2.  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 Extended C – 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  F ext = 0 then  P = 0 P 0 = P f 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.

3. P ROOF OF THE C ONSERVATION OF L INEAR M OMENTUM  Recall Newton's 2nd law (p-form): Topic 2.2 Extended C – Conservation of linear momentum F net = ptpt Newton's Second Law (p-form)  We begin by dividing the net force into all of its individual forces - both external and internal: F net = F 1,ext +...+ F n,ext + F 1,int +...+ F m,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 F net = 0 so that 0 = ptpt. Thus  P = 0. //

4. Topic 2.2 Extended C – 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 P f = P i mv f + MV f = mv i + MV i.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

5. Topic 2.2 Extended C – 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 = W nc 4 m K – K 0 + ∆U = -f k d 0 0 (m+M)v 0 2 = μ k Nd 1212 (m+M)v 0 2 = μ k (m+M)gd 1212 μ k = v 0 2 2gd μ k = 1.49 2 2(10)(4) μ k = 0.027

6. Topic 2.2 Extended C – 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 – K 0 (m+M)v f 2 – mv i 2 1212 ∆K = 1212 (4.02)1.49 2 – (0.02)300 2 1212 ∆K = 1212 ∆K = - 895.53 J