
Our solution yields post-collision velocities of 3.96 m/s for Object A and 5.89 m/s for Object B.
vA' = 3.96 m/s
vB' = 5.89 m/s
Momentum Conservation in Two Dimensions
As we have seen here in Lesson 3, momentum conservation is a powerful book keeping tool for analyzing two-dimensional collisions and predicting post-collision velocities of the colliding objects. It works for hit-and-stick, right angle collisions. It works for glancing collision in which one object is initially at rest. It works for collisions in which both objects are moving before the collision at acute angles to one another. It works for any collision.