Lesson 3 of this chapter of our Tutorial has focused on situations involving twodimensional collisions. We have learned that for collisions occurring in an isolated system the total system momentum is conserved. The unique aspect of twodimensional collisions is that we can analyze the xdimension and the ydimension independently of one another and observe that the total momentum in the xdirection is the same before the collision as it is after the collision. Similarly, the total momentum in the ydirection 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 hitandstick, right angle collisions. On this page of Lesson 3 we will analyze hitandbounce, 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 beforeandafter 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 postcollision 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 ymomentum 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.
m_{A}: mass of object A
m_{B}: mass of object B
v_{A}: velocity of object A before the collision
v_{B}: velocity of object B before the collision (0 m/s in our scenario)
v_{A}': velocity of object A after the collision
v_{B}': velocity of object B after the collision
θ_{A}: angle between the horizontal and the postcollision direction of object A
θ_{B}: angle between the horizontal and the postcollision direction of object B
We can redraw our diagram with the symbols included. For clarity sake, we have included the xy axis convention in the diagram.
The total system momentum before the collision in each dimension is represented by the following expressions:
xdimension: m_{A}•v_{A} + m_{B}•v_{B} = m_{A}•v_{A} (v_{B} is 0 m/s; the 2nd term cancels)
ydimension: 0 kg•m/s
We must consider vector components for the aftercollision analysis. Each object is moving at an angle to the axes. As such, each object has both an x and a ycomponent of momentum. The values can be determined by using the sine and cosine functions. The expressions for the momentum values are m_{A}•v_{A}' and m_{B}•v_{B}'. By multiplying these values by the cosine and sine of the given angles, the x and ycomponents 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 
AfterCollision Momentum 
xdimension 
m_{A}•v_{A} 
m_{A}•v_{A}'•cosθ_{A} + m_{B}•v_{B}'•cosθ_{B} 
ydimension 
0 
m_{B}•v_{B}'•cosθ_{B}  m_{A}•v_{A}'•cosθ_{A} 
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 negativedirection of motion by subtracting the value for Object A's momentum from that of Object B.
Now we can conduct our aftercollision 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.
xdimension: m_{A}•v_{A} = m_{A}•v_{A}'•cosθ_{A} + m_{B}•v_{B}'•cosθ_{B}
ydimension: 0 = m_{B}•v_{B}'•cosθ_{B}  m_{A}•v_{A}'•cosθ_{A}
The equation for the ydimension 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 0.50kg puck (Puck A) moving rightward at 82 cm/s collides with a 0.25kg stationary puck (Puck B) on an air table. After the collision, the 0.50kg puck travels in a direction that is 23.6° from the original line of motion and the 0.25kg puck travels in a direction that is 51.7° from the original line of motion. Determine their postcollision velocities.
Before the collision, the system momentum is possessed by Puck A. It's value is
p_{systembefore} = p_{A} = m_{A}•v_{A} = (0.50 kg)•(82 cm/s) = 41 kg•cm/s, Rightward
This momentum is xmomentum. There is no ymomentum. 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 ...
xdimension: 41 = (0.50)•v_{A}'•cos(23.6°) + (0.25)•v_{B}'•cos(51.7°)
ydimension: 0 = (0.25)•v_{B}'•sin(51.7°)  (0.50)•v_{A}'•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 ydimension equation to write an expression for v_{A}' as a function of v_{B}' and then taking that expression for v_{A}' and substituting it into the xdimension equation in order to solve for v_{B}'. Once we determine v_{B}', we can use our original expression to solve for v_{A}'. Here it goes ... stepbystep:
From the ydimension equation:
(0.25)•v_{B}'•sin(51.7°) = (0.50)•v_{A}'•sin(23.6°)
v_{A}' = (0.25)•v_{B}'•sin(51.7°) / [(0.50)•sin(23.6°)]
v_{A}' = (0.980115232 ...)•v_{B}'
(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 v_{A}' into the xdimension equation, reducing that equation to an equation with a single unknown.
41 = (0.50)•[(0.980115232 ...)•v_{B}']•cos(23.6°) + (0.25)•v_{B}'•cos(51.7°)
Now we will perform careful algebraic steps to solve for v_{B}'.
41 = (0.449070535 ...)•v_{B}' + (0.154944757 ...)•v_{B}'
41 = (0.604015293 ...)•v_{B}'
v_{B}' = 67.87907604 ... cm/s
Now that we have determined the value of v_{B}', we will substitute it back into our original equation for v_{A}' in order to solve for v_{A}'.
v_{A}' = (0.980115232 ...)•v_{B}'
v_{A}' = (0.980115232 ...)•(67.87907604 ...)
v_{A}' = 66.52931636 ... cm/s
Our solution yields postcollision velocities of 67 cm/s for Puck A and 68 cm/s for Puck B.
v_{A}' = 67 cm/s
v_{B}' = 68 cm/s
Example 2  Billiards Table Collision
A 161gram billiard ball (Ball A) moving rightward at 1.50 m/s collides with a 169gram 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 postcollision velocities.
