1. PRINCIPLE OF LINEAR IMPULSE AND MOMENTUM AND CONSERVATION OF LINEAR MOMENTUM FOR SYSTEMS OF PARTICLES Today’s Objectives: Students will be able to: 1.Apply the principle of linear impulse and momentum to a system of particles. 2.Understand the conditions for conservation of momentum. In-Class Activities: Applications Linear Impulse and Momentum for a System of Particles Conservation of Linear Momentum Group Problem Solving

2. APPLICATIONS Does the release velocity of the ball depend on the mass of the ball? As the wheels of this pitching machine rotate, they apply frictional impulses to the ball, thereby giving it linear momentum in the direction of Fdt and F ’dt. The weight impulse, W  t is very small since the time the ball is in contact with the wheels is very small.

3. APPLICATIONS (continued) This large crane-mounted hammer is used to drive piles into the ground. Conservation of momentum can be used to find the velocity of the pile just after impact, assuming the hammer does not rebound off the pile. If the hammer rebounds, does the pile velocity change from the case when the hammer doesn’t rebound ? Why ? In the impulse-momentum analysis, do we have to consider the impulses of the weights of the hammer and pile and the resistance force ? Why or why not ?

4. PRINCIPLE OF LINEAR IMPULSE AND MOMENTUM FOR A SYSTEM OF PARTICLES (Section 15.2) The linear impulse and momentum equation for this system only includes the impulse of external forces. mi(vi)2mi(vi)2 dt FiFi mi(vi)1mi(vi)1 t2t2 t1t1   For the system of particles shown, the internal forces f i between particles always occur in pairs with equal magnitude and opposite directions. Thus the internal impulses sum to zero.

5. For a system of particles, we can define a “fictitious” center of mass of an aggregate particle of mass m tot, where m tot is the sum (  m i ) of all the particles. This system of particles then has an aggregate velocity of v G = (  m i v i ) / m tot. MOTION OF THE CENTER OF MASS The motion of this fictitious mass is based on motion of the center of mass for the system. The position vector r G = (  m i r i ) / m tot describes the motion of the center of mass.

6. CONSERVATION OF LINEAR MOMENTUM FOR A SYSTEM OF PARTICLES (Section 15.3) When the sum of external impulses acting on a system of objects is zero, the linear impulse-momentum equation simplifies to  m i (v i ) 1 =  m i (v i ) 2 This equation is referred to as the conservation of linear momentum. Conservation of linear momentum is often applied when particles collide or interact. When particles impact, only impulsive forces cause a change of linear momentum. The sledgehammer applies an impulsive force to the stake. The weight of the stake is considered negligible, or non-impulsive, as compared to the force of the sledgehammer. Also, provided the stake is driven into soft ground with little resistance, the impulse of the ground acting on the stake is considered non-impulsive.

7. EXAMPLE I Given:M = 100 kg, v i = 20j (m/s) m A = 20 kg, v A = 50i + 50j (m/s) m B = 30 kg, v B = -30i – 50k (m/s) An explosion has broken the mass m into 3 smaller particles, a, b and c. Find:The velocity of fragment C after the explosion. Plan:Since the internal forces of the explosion cancel out, we can apply the conservation of linear momentum to the SYSTEM. z x y M vivi vCvC C vBvB B vAvA A =

8. mv i = m A v A + m B v B + m C v C 100(20j) = 20(50i + 50j) + 30(-30i-50k) + 50(v cx i + v cy j + v cz k) Equating the components on the left and right side yields: 0 = 1000 – 900 + 50(v cx )v cx = -2 m/s 2000 = 1000 + 50 (v cy )v cy = 20 m/s 0 = -1500 + 50 (v cz )v cz = 30 m/s So v c = (-2i + 20j + 30k) m/s immediately after the explosion. EXAMPLE I (continued) Solution:

9. EXAMPLE II Find:The speed of the car A after collision if the cars collide and rebound such that B moves to the right with a speed of 2 m/s. Also find the average impulsive force between the cars if the collision place in 0.5 s. Plan: Use conservation of linear momentum to find the velocity of the car A after collision (all internal impulses cancel). Then use the principle of impulse and momentum to find the impulsive force by looking at only one car. Given:Two rail cars with masses of m A = 20 Mg and m B = 15 Mg and velocities as shown.

10. EXAMPLE II (continued) Conservation of linear momentum (x-dir): m A (v A1 ) + m B (v B1 ) = m A (v A2 )+ m B (v B2 ) 20,000 (3) + 15,000 (-1.5) = (20,000) v A2 + 15,000 (2) v A2 = 0.375 m/s The average force is ∫ F dt = 52,500 N·s = F avg (0.5 sec); F avg = 105 kN Solution: Impulse and momentum on car A (x-dir): m A (v A1 )+ ∫ F dt = m A (v A2 ) 20,000 (3) - ∫ F dt = 20,000 (0.375) ∫ F dt = 52,500 N·s

11. GROUP PROBLEM SOLVING Find:The ramp’s speed when the crate reaches B. Plan: Use the energy conservation equation as well as conservation of linear momentum and the relative velocity equation (you thought you could safely forget it?) to find the velocity of the ramp. Given:The free-rolling ramp has a weight of 120 lb. The 80 lb crate slides from rest at A, 15 ft down the ramp to B. Assume that the ramp is smooth, and neglect the mass of the wheels.

12. GROUP PROBLEM SOLVING (continued) Energy conservation equation: 0 + 80 (3/5) (15) = 0.5 (80/32.2)(v B ) 2 + 0.5 (120/32.2)(v r ) 2 Solution: Since v B = v r + v B/r  -v Bx i + v By j = v r i + v B/r (−4/5 i −3/5 j)  -v Bx = v r − (4/5) v B/r (2) v By = − (3/5) v B/r (3) + To find the relations between v B and v r, use conservation of linear momentum: → 0 = (120/32.2) v r − (80/32.2) v Bx  v Bx =1.5 v r (1) Eliminating v B/r from Eqs. (2) and (3) and Substituting Eq. (1) results in v By =1.875 v r

13. GROUP PROBLEM SOLVING (continued) Then, energy conservation equation can be rewritten ; 0 + 80 (3/5) (15) = 0.5 (80/32.2)(v B ) 2 + 0.5 (120/32.2)(v r ) 2 0 + 80 (3/5) (15) = 0.5 (80/32.2) [(1.5 v r ) 2 +(1.875 v r ) 2 ] + 0.5 (120/32.2) (v r ) 2 720 = 9.023 (v r ) 2 v r = 8.93 ft/s