Along with the conservation of energy, conservation of momentum makes the second of two pillars used in physics.

Here's our statement of the conservation of linear momentum:

Topic 2.2 ExtendedC – Conservation of linear momentum

The total linear momentum of a system is conserved if the net external force acting on the system is zero.

FYI: Another type of momentum is angular momentum. This type of momentum is not considered in this chapter.

In mathematical symbols we have

If Fext = 0 thenP = 0P0 = Pf

Conservation of Linear Momentum

FYI: We use a capital P to represent the total momentum of a system of more than one particle. Lower case p is used for single particles.

PROOF OF THE CONSERVATION OF LINEAR MOMENTUMRecall Newton's 2nd law (p-form):

Topic 2.2 ExtendedC – Conservation of linear momentum

Fnet =pt

Newton's Second Law (p-form)

We begin by dividing the net force into all of its individual forces - both external and internal:

Fnet = F1,ext +...+ Fn,ext + F1,int +...+ Fm,int

From Newton's 3rd law, every internal force has an equal and opposite reaction force. Therefore, the sum of the internal forces is zero.

FYI: In any system, the internal forces will sum to ZERO.

Now if the external forces also sum to zero, we have

Fnet = 0 so that 0 = p

t. Thus P = 0. //

Topic 2.2 ExtendedC – Conservation of linear momentum

Here’s a sample problem:Suppose a .02-kg bullet traveling horizontally at 300 m/s strikes a 4-kg block of wood resting on a wood floor. (a) How fast is the block/bullet combo moving immediately after collision?

If we consider the bullet-block combo as our system, there are no external forces in the x-direction. Thus

Pf = Pi

mvf + MVf = mvi + MVi

.02v + 4v = (.02)(300) + 4(0)

4.02v = 6

v = 1.49 m/s

the bullet and the block move at the same speed after collision

Topic 2.2 ExtendedC – Conservation of linear momentum

Here’s a sample problem:Suppose a .02-kg bullet traveling horizontally at 300 m/s strikes a 4-kg block of wood resting on a wood floor. (b) If the block/bullet combo slides 4 m before coming to a stop, what is the coefficient of kinetic friction between the block and the floor?

We can use conservation of energy to solve this problem:

∆K + ∆U = Wnc

4 m

K – K0 + ∆U = -fkd0 0

(m+M)v02 = μkNd

12(m+M)v0

2 = μk(m+M)gd12

μk =v0

2

2gd

μk =1.492

2(10)(4)

μk = 0.027

Topic 2.2 ExtendedC – Conservation of linear momentum

Here’s a sample problem:Suppose a .02-kg bullet traveling horizontally at 300 m/s strikes a 4-kg block of wood resting on a wood floor. (c) How much energy is lost during the collision?

The energy lost is kinetic:

∆K = K – K0

(m+M)vf2 – mvi

212

∆K = 12

(4.02)1.492 – (0.02)300212

∆K = 12

∆K = -895.53 J