Before the collision, the system momentum is possessed by Ball A. It's value is
p_{systembefore} = p_{A} = m_{A}•v_{A} = (161 g)•(1.5 m/s) = 241.5 g•m/s, Rightward
This momentum is xmomentum. There is no ymomentum 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 ...
xdimension: 241.5 = (161)•v_{A}'•cos(65.3°) + (169)•v_{B}'•cos(28.0°)
ydimension: 0 = (169)•v_{B}'•sin(28.0°)  (161)•v_{A}'•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 ydimension equation to write an expression for v_{A}' as a function of v_{B}' and then taking that expression for v_{A}' and substituting it into the xdimension equation in order to solve for v_{B}'. Once we determine v_{B}', we can use our original expression to solve for v_{A}'. Here it goes ... stepbystep:
From the ydimension equation:
(161)•v_{A}'•sin(65.3°) = (169)•v_{B}'•sin(28.0°)
v_{A}' = (169)•v_{B}'•sin(28.0°) / [(161)•v_{B}'•sin(65.3°)]
v_{A}' = (0.542426974 ...)•v_{B}'
(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 v_{A}' into the xdimension equation, reducing that equation to an equation with a single unknown.
241.5 = (161)•[(0.542426974 ...)•v_{B}']•cos(65.3°) + (169)•v_{B}'•cos(28.0°)
Now we will perform careful algebraic steps to solve for v_{B}'.
241.5 = (36.49264201 ...)•v_{B}' + (149.2181432 ...)•v_{B}'
241.5 = (185.7107852 ...)•v_{B}'
v_{B}' = 1.300409127 ... m/s
Now that we have determined the value of v_{B}', we will substitute it back into our original equation for v_{A}' in order to solve for v_{A}'.
v_{A}' = (0.542426974 ...)•v_{B}'
v_{A}' = (0.542426974 ...)•(1.300409127 ...)
v_{A}' = 0.705376987 ... m/s
Our solution yields postcollision velocities of 0.71 m/s for Ball A and 1.30 m/s for Ball B.
v_{A}' = 0.71 m/s
v_{B}' = 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 aftercollision velocities if we know enough about the before collision parameters (mass, velocity, and direction of both objects) and the postcollision directions of both objects. To illustrate the power of this book keeping tool, consider the final example problem.
A 2.00kg object (A) moving East at 5.00 m/s collides with a 1.00kg 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 postcollision velocities.
The solution begins by determining the total x and total ymomentum of the system before the collision. In this situation, Object B is moving in both the x and the ydimension. No problem! We know vectors. We will begin by using sine and cosine to determine the total x and total ymomentum of the system:
xdimension: (2.00)•(5.00) + (1.00)•(4.00)•cos(45.7°) = 10.00 + 2.793661142 ... = 12.793661142 ..., East
ydimension: (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 ymomentum and both contribute to the total system momentum in each direction. We can write the following momentum conservation equations for the two dimensions:
xdimension: 12.793661142 ...= (2.00)•v_{A}'•cos(28.3°) + (1.00)•v_{B}'•cos(8.7°)
ydimension: 2.862770935 ...(South) = (2.00)•v_{A}'•sin(28.3°)  (1.00)•v_{B}'•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 ydimension equation to write an expression for v_{A}' as a function of v_{B}' and then taking that expression for v_{A}' and substituting it into the xdimension equation in order to solve for v_{B}'. Once we determine v_{B}', we can use our original expression to solve for v_{A}'. Here it goes ... stepbystep:
From the ydimension equation:
2.862770935 ...(South) = (2.00)•v_{A}'•sin(28.3°)  (1.00)•v_{B}'•sin(8.7°)
(2.00)•v_{A}'•sin(28.3°) = 2.862770935 ... + (1.00)•v_{B}'•sin(8.7°)
(0.948176418 ...)•v_{A}' = 2.862770935 ... + (0.15126082 ...)•v_{B}'
v_{A}' = 3.019238699 ... + 0.159528139 ... • v_{B}'
Now we will substitute this expression for v_{A}' into the xdimension equation, reducing that equation to an equation with a single unknown.
12.793661142 ... = (2.00)•[3.019238699 ... + 0.159528139 ... • v_{B}']•cos(28.3°) + (1.00)•v_{B}'•cos(8.7°)
Now we will perform careful algebraic steps to solve for v_{B}'.
12.793661142 ... = 5.316742599 ... + (0.280921827...)•v_{B}' + (0.988493886...)•v_{B}'
7.476918541 ... = (1.269415713 ...)•v_{B}'
v_{B}' = (7.476918541 ...) / (1.269415713 ...)
v_{B}' = 5.890047259 ... m/s
Now that we have determined the value of v_{B}', we will substitute it back into our original equation for v_{A}' in order to solve for v_{A}'.
v_{A}' = 3.019238699 ... + 0.159528139 ... • v_{B}'
v_{A}' = 3.019238699 ... + 0.159528139 ... • (5.890047259 ...)
v_{A}' = 3.019238699 ... + 0.939628277
v_{A}' = 3.958866977 ... m/s
Our solution yields postcollision velocities of 3.96 m/s for Object A and 5.89 m/s for Object B.
v_{A}' = 3.96 m/s
v_{B}' = 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 twodimensional collisions and predicting postcollision velocities of the colliding objects. It works for hitandstick, 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.