Focusing on Newton's 3rd Law which states that "when object A exerts a force on object B then object B exerts an equal but opposite force on object A," we can state that when two objects collide, the total momentum present before a collision ALWAYS equals the total momentum present after a collision - that is, momentum is ALWAYS conserved during collisions.

One-dimensional collisions.

Consider two balls rolling towards each other. 

The 3 kg mass has a positive momentum of 6 kg m/sec and the 2 kg mass has a negative momentum of 6 kg m/sec. When they collide, their momentums will cancel, and they will either both with stop dead in their tracks, or rebound with .

Two-dimensional collisions.

Moreover, objects traveling at a diagonal have momentum in each of two dimensions: x and y.  Hence, momentum vectors have to be resolved when working 2-D collisions.   Consider, for example, the following system.

.

x components y components BEFORE COLLISION 2(-3) = -6 kg m/sec 0 0 3(2) = 6 kg m/sec AFTER COLLISION - 5 vc cos q 5 vc sin q ANALYSIS conservation of momentum -6 + 0 = - 5 vc cos q 0 + 6 = 5 vc sin q vc cos q = 6/5 vc sin q = 6/5 tan q = 1 q = 45º vc cos (45) = 6/5 vc (0.707) = 1.2 vc = 1.2 / 0.707 vc = 1.70 m/sec

The relationship between conservation of energy and conservation of momentum is an important one.  During any collision, momentum is conserved.  However, sometimes, kinetic energy is also conserved during a collision. That type of collision is called a perfectly elastic collision.  In theory, molecules of ideal gases collide elastically as do photons and electrons.  In many problems, they will state that a steel ball or a "superball" bounces elastically off a surface. The implication is that the ball will return to the same height from which it is released.  That is, its original gravitational potential energy is transformed into kinetic energy which remains constant throughout the collision allowing the ball to return to its original height. This type of collision is rare and seldom actually occurs.  A means of measuring the elasticity of a collision is called the coefficient of restitution, e. The value of e equals the ratio of the  relative velocities of the objects as they move away from each other to the relative velocities when they were approaching each other. If e = 1, then the objects must remain separated (that is, bounce apart) and leave the collision with the same total amount of KE as they had in their original approach. Their collision is said to be perfectly elastic. If e = 0, then the two objects must stick together and the collision is said to be perfectly inelastic.  Most collisions are somewhere in-between, hovering closer to e-values less than 0.5 in which the objects bounce apart but lose energy to heat, light, sound, or structural deformation.

But even if collisions are inelastic, we often use conservation of energy methods to analyze the behaviors of the objects involved. We do this by considering the energy content of the system before the objects collide and then again after they collide - just NOT during the collision. Given below are some classic examples.

[ballistic pendulum] A 10 gram bullet is fired from a rifle at a speed of 700 m/sec into a 1.50 kg block of a ballistic pendulum suspended by a string 4 meters long. After the collision, through what vertical distance does the block rise?

During the collision we can only use conservation of momentum.

After the collision, the block and bullet are now behaving as one object. We can use conservation of energy to determine how high the pendulum swings.

To determine how much KE is lost, compare the total KE before the collision to the total KE afterwards.

This collision lost 2434 J of energy, or 99.3% of the bullet's original energy!

[table projectile] A block having a mass of 800 grams is at rest near the edge of a frictionless table.  A second block, of mass 600 grams, is sliding towards it at a speed of 4 m/sec.  If the table is 1 meter above the floor, and the blocks stick together after the collision, how far from the end of the table will they strike the floor?

During the collision we can only use conservation of momentum.

After the collision, the two blocks now behave as one object. We can use kinematics to determine how far from the base of the table that they strike the floor.

[impact collision] A block having a mass of 600 grams is sliding towards the edge of a frictionless table at a speed of 4 m/sec. On the floor, at the base of the table is a 200 gram cart, initially at rest. The table is 1 meter above the floor. (1) How far from the base of the table should the cart be placed? (2) If the block sticks to the cart after the collision, how fast will the block-cart combination begin to move across the floor?

The first part of this problem is simply that of a two-dimensional projectile. We will use an H|V chart to determine the range of 600 gram block so that we can place the cart.

When the block collides with the cart, the resulting motion will be horizontal across the floor. Therefore, we will examine conservation of momentum along the x-axis to obtain the speed that they slip across the floor.

[skid marks] A 2000 kg car moving with an unknown speed strikes a 1000 kg car that is initially at rest. After the collision, the two cars stick together and skid to a stop on asphalt, where the friction coefficent is 0.65 The police measure the skid marks to be 40 meters long, determine the speed of the 2000 kg car just before the collision.

Initially, this is a momentum problem involving a one-dimensional collision.  That process will determine how fast the two cars begin their skid.  Then it becomes a work-energy problem as the work done by friction brings them to a stop.

Now we will use the work-energy theorem to determine the velocity immediately after the collision.