Lecture 13 Goals: Chapter 9 Employ conservation of momentum in 2D Examine forces over time (aka Impulse) Chapter 10 Understand the relationship between motion and energy Assignments: HW5, due tomorrow, HW6 due Tuesday, Oct. 19th For Wednesday, Read all of Chapter 10 1

Exercise: Momentum is a Vector (!) quantity Let a 2 kg block start at rest on a 30° incline and slide vertically a distance 5.0 m and fall a distance 7.5 m into the 10 kg cart System….block and cart only (assume Earth has infinite mass) Px is conserved once the block leaves the slide Py is not conserved (if “system” is block and cart only) 2 kg 5.0 m 30° What is the final velocity of the cart? 10 kg 7.5 m

Exercise Momentum is a Vector (!) quantity 1) ai = g sin 30° = 5 m/s2 2) d = 5 m / sin 30° = ½ ai Dt2 10 m = 2.5 m/s2 Dt2 2s = Dt v = ai Dt = 10 m/s vx= v cos 30° = 8.7 m/s x-direction: No net force so Px is conserved y-direction: vy of the cart + block will be zero and we can ignore vy of the block when it lands in the cart. i j N 5.0 m mg 30° 30° Initial Final Px: MVx + mvx = (M+m) V’x M 0 + mvx = (M+m) V’x V’x = m vx / (M + m) = 2 (8.7)/ 12 m/s V’x = 1.4 m/s 7.5 m y x

Home Exercise Inelastic Collision in 1-D with numbers Do not try this at home! ice (no friction) Before: A 4000 kg bus, twice the mass of the car, moving at 30 m/s impacts the car at rest. What is the final speed after impact if they move together?

Home exercise Inelastic Collision in 1-D v = 0 ice M = 2m V0 (no friction) initially vf =? finally 2Vo/3 = 20 m/s

A perfectly inelastic collision in 2-D Consider a collision in 2-D (cars crashing at a slippery intersection...no friction). V v1 q m1 + m2 m1 m2 v2 before after If no external force momentum is conserved. Momentum is a vector so px, py and pz

A perfectly inelastic collision in 2-D If no external force momentum is conserved. Momentum is a vector so px, py and pz are conseved V v1 m1 + m2 q m1 m2 v2 before after x-dir px : m1 v1 = (m1 + m2 ) V cos q y-dir py : m2 v2 = (m1 + m2 ) V sin q

Elastic Collisions Elastic means that the objects do not stick. There are many more possible outcomes but, if no external force, then momentum will always be conserved Start with a 1-D problem. Before After

Billiards Consider the case where one ball is initially at rest. after before pa q pb vcm Pa f F The final direction of the red ball will depend on where the balls hit.

Billiards: Without external forces, conservation of momentum (and energy Ch. 10 & 11) x-dir Px : m vbefore = m vafter cos q + m Vafter cos f y-dir Py : 0 = m vafter sin q + m Vafter sin f before after pafter q pb Pafter f F

Explosions: A collision in reverse A two piece assembly is hanging vertically at rest at the end of a 3.0 m long massless string. The mass of the two pieces are 60 and 20 kg respectively. Suddenly you observe that the 20 kg is ejected horizontally at 30 m/s. The time of the “explosion” is short compared to the swing of the string. Does the tension in the string increase or decrease after the explosion? Before After

Explosions: A collision in reverse A two piece assembly is hanging vertically at rest at the end of a 3.0 m long massless string. The mass of the two pieces are 60 and 20 kg respectively. Suddenly you observe that the 20 kg is ejected horizontally at 30 m/s. Decipher the physics: 1. The green ball recoils in the –x direction (3rd Law) and, because there is no net force in the x-direction the x-momentum is conserved. 2. The motion of the green ball is constrained to a circular path…there must be centripetal (i.e., radial acceleration) Before After

Explosions: A collision in reverse A two piece assembly is hanging vertically at rest at the end of a 3.0 m long massless string. The mass of the two pieces are 60 & 20 kg respectively. Suddenly you observe that the 20 kg mass is suddenly ejected horizontally at 30 m/s. Cons. of x-momentum Px before= Px after = 0 = - M V + m v V = m v / M = 20*30/ 60 = 10 m/s Tbefore = Weight = (60+20) x 10 N = 800 N SFy = m ay (radial) = M V2/r = T – Mg T = Mg + MV2 /r = 600 N + 60x(10)2/3 N = 2900 N Before After

Impulse (A variable force applied for a given time) Gravity: At small displacements a “constant” force t Springs often provide a linear force (-k x) towards its equilibrium position (Chapter 10) Collisions often involve a varying force F(t): 0  maximum  0 We can plot force vs time for a typical collision. The impulse, J, of the force is a vector defined as the integral of the force during the time of the collision.

