PhysicsLAB Resource Lesson
Linear Momentum

Printer Friendly Version
Momentum is defined as the product of an object's mass and its velocity. Since velocity is a vector quantity and mass is a scalar quantity, momentum's vector nature is dependent on the vector properties of the object's velocity. If an object is moving in a positive direction, then its momentum is considered to be positive. Momentum can be represented by the variable p and has units of kg m/sec.
p = mv
Since momentum is a vector, all objects in a closed system can individually have momentum but the system as a whole can have a net momentum of zero. Consider, for example, the following system:
the 3 kg mass has a positive momentum of 6 kg m/sec;
the 2 kg mass has a negative momentum of 6 kg m/sec;
each mass is moving, but the total momentum of the system equals 0.
Derivations based on Newton's Laws
To understand the properties of momentum, we must first re-examine Newton's 2nd and 3rd Laws.
Newton's 2nd Law
net F = ma
net F = m (vf - vo) / t
(net F)t = m (vf - vo)
(net F) t = mvf - mvo
(net F)t = pf - po
(net F)t = Δp
impulse = the change in momentum
Newton's 3rd Law
FAB t = - FBA t
mBvfB - mBvoB = - (mAvfA- mAvoA)
mBvfB + mAvfA = mAvoA + mBvoB
Σpf = Σpo
The Law of Conservation of Momentum states the sum of the momenta before a collision equals the sum of the momenta after a collision
Impulse can be represented by the variable J and has units of N sec. Since the impulse an object receives equals the change in its momentum, the units for impulse and momentum must be equivalent.
N sec kg m/sec

[kg m/sec2] sec
kg m/sec
Note that in the equation for impulse, if the amount of momentum being changed remains constant, then the net force can be maximized by decreasing the time of impact. That is, a short jab is much more effective than a long, drawn-out swing when striking someone for maximum force. Conversely, when a large amount of momentum has to be lost doing so over a longer period of time decreases the net force applied to the object. For example, when you jump off of a high tower bending your knees when you land decreases the impact force by increasing the time of impact. Moreover, if the amount of force remains constant then the change in momentum can be maximized by increasing the time of impact. For example, in sports, coaches are always encouraging athletes to "follow through." This allows the force to act on the ball for a greater amount of time therefore increasing the time of impact and the net impulse delivered to the ball. Subsequently, the ball receives a greater change in its momentum and has a greater final velocity.
Accompanying this restatement for Newton's 2nd law are two important graphical relationships.
Conservation of Momentum
During any collision, momentum is conserved as a consequence of Newton's 3rd Law - the Law of Action-Reaction. What this means is that the total momentum before a collision is always equal to the total momentum after a collision.
Σpo = Σpf
There are two general categories of collisions: elastic and inelastic. Elastic collisions occur when objects bounce off each other AND kinetic energy is conserved during the collision. These types of collisions are more "ideal" in nature and are difficult to observe. Usually students encounter discussions of elastic collisions when working problems with collisions between objects and ideal springs as well as collisions on the subatomic level between photons and electrons. They are also one of the fundamental assumptions when deriving laws of ideal gases. In our everyday world, the majority of collisions are inelastic.
Inelastic collisions occur when objects collide and energy is lost during the collision. That is, ΣKEbefore > ΣKEafter. During perfectly inelastic collisions the objects stick together during the collision and leave as one single mass. The degree to which a collision is considered to be elastic or inelastic is measured by a quantity called the coefficient of restitution, e, which is defined as the ratio of the (relative velocities of recession) /(relative velocities of approach) for the two objects involved in the collision. When e = 1, the collision is perfectly elastic; when e = 0 is it perfectly inelastic.
e = (vf2 - vf1) / (vo1 - vo2)
Recoil velocities are considered in problems involving explosions: a gun is fired and the bullet goes in one direction while the gun "recoils" in the opposite direction, or when a canon is fired, or a coiled spring is released. Usually the total momentum before the collision is zero (the system is at rest) and then the objects travel off in opposite directions with a total final momentum that must also add to zero.
Let's look at a few examples.
 Suppose that a 3-kg mass moving at 5 m/sec strikes a stationary 1-kg mass whereupon they stick other. What will be the final velocity of the combined mass, vc, after this inelastic collision?

 Suppose that a 3-kg mass moving at 2 m/sec strikes a 1-kg mass moving towards it at 10 m/sec whereupon they stick together. What will be the final velocity of the combined mass, vc, after this inelastic collision?

 Suppose a 1-kg pistol containing a 10-gram bullet is resting on a table when it is accidentally discharged. If the bullet has a muzzle velocity of 150 m/sec, how fast will the pistol recoil?

 How do the impulses an object receives compare when (1) it strikes a wall and sticks to it, versus (2) it rebounds elastically off of the wall?

When objects travel along diagonal paths, they have momentum in each of the x and y dimensions. Consequently momentum vectors must be resolved into their x- and y-components when working two-dimensional collisions.

Consider the following scenario.

 A 2-kg mass traveling left at 3 m/sec along the positive x-axis and a 3-kg mass traveling upwards along the positive y-axis at 2 m/sec collide with each other. If the two masses stick together during the collision, how fast and in what direction will they leave the collision?


Related Documents

Copyright © 1997-2014
Catharine H. Colwell
All rights reserved.
Application Programmer
    Mark Acton