1. Chapter 9 Linear Momentum and Collisions EXAMPLES

2. Example 9.1 The Archer The archer of mass 60kg is standing on a frictionless surface (ice). He fires a 0.50kg arrow horizontally at 50 m/s. With what velocity does the archer move across the ice after firing the arrow? Can we use: Newton’s Second Law ? NO No information about F or a Energy? NO No information about work or energy Momentum? YES The System will be: the archer with bow (particle 1) and the arrow (particle 2)

3. Example 9.1 The Archer, final ΣF x = 0, so it is isolated in terms of momentum in the x-direction Total momentum before releasing the arrow is 0: p 1i + p 2i = 0 The total momentum after releasing the arrow is p 1f + p 2f = 0 m 1 v 1f + m 2 v 2f = 0  v 1f = – (m 2 /m 1 )v 2f  v 1f = – (0.50kg/60kg )(50.0î)m/s  v 1f = –0.42 î m/s The archer will move in the opposite direction of the arrow after the release Agrees with Newton’s Third Law Because the archer is much more massive than the arrow, his acceleration and velocity will be much smaller than those of the arrow

4. Example 9.2 Railroad Cars Collide (Perfectly Inelastic Collision) Initial Momentum = Final Momentum (One-Dimension) If: m 1 v 1 +m 2 v 2 = (m 1 + m 2 )v’ Given: v 2 = 0, v’ 1 = v’ 2 = v’ = ? v’ = m 1 v 1 /(m 1 + m 2 ) = 240,000/20,000m/s  v’ = 12 m/s m 1 = 10,000kg m 2 = 10,000kg m 1 + m 2 = 20,000 kg

5. Example 9.3 How Good Are the Bumpers? Mass of the car is 1500 kg. The collision lasts 0.105s. Find:  p = I and the average force (F avg ) exerted on the car.

6. Example 9.4 Tennis Ball Hits the Wall  p = I ? Given to the ball. If m = 0.060 kg and v = 8.00 m/s  p = I : change in momentum  wall. Momentum ∕∕ to wall doesn’t change. Impulse will be  wall. Take + direction toward wall,  p = I = m  v = m (v f – v i )   p = I = m(–vsin45 – vsin45)  p = I = –2mvsin45 = –2.1 N.s Impulse on wall is in opposite direction: 2.1 N.s v f = –vsin45 v i = vsin45

7. Example 9.5 Explosion as a Collision Initial Momentum = Final Momentum (One-Dimension) m 1 v 1 +m 2 v 2 = m 1 v’ 1 + m 2 v’ 2 Initially: v = 0 Explodes! Finally: mv = 0 = m 2 v’ 2 + m 1 v’ 1 Given: m 1, m 2, v’ 2, you may compute v’ 1  v’ 1 = – (m 2 /m 1 )v’ 2 m v = 0

8. Example 9.6 Rifle Recoil Momentum Before = Momentum After m 1 v 1 +m 2 v 2 = m 1 v’ 1 + m 2 v’ 2 Given: m B = 0.02 kg, m R = 5.00 kg, v’ B = 620 m/s 0 = m B v’ B + m R v’ R v’ R = – m B v’ B /m R = – (0.02)(620)/5.00 m/s = – 2.48 m/s (to the left, of course!)

9. Example 9.7 Ballistic Pendulum Perfectly inelastic collision – the bullet is embedded in the block of wood Momentum equation will have two unknowns Use conservation of energy from the pendulum to find the velocity just after the collision Then you can find the speed of the bullet

10. Example 9.7 Ballistic Pendulum, final Before: Momentum Conservation: After: Conservation of energy: Solving for v B :  Replacing v B into 1st equation and solving for v 1A :

11. Example 9.8 Collision at an Intersection Mass of the car m c = 1500kg Mass of the van m v = 2500kg Find v f if this is a perfectly inelastic collision (they stick together). Before collision: The car’s momentum is: Σp xi = m c v c  Σp xi = (1500)(25) = 3.75x10 4 kg·m/s The van’s momentum is: Σp yi = m v v v  Σp yi = (2500)(20) = 5.00x10 4 kg·m/s After collision: both have the same x- and y-components: Σp xf = (m c + m v )v f cos  Σp yf = (m c + m v )v f sin 

12. Example 9.8 Collision at an Intersection, final Because the total momentum is both directions is conserved: Σp xf = Σp xi  3.75x10 4 kg·m/s = (m c + m v )v f cos  = 4000 v f cos  (1) Σp yf = Σp yi  5.00x10 4 kg·m/s = (m c + m v )v f sin  = 4000v f sin  (2) Dividing Eqn (2) by (1) 5.00/3.75 =1.33 = tan    = 53.1° Substituting  in Eqn (2) or (1)  5.00x10 4 kg·m/s = 4000v f sin53.1°  v f = 5.00x10 4 /(4000sin53.1° )  v f = 15.6m/s

13. Example 9.9 Center of Mass (Simple Case) Both masses are on the x-axis The center of mass (CM) is on the x-axis One dimension x CM = (m 1 x 1 + m 2 x 2 )/M M = m 1 +m 2 x CM ≡ (m 1 x 1 + m 2 x 2 )/(m 1 +m 2 ) The center of mass is closer to the particle with the larger mass If: x 1 = 0, x 2 = d & m 2 = 2m 1 x CM ≡ (0 + 2m 1 d)/(m 1 +2m 1 )  x CM ≡ 2m 1 d/3m 1  x CM = 2d/3

14. Example 9.10 Three Guys on a Raft A group of extended bodies, each with a known CM and equivalent mass m. Find the CM of the group. x CM = (Σm i x i )/Σm i x CM = (mx 1 + mx 2 + mx 3 )/(m+m+m)  x CM = m(x 1 + x 2 + x 3 )/3m = (x 1 + x 2 + x 3 )/3  x CM = (1.00m + 5.00m + 6.00m)/3 = 4.00m

15. Example 9.11 Center of Mass of a Rod Find the CM position of a rod of mass M and length L. The location is on the x-axis (A). Assuming the road has a uniform mass per unit length λ = M/L (Linear mass density) From Eqn 9.32 But: λ = M/L

16. Example 9.11 Center of Mass of a Rod, final. (B). Assuming now that the linear mass density of the road is no uniform: λ =  x The CM will be: But mass of the rod and  are related by:  The CM will be:

17. Examples to Read!!! Example 9.2 (page 239) Example 9.5 (page 247) Example 9.10 (page 256) Material from the book to Study!!! Objective Questions: 7-8-13 Conceptual Questions: 3-5-6 Problems: 1-9-11-15-25-26-27-37-40-65 Material for the Final Exam