Guide 9-1a. Solving Conservation of Momentum Problems in One Dimension

The law of conservation of momentum is the second of the major conservation laws that we'll use this year. The law in its most general form is written in a way that looks similar to the law of conservation of energy.

This looks suspiciously like the impulse-momentum theorem. However, there's an important difference. The impulse-momentum theorem applies to a single object. The object could, for example, be a ball colliding with the floor. In that case, F_net Δt = Δp, where F_net is the net force on the ball, and Δp is the change of momentum of the ball. The equation above applies to a system of objects. The system could, for example, be two colliding objects. In that case, F_net,ext represents the net, external force acting on the system, and Δp_net represents the total change of momentum of all the objects in the system. We list important points below.

In all of the problems that you will do with conservation of momentum in this course, you will be able to define the system in such a way that F_net,ext = 0. In that case, the law of conservation of momentum can be written as follows.

We will simplify the notation by letting upper case P represent the net momentum of the system. Therefore, we have for an isolated system the following.

In two dimensions, the vector equation becomes two component equations:

P_ix = P_fx

P_iy = P_fy

For 1-dimensional problems, we omit the x or y subscript.

The general plan of 1-dimensional conservation of momentum problem is similar to that of a conservation of energy problem with these exceptions: One selects the system so that the net, external force on the system is 0. Momentum is treated as a vector. Therefore, direction matters.

Now for some examples.

Example 1:  Two ice skaters push off against one another starting from a stationary position. The 45-kg skater acquires a velocity of 0.375 m/s to the right. What velocity does the 60-kg skater acquire?

Given: v1i = v2i = 0 v2f = +0.375 m/s m1 = 60 kg m2 = 45 kg Goal: Find v1f System:  ice skaters External forces: friction (assumed to be negligible), normal and weight add to 0 Initial state:  skaters motionless ready to push apart Final state:  skaters moving in opposite directions		The skaters exert internal forces on each other, but the external forces of normal and weight balance.  We're assuming no friction.  Actually, if we look at the situation immediately after they push off and haven't separated far, then any influence due to friction will be minimal. Therefore, we've selected the system so that Fnet,ext = 0. In the diagram, note that: +x is to the right. The skaters are denoted 1 and 2 to distinguish them. The initial and final velocities have double subscripts.  One subscript is for the object and the other is for the state.  (The textbook inserts a comma between the subscripts.)
Solution:
	We use upper-case P to denote the total momentum of the system and lower-case p to denote the momentum of a single object.  The total momentum is the sum of the individual momenta of the objects in the system.
	The total initial momentum of the system is 0, because the skaters are motionless.
	Substitute the definition of momentum .
	Solve for the unknown.
	Substitute values and units and reduce.  The sign of the result is negative, indicating motion to the left as expected.  The magnitude of the velocity of skater 1 is less than skater 2.  This is also expected, because skater 1 has the greater mass.

Example 2:  A 0.60-kg glider traveling at 8.0 m/s on a level air track undergoes a head-on collision with a 0.20-kg mass traveling toward it at 4.0 m/s. The two gliders stick in the collision.  What is the velocity of the combined gliders after the collision?

Given: v1i = +8.0 m/s v2i = -4.0 m/s m1 = 0.60 kg m2 = 0.20 kg Goal: Find vf System:  gliders External forces: friction with track (assumed to be negligible), normal and weight add to 0 Initial state:  gliders moving toward each other before collision Final state:  gliders moving together after collision		We have to be careful to assign the correct signs to the initial velocities. There is only one final velocity, so a single subscript is all that is needed on that velocity. The direction of vf isn't given, but we're guessing it's to the right.  The sign of the final result will tell us whether that choice is correct.
Solution:
	We proceed as in the last example setting the total initial and final momenta equal to each other and then writing each as the sum of the momenta of the objects in the system.
	The mass of the combined gliders is m1 + m2.  They have the same final velocity.
	Solve for the unknown.
	Substitute values and units and reduce.  The sign of the result is positive, indicating motion to the right as we had guessed.

Example 3.  A glider with mass m and speed v moving along a horizontal, frictionless air track collides with a stationary glider with a mass m/3. After the collision the first glider has a speed v/2. What is the velocity of the second glider? 

Given: v1i = v v2i = 0 v1f  = v/2 m1 = m m2 = m/3 Goal: Find v2f System:  gliders External forces: friction with track (assumed to be negligible), normal and weight add to 0 Initial state:  gliders before collision Final state:  gliders after collision		In order to solve this problem, we need to be given both masses and 3 velocities.  That leaves the fourth velocity as the only unknown.  Later, we'll see how to solve problems like this when both of the final velocities are unknown.
Solution:
	After the collision, both gliders are moving to the right.  Glider 1 is slower than glider 2.  Therefore, the gliders separate after the collision.

Go on to read about elastic and inelastic collisions.

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