Force and Impulse (A variable force applied for a given time) J a vector that reflects momentum transfer t ti tf t F Impulse J = area under this curve ! (Transfer of momentum !) Impulse has units of Newton-seconds

Force and Impulse Two different collisions can have the same impulse since J depends only on the momentum transfer, NOT the nature of the collision. F t same area F t t t t big, F small t small, F big

Average Force and Impulse t Fav F Fav t t t t big, Fav small t small, Fav big

Exercise Force & Impulse Two boxes, one heavier than the other, are initially at rest on a horizontal frictionless surface. The same constant force F acts on each one for exactly 1 second. Which box has the most momentum after the force acts ? F light heavy heavier lighter same can’t tell

Boxing: Use Momentum and Impulse to estimate g “force”

Back of the envelope calculation (1) marm~ 7 kg (2) varm~7 m/s (3) Impact time t ~ 0.01 s  Question: Are these reasonable?  Impulse J = p ~ marm varm ~ 49 kg m/s    Favg ~ J/t ~ 4900 N (1) mhead ~ 7 kg  ahead = F / mhead ~ 700 m/s2 ~ 70 g ! Enough to cause unconsciousness ~ 35% of a fatal blow Only a rough estimate!

Woodpeckers abeak ~ 600 - 1500 g How do they survive? During "collision" with a tree  abeak ~ 600 - 1500 g   How do they survive? Jaw muscles act as shock absorbers Straight head trajectory reduces damaging rotations (rotational motion is very problematic)

Discussion Exercise The only force acting on a 2.0 kg object moving along the x-axis. Notice that the plot is force vs time. If the velocity vx is +2.0 m/s at 0 sec, what is vx at 4.0 s ? Dp = m Dv = Impulse m Dv = J0,1 + J1,2 + J2,4 m Dv = (-8)1 N s + ½ (-8)1 N s + ½ 16(2) N s m Dv = 4 N s Dv = 2 m/s vx = 2 + 2 m/s = 4 m/s

Newton’s Laws rearranged (Ch. 10) From motion in the y-dir, Fy = m ay and let ay be constant y(t) = y0 + vy0 Dt + ½ ay Dt2  Dy = y(t)-y0= vy0 Dt + ½ ay Dt2 vy (t) = vy0 + ay Dt  Dt = (vy - vy0) / ay and eliminating Dt yields 2 ay Dy= (vy2 - vy02 ) which is nothing new or if ay= -g -mg Dy= ½ m (vy2 - vy02 )

-mg (yf – yi) = ½ m ( vyf2 -vyi2 ) “Energy” defined -mg Dy= ½ m (vy2 - vy02 ) -mg (yf – yi) = ½ m ( vyf2 -vyi2 ) A relationship between y- displacement and change in the y-speed squared Rearranging to give initial on the left and final on the right ½ m vyi2 + mgyi = ½ m vyf2 + mgyf We now define mgy as the “gravitational potential energy”

Energy ½ m vi2 + mgyi = ½ m vf2 + mgyf Notice that if we only consider gravity as the external force then the x and z velocities remain constant To ½ m vyi2 + mgyi = ½ m vyf2 + mgyf Add ½ m vxi2 + ½ m vzi2 and ½ m vxf2 + ½ m vzf2 ½ m vi2 + mgyi = ½ m vf2 + mgyf where vi2 ≡ vxi2 +vyi2 + vzi2 ½ m v2 terms are defined to be kinetic energies (A scalar quantity of motion)

Lecture 13 Assignment: HW6 up soon For Wednesday: Read all of chapter 10